Question 1

If x is a whole number, then the expression x²(x² − 1) is always divisible by which of the following numbers?

Accepted Answer

Introduction / Context: This problem tests your understanding of divisibility properties and factorisation for algebraic expressions. You must analyse x²(x² − 1) for whole number values of x and determine by which number it is always divisible. Instead of checking individual values one by one, the key is to factor the expression and use properties of consecutive integers, parity and divisibility by 3. This is a classic number theory style question in aptitude exams. Given Data / Assumptions: x is a whole number (0, 1, 2, 3, and so on). Expression: x²(x² − 1). We must decide whether it is always divisible by 4, 8, 12, 16 or by none of these. Whole numbers include 0, and 0 is divisible by any positive integer. Concept / Approach: Start by factoring the expression algebraically. Note that x² − 1 is a difference of squares, so x² − 1 = (x − 1)(x + 1). Thus x²(x² − 1) = x²(x − 1)(x + 1). This product contains the three consecutive integers x − 1, x and x + 1, and one extra factor of x. The product of three consecutive integers is always divisible by 6 because among any three consecutive integers there is at least one multiple of 2 and one multiple of 3. The extra factor of x ensures additional powers of 2 in many cases. We will check divisibility by 12, 8 and 4 in turn. Step-by-Step Solution: Write x²(x² − 1) as x²(x − 1)(x + 1). Observe that (x − 1), x and (x + 1) are three consecutive integers. Among three consecutive integers, one is always a multiple of 3, so the product is divisible by 3. Also, among these three integers at least one is even, so the product is divisible by 2. Therefore x(x − 1)(x + 1) is always divisible by 6. Now consider divisibility by 4. If x is even, then x² has a factor of 4 so the whole product is divisible by 4. If x is odd, then both (x − 1) and (x + 1) are even and consecutive multiples of 2, so one of them is a multiple of 4. Thus the product x²(x − 1)(x + 1) is always divisible by 4 in either case. Since the expression is always divisible by both 3 and 4, it is always divisible by lcm(3, 4) = 12. However it is not always divisible by 8; for example when x = 2 the value is 4 * 3 = 12, which is not a multiple of 8. Verification / Alternative check: Test a few small whole numbers to confirm. For x = 0 the expression is 0, divisible by 12. For x = 1 we get 1²(1² − 1) = 1 * 0 = 0, again divisible by 12. For x = 2 we get 4 * 3 = 12, exactly a multiple of 12. For x = 3 we get 9 * 8 = 72, and 72 / 12 = 6. For x = 4 we get 16 * 15 = 240, and 240 / 12 = 20. These checks support our general reasoning. Why Other Options Are Wrong: 4: The expression is indeed always divisible by 4, but this is not the largest guaranteed divisor among the options. 8: For x = 2 the value is 12, which is not divisible by 8, so 8 fails. 16: For x = 2 or x = 3 the resulting values are not multiples of 16. None of these: This is incorrect because we have shown that 12 always divides x²(x² − 1). Common Pitfalls: A common error is to test only one or two values of x and then guess a pattern without algebraic justification. Another mistake is to overlook the factorisation x² − 1 = (x − 1)(x + 1) and therefore miss the property of three consecutive integers. For divisibility questions, always look for factorisations, parity patterns and use of least common multiples, rather than relying on trial alone. Final Answer: For any whole number x, the expression x²(x² − 1) is always divisible by 12.

Question 2

Find the total number of positive factors of 9321.

Accepted Answer

Introduction / Context: This question checks your ability to factorise a composite number into primes and then use the prime factorisation to count the total number of its positive divisors. This is a standard number theory technique useful in many aptitude and competitive exams. The key is to express the number as a product of powers of primes and then apply a simple rule for counting factors. Given Data / Assumptions: The given number is 9321. We are interested in positive factors only. We assume basic knowledge of divisibility tests and prime numbers. The factor counting formula for a number in the form p^a * q^b * r^c is (a + 1)(b + 1)(c + 1). Concept / Approach: First, we check whether 9321 is divisible by small primes such as 2, 3, 5, 7, 11, 13 and so on. Divisibility by 3 is quickly tested using the sum of digits. After we find one prime factor, we divide the number by that factor and continue to factor the quotient. Once we have the full prime factorisation, we apply the exponent based rule to count all positive factors. This method is far more efficient than listing every divisor by trial. Step-by-Step Solution: Given number N = 9321. Check divisibility by 3: sum of digits = 9 + 3 + 2 + 1 = 15, which is divisible by 3, so 9321 is divisible by 3. Compute 9321 / 3 = 3107. Now factor 3107. Check divisibility by small primes: 3107 is not divisible by 3, 5 or 7. Test 13: 13 * 239 = 3107, so 3107 = 13 * 239. Check 239 for primality. It is not divisible by 2, 3, 5, 7, 11 or 13, so 239 is prime. Therefore, prime factorisation of 9321 is 3^1 * 13^1 * 239^1. Number of positive factors = (1 + 1)(1 + 1)(1 + 1) = 2 * 2 * 2 = 8. Verification / Alternative check: We can list the factors explicitly to verify. From primes 3, 13 and 239, every factor is a product of a subset of these primes. The factors are 1, 3, 13, 39, 239, 3 * 239 = 717, 13 * 239 = 3107 and 3 * 13 * 239 = 9321. Counting them gives exactly 8 factors, matching the result from the exponent formula. This confirms that the factorisation and counting are both correct. Why Other Options Are Wrong: 4 and 5: These would correspond to too few factors and would imply that the number has fewer distinct prime factors or lower exponents than it actually has. 6 and 7: These are closer but still do not match the complete combination count from three distinct primes. Only 8 is consistent with the prime factorisation 3 * 13 * 239. Common Pitfalls: Typical mistakes include stopping factorisation too early, miscalculating a quotient, or forgetting to check whether the last factor is prime. Another error is misusing the factor counting formula, for example adding exponents instead of adding one to each exponent and multiplying. Always write the prime factorisation clearly with exponents, then apply the formula carefully. Final Answer: The number 9321 has 8 positive factors.

Question 3

For a non zero number, what is the product of the number and its multiplicative inverse (reciprocal)?

Accepted Answer

Introduction / Context: This conceptual question checks your understanding of the idea of a multiplicative inverse, also known as a reciprocal. The multiplicative inverse of a non zero number is defined so that when you multiply the number by its inverse, the result is the multiplicative identity. Understanding this idea is important in algebra, fractions, and many areas of higher mathematics, because it underlies division and solving equations. Given Data / Assumptions: We are working with a non zero real number, call it a. The multiplicative inverse of a is another number that we can denote as 1/a. The question asks for the product a * (1/a). Zero is excluded because its multiplicative inverse does not exist. Concept / Approach: By definition, the multiplicative inverse of a non zero number a is the number that, when multiplied by a, gives the multiplicative identity, which is 1 in standard arithmetic. For any non zero real number a, its reciprocal is 1/a. This definition is built into how we understand division: dividing by a is the same as multiplying by 1/a. Therefore, when you multiply a by 1/a, the a in the numerator and denominator cancel, leaving 1. This is independent of whether a is positive or negative, as long as it is not zero. Step-by-Step Solution: Let a be a non zero number. Its multiplicative inverse is 1/a. Compute the product: a * (1/a). Since a is non zero, the factor a in the numerator cancels with a in the denominator. So a * (1/a) = 1. Therefore, the product of a non zero number and its multiplicative inverse is always exactly 1. Verification / Alternative check: Test with a few concrete numbers. If a = 5, then the multiplicative inverse is 1/5, and 5 * (1/5) = 1. If a = -3, the inverse is -1/3, and (-3) * (-1/3) = 1 as well. Even for fractions, such as a = 2/7, its inverse is 7/2, and (2/7) * (7/2) = 1. In every non zero case the product is 1, confirming the general rule. Why Other Options Are Wrong: 0: This would be the result if one factor were zero, but the definition excludes zero because it has no multiplicative inverse. -1: This would require the inverse to be the negative of 1/a, which is not the definition of multiplicative inverse. Infinity: This is not a meaningful result in standard real number arithmetic for this operation. Cannot be determined: In fact, it is completely determined by the definition of multiplicative inverse. Common Pitfalls: Some learners confuse multiplicative inverse with additive inverse. The additive inverse of a is the number that sums with a to give 0, namely −a. The multiplicative inverse instead relates to multiplication and gives a product of 1. Keeping these two ideas separate is essential: additive inverse leads to zero, multiplicative inverse leads to one. Remembering that division by a non zero number is multiplication by its reciprocal also reinforces this concept. Final Answer: For any non zero number, the product of the number and its multiplicative inverse is 1.

Question 4

Evaluate 31 × 31 × 31 (that is, 31³).

Accepted Answer

Introduction / Context: This question is a straightforward computation of a cube of an integer. You are asked to evaluate 31 multiplied by itself three times, which is written as 31³. Calculating cubes accurately is useful in many algebra and arithmetic problems, and it also helps build confidence with mental or written multiplication techniques. Given Data / Assumptions: We need to compute 31 × 31 × 31. Standard integer arithmetic applies with no approximations. The options given are close in value, so accurate calculation is important. Concept / Approach: To compute 31³, it is efficient to first find 31² and then multiply the result by 31 again. Squaring 31 can be done using the identity (30 + 1)² = 30² + 2 * 30 * 1 + 1². Once 31² is obtained, a second multiplication by 31 yields the cube. Writing down each step clearly reduces the chance of arithmetic slips, especially when dealing with similar looking numbers in the answer choices. Step-by-Step Solution: First compute 31² = 31 × 31. Use (30 + 1)² = 30² + 2 * 30 * 1 + 1² = 900 + 60 + 1 = 961. So 31² = 961. Now compute 31³ = 31 × 31² = 31 × 961. Multiply 961 by 31: 961 × 30 = 28830 and 961 × 1 = 961. Add these: 28830 + 961 = 29791. Therefore, 31³ = 29791. Verification / Alternative check: As a quick check, we know that 30³ = 27000 and 32³ = 32768. Since 31 lies between 30 and 32, 31³ should be between 27000 and 32768, which 29791 is. Additionally, we can multiply in a different order, say 31 × 31 = 961 and then 961 × 31 again with a calculator or independent calculation. Getting the same result 29791 from two routes confirms that the computation is correct. Why Other Options Are Wrong: 27991 and 29719: These are close in magnitude but result from incorrect intermediate additions or multiplications. 29971: Another near miss that can arise from reversing digits or minor arithmetic mistakes. 31000: This is a rounded guess and not the exact cube. Common Pitfalls: Students sometimes misplace digits when adding partial products, especially with similar numbers like 28830 and 961. Others confuse squaring and cubing, computing 31² instead of 31³. A reliable method is to break the multiplication into two stages, double check the square, then multiply that result by 31 with careful alignment of place values. Taking a moment to estimate the range of the answer also helps catch obviously wrong results. Final Answer: The value of 31 × 31 × 31 is 29791.

Question 5

What is the remainder when 15²³ + 23²³ is divided by 19?

Accepted Answer

Introduction / Context: This is a modular arithmetic question involving large powers. Directly computing 15²³ and 23²³ is impossible in an exam setting, so you must use properties of remainders and congruences. The key ideas are reducing bases modulo 19 and exploiting patterns in powers, especially when numbers are negatives of each other modulo the divisor. Such problems test deeper understanding of number theory techniques. Given Data / Assumptions: Expression: 15²³ + 23²³. Divisor: 19. We seek the remainder when the expression is divided by 19. All calculations are done using integer arithmetic and modular congruences. Concept / Approach: We work modulo 19. First reduce the bases 15 and 23 modulo 19. Notice that 23 is congruent to 4 modulo 19 and 15 is congruent to −4 modulo 19 because 15 + 4 = 19. That means the two bases are negatives of each other in this modular system. Then, powers of these bases can be related: (−4)²³ = −4²³ when the exponent is odd. Since 23 is an odd exponent, 15²³ ≡ (−4)²³ ≡ −4²³ modulo 19, and 23²³ ≡ 4²³ modulo 19. Adding these two expressions leads to cancellation. Step-by-Step Solution: Work modulo 19. Reduce 15 mod 19: 15 ≡ −4 (since 15 + 4 = 19). Reduce 23 mod 19: 23 ≡ 4 (since 23 − 19 = 4). Therefore, 15²³ ≡ (−4)²³ and 23²³ ≡ 4²³ (mod 19). Because 23 is odd, (−4)²³ = −4²³. So 15²³ ≡ −4²³ and 23²³ ≡ 4²³ (mod 19). Now add them: 15²³ + 23²³ ≡ (−4²³) + 4²³ ≡ 0 (mod 19). Hence the remainder when 15²³ + 23²³ is divided by 19 is 0. Verification / Alternative check: We can verify the reasoning by checking with smaller exponents. For example, consider 15¹ + 23¹ modulo 19. We get 15 + 23 = 38, and 38 / 19 gives remainder 0. With exponent 3 we have 15³ + 23³. Modulo 19, 15 ≡ −4 and 23 ≡ 4, so (−4)³ + 4³ = −64 + 64 = 0, again giving remainder 0. This consistent pattern for odd exponents supports the same conclusion for exponent 23, since the modular reasoning depends only on parity of the exponent, not its exact size. Why Other Options Are Wrong: 4, 3 and 1: These values might arise if you make mistakes in reducing the bases or overlook that the exponents are odd, but they do not satisfy the exact modular relationships described. 7: This is just another incorrect remainder with no basis in the given congruences. Common Pitfalls: A common error is attempting to compute large powers directly or using a calculator, which is impractical and unnecessary. Another mistake is forgetting that (−a) raised to an odd power is negative, while raised to an even power is positive. In this problem, misidentifying the parity of 23 leads to wrong signs and incorrect cancellation. Always reduce bases modulo the divisor early, look for symmetry or opposite numbers, and carefully apply power rules for negative bases. Final Answer: The remainder when 15²³ + 23²³ is divided by 19 is 0.

Question 6

Find the next number in the series: 2, 3, 4, 8, 9, 10, 11, 16, 17, 18, 19, 20, ...

Accepted Answer

Introduction / Context: This is a number series question where you must identify the underlying pattern that generates the given sequence. Unlike simple arithmetic or geometric progressions, the series 2, 3, 4, 8, 9, 10, 11, 16, 17, 18, 19, 20, ... uses blocks of consecutive numbers separated by gaps. The challenge is to recognise how the blocks and the gaps are growing and then extend the pattern to find the next term. Given Data / Assumptions: The series is: 2, 3, 4, 8, 9, 10, 11, 16, 17, 18, 19, 20, ... We are asked to find the next number after 20. The pattern may involve groups of consecutive numbers combined with skipped intervals. Concept / Approach: Group the numbers into blocks of consecutive integers and observe the sizes of those blocks and the gaps in between. The first block is 2, 3, 4 (three consecutive numbers). Then there is a gap of three numbers 5, 6, 7 which do not appear. The next block is 8, 9, 10, 11 (four consecutive numbers), followed by a gap 12, 13, 14, 15 of four missing numbers. Then we see a block of five consecutive numbers: 16, 17, 18, 19, 20. This pattern suggests that each time, the series skips k numbers and then includes k + 1 consecutive numbers, where k increases as we move forward. Step-by-Step Solution: First block: 2, 3, 4 (three consecutive numbers). Numbers 5, 6, 7 are skipped (a gap of three numbers). Second block: 8, 9, 10, 11 (four consecutive numbers). Numbers 12, 13, 14, 15 are skipped (a gap of four numbers). Third block: 16, 17, 18, 19, 20 (five consecutive numbers). By continuing the pattern, after a block of five numbers, the next gap should contain five numbers: 21, 22, 23, 24, 25 are skipped. The next block must therefore be six consecutive numbers starting from 26. Thus, the next term after 20 is 26. Verification / Alternative check: We can verify the logic by writing the structure explicitly: skip 3, write 3; skip 4, write 4; skip 5, write 5; and so on. The observed data matches skip 3 (5 to 7) and write 3 (2 to 4), then skip 4 (12 to 15) and write 4 (8 to 11), then write 5 (16 to 20). Following this systematic pattern, the next operation must be skip 5 (21 to 25) and write 6 (26 to 31). The first number of that new block is clearly 26, confirming the answer. Why Other Options Are Wrong: 27, 24, 25 and 30: These are numbers close to 26, but they do not respect the rule that the next visible number must start immediately after skipping 21 to 25. 27 would be the second number in the next block, not the first. Common Pitfalls: Many learners first attempt to look for a simple difference pattern between successive terms, which quickly breaks down in this sequence. Others try to link the series to squares or primes without success. When a series feels irregular, it often helps to group numbers into segments and look for regularity in group lengths or gaps. Drawing a number line and marking present and missing numbers also makes the pattern more visible. Final Answer: The next number in the series is 26.

Question 7

Find the next number in the series: 141, 137, 146, 130, 155, 119, ...

Accepted Answer

Introduction / Context: This is a number series problem that involves observing changes between consecutive terms. The series 141, 137, 146, 130, 155, 119, ... may look irregular at first, but the differences between terms reveal a clear pattern. Spotting such patterns is a key skill in aptitude exams and helps strengthen numerical reasoning and pattern recognition. Given Data / Assumptions: The series given is 141, 137, 146, 130, 155, 119, ... We must find the next term after 119. The pattern is likely encoded in the sequence of differences between consecutive terms. Concept / Approach: First compute the difference between each pair of consecutive terms. If these differences themselves follow a recognizable pattern, such as squares, alternating signs or constant changes, we can use that pattern to find the next difference and thus the next term. For more complex series, it is often helpful to look at the absolute values of differences separately from their signs. Step-by-Step Solution: Compute the differences: 137 − 141 = −4. 146 − 137 = +9. 130 − 146 = −16. 155 − 130 = +25. 119 − 155 = −36. So the sequence of differences is −4, +9, −16, +25, −36. Look at absolute values: 4, 9, 16, 25, 36. These are perfect squares: 2², 3², 4², 5², 6². The sign pattern alternates: negative, positive, negative, positive, negative. Following this pattern, the next square should be 7² = 49 with a positive sign. So the next difference should be +49. Therefore, the next term = 119 + 49 = 168. Verification / Alternative check: We can reconstruct the series using the identified rule: start at 141, subtract 2² to get 137, add 3² to get 146, subtract 4² to get 130, add 5² to get 155, subtract 6² to get 119, and then add 7² to reach 168. This sequence exactly matches the given terms and the next computed term, confirming that the pattern is consistent and correctly applied. Why Other Options Are Wrong: 147, 162, 182 and 140: None of these correspond to adding 49 to 119, so they do not follow the discovered pattern of alternating plus and minus perfect squares. Any other value would break the neat progression of squared differences and alternating signs. Common Pitfalls: A frequent mistake is to search for a single arithmetic or geometric progression, which does not exist in the raw terms. Another is to check only the first few differences and give up when they do not appear constant. Instead, it is helpful to look at patterns in the differences themselves, such as squares, cubes or simple sequences with sign changes. Recognising that 4, 9, 16, 25 and 36 are consecutive squares is the key insight here. Final Answer: The next number in the series is 168.

Question 8

What is the smallest number by which 2880 must be divided so that the result is a perfect square?

Accepted Answer

Introduction / Context: This question involves prime factorisation and the condition for a number to be a perfect square. A perfect square has all prime factors raised to even powers. You are asked to find the smallest positive integer that can divide 2880 so that the quotient becomes a perfect square. This is a common type of question in number theory sections of aptitude tests. Given Data / Assumptions: The given number is 2880. We must divide 2880 by a positive integer k so that 2880 / k is a perfect square. We seek the smallest such k. Concept / Approach: First factor 2880 into prime factors and express it as a product of prime powers. For the quotient to be a perfect square, every prime in its factorisation must have an even exponent. Dividing by k will remove some prime factors from 2880. The strategy is to remove the minimal prime factor power that turns all exponents even. We then check which small option matches that required k. Step-by-Step Solution: Factor 2880: start with 2880 = 288 * 10. Factor 288 = 32 * 9 = 2^5 * 3^2. Factor 10 = 2 * 5. So 2880 = 2^5 * 3^2 * 2 * 5 = 2^6 * 3^2 * 5. Write 2880 as 2^6 * 3^2 * 5^1. For a perfect square, the exponents of all primes must be even. Here the exponents of 2 and 3 are 6 and 2, both even, but the exponent of 5 is 1, which is odd. To make all exponents even, we need to remove one factor of 5 by dividing by 5. Then the quotient is (2^6 * 3^2 * 5) / 5 = 2^6 * 3^2. This simplifies to (2^3 * 3)^2 = 24^2, which is a perfect square. Verification / Alternative check: Compute the quotient explicitly: 2880 / 5 = 576. Now check whether 576 is a perfect square. We know 24 * 24 = 576, so 576 is indeed a perfect square. If we tried a smaller divisor, for example 2 or 3, the quotient would still have an odd exponent on one of the prime factors and would not be a perfect square. Thus 5 is the smallest divisor that works. Why Other Options Are Wrong: 2: 2880 / 2 = 1440, whose prime factorisation still includes 5^1, so it is not a perfect square. 3: 2880 / 3 = 960, which has 2^6 * 3^1 * 5, with an odd exponent on 3. 4: 2880 / 4 = 720, with factorisation 2^4 * 3^2 * 5, still having 5^1. 6: 2880 / 6 = 480, with prime factorisation 2^5 * 3^1 * 5, again not all exponents even. Common Pitfalls: Some students try to test each option by long division and then guess whether the quotient is a square, which is slow and error prone. Others forget that only the parity (evenness or oddness) of each exponent matters. A systematic approach is to factor the original number, look at exponents of each prime, and then decide which minimal prime powers must be removed or added. This technique applies to both making perfect squares and perfect cubes. Final Answer: The smallest number by which 2880 must be divided to obtain a perfect square is 5.

Question 9

In triangle ABC, angle A is the smallest angle and A + 7B = 112°. What is the possible range of values of angle C (in degrees)?

Accepted Answer

Introduction / Context: This problem combines angle sum properties of a triangle with inequality reasoning. You are given a linear relation between two angles, A and B, namely A + 7B = 112°, and the information that A is the smallest angle in triangle ABC. You must deduce the possible range of values for the third angle C. Such questions test both algebraic manipulation and geometric reasoning about angle sizes and ordering. Given Data / Assumptions: Triangle ABC with angles A, B and C. A + B + C = 180° for any triangle. A + 7B = 112°. A is the smallest of the three angles, so A < B and A < C. Angles are positive and strictly less than 180°. Concept / Approach: Start by expressing A in terms of B using A + 7B = 112°. Then express C in terms of B using the triangle angle sum A + B + C = 180°. This gives C as a linear function of B. Next, use the conditions that A is the smallest angle to derive inequalities for B, which then translate into inequalities for C. Because A, B and C must all be positive, we also include these basic constraints when finding the range of B and therefore the range for C. Step-by-Step Solution: From A + 7B = 112°, express A as A = 112° − 7B. Using A + B + C = 180°, substitute A: (112° − 7B) + B + C = 180°. Simplify: 112° − 6B + C = 180° ⇒ C = 68° + 6B. Since angles are positive, A = 112° − 7B > 0 ⇒ B < 16°. Also, A is the smallest angle, so A < B ⇒ 112° − 7B < B. This gives 112° < 8B ⇒ B > 14°. Thus B lies between 14° and 16°: 14° < B < 16°. Now express C in terms of B: C = 68° + 6B. For B just above 14°, C is slightly greater than 68° + 84° = 152°. For B just below 16°, C is slightly less than 68° + 96° = 164°. Thus C lies strictly between 152° and 164°. In interval notation, 152° < C < 164°. Verification / Alternative check: Pick a sample value such as B = 15°. Then A = 112° − 7 * 15° = 112° − 105° = 7°, and C = 180° − A − B = 180° − 7° − 15° = 158°. Here A = 7°, B = 15°, C = 158°. Clearly A is the smallest, and C lies between 152° and 164°. Trying B = 14.5° gives A = 112° − 7 * 14.5° = 112° − 101.5° = 10.5° and C = 180° − 10.5° − 14.5° = 155°. Again C is within the same band. These checks support the derived range. Why Other Options Are Wrong: 138 to 164 and 125 to 147: These ranges are too wide or shifted; they include values of C that would violate A being the smallest angle under the given equation. 148 to 158: This interval is narrower than the true possible range and excludes valid values near 152° and 164°. 130 to 150: This is entirely too low for C based on C = 68° + 6B with B > 14°. Common Pitfalls: Learners sometimes forget to enforce the condition that A is the smallest angle, using only positivity and the angle sum rule. This leads to incomplete or incorrect ranges for B and C. Another mistake is to treat the derived bounds as inclusive without checking whether equality is allowed. Here B cannot equal exactly 14° or 16° because that would make A equal to B or zero, both of which break the conditions. Therefore the realistic answer is a range that begins just above 152° and ends just below 164°, best represented by 152° to 164° in the option set. Final Answer: The possible range of values for angle C is approximately 152° to 164°.

Question 10

If i is the imaginary unit with i² = -1, which expression is equivalent to i²³³?

Accepted Answer

Introduction / Context: This question involves complex numbers and powers of the imaginary unit i. The imaginary unit is defined by the property i² = -1. Powers of i follow a simple repeating pattern with period 4, which makes evaluating large exponents like i²³³ straightforward once you know the cycle. Such questions commonly appear in algebra sections to test familiarity with complex number basics. Given Data / Assumptions: i is the imaginary unit, so i² = -1. We must compute i²³³. Arithmetic is in the usual complex number system. Concept / Approach: Powers of i repeat in a cycle of length 4: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then i⁵ = i again, and so on. Therefore, to evaluate i raised to a large power, we reduce the exponent modulo 4 and identify which one of these four values corresponds to the remainder. The exponent 233 can be divided by 4 to find this remainder. Step-by-Step Solution: List the basic cycle: i¹ = i, i² = -1, i³ = -i, i⁴ = 1. The cycle repeats every 4 powers. Compute 233 divided by 4: 4 * 58 = 232 with remainder 1. Thus 233 = 4 * 58 + 1. So i²³³ = i^(4 * 58 + 1) = (i⁴)⁵⁸ * i¹. Since i⁴ = 1, (i⁴)⁵⁸ = 1⁵⁸ = 1. Therefore i²³³ = 1 * i = i. Verification / Alternative check: We can test the method on smaller exponents. For example, i⁵ = i^(4 + 1) = i⁴ * i = 1 * i = i, which matches the cycle. For i⁹, we get i^(8 + 1) = (i⁴)² * i = 1 * i = i again. Each time, reducing the exponent modulo 4 and using the remainder as an index in the cycle works. Since 233 leaves a remainder of 1 when divided by 4, the correct value must be the same as i¹, which is i. Why Other Options Are Wrong: 1: This would correspond to exponents that are multiples of 4, such as i⁴, i⁸, i¹² and so on. -1: This value appears when the exponent is congruent to 2 modulo 4, for example i², i⁶, i¹⁰. -i: This occurs when the exponent is congruent to 3 modulo 4, such as i³ or i⁷. 0: No non zero power of i equals zero. Common Pitfalls: Students sometimes try to expand i²³³ step by step, which is unnecessary and impossible to manage. Others misremember the cycle or confuse it with other periodic patterns. A reliable approach is always to reduce the exponent modulo 4 because i⁴ = 1 resets the pattern. Writing down the four term cycle and matching the remainder directly to one of i, -1, -i or 1 keeps the process clear and error free. Final Answer: The value of i²³³ is i.

Question 11

If n is a natural number, then the expression 6n² + 6n is always divisible by which of the following?

Accepted Answer

Introduction / Context: This question focuses on divisibility and factorisation. You are given an expression 6n² + 6n where n is a natural number and asked which numbers always divide the value of this expression. Instead of plugging in many values of n, the correct strategy is to factor the expression and reason about the divisibility properties of the factors, especially for evenness and multiples of 3. Given Data / Assumptions: n is a natural number (1, 2, 3, and so on). Expression: 6n² + 6n. We need to determine whether it is always divisible by 6, by 12, or by 18. Concept / Approach: Factor out common terms from the expression. The greatest common factor is 6n, so 6n² + 6n = 6n(n + 1). Because n and n + 1 are consecutive integers, their product has predictable divisibility properties. Specifically, among two consecutive integers one must be even, and among n and n + 1, at least one of them may be a multiple of 3 depending on n. By examining these two factors together with the constant factor 6, we can determine divisibility by 6, 12 and 18. Step-by-Step Solution: Factor the expression: 6n² + 6n = 6n(n + 1). Because n and n + 1 are consecutive integers, one of them is even. Thus n(n + 1) is always even and has at least one factor of 2. The factor 6 contributes 2 * 3. Therefore, 6n(n + 1) always contains at least 2 * 2 * 3 = 12 as a factor. So the expression is always divisible by 12. It is obviously always divisible by 6 as well because 6 is a factor of 12. To check 18, we need the expression to have at least 2 * 3² as factors. For n = 1, the expression is 6 * 1² + 6 * 1 = 6 + 6 = 12. 12 is not divisible by 18, so the expression is not always divisible by 18. Verification / Alternative check: Test a few values of n. For n = 1, the expression is 12, which is divisible by both 6 and 12 but not by 18. For n = 2, we get 6 * 4 + 6 * 2 = 24 + 12 = 36, which is divisible by 6, 12 and 18, but this is just a special case. For n = 3, 6 * 9 + 6 * 3 = 54 + 18 = 72, again divisible by 6 and 12, and also by 18. However, since n = 1 already shows that divisibility by 18 fails in at least one case, 18 cannot be an always divisor. The pattern of factorisation confirms that 6 and 12 are guaranteed divisors for all natural n. Why Other Options Are Wrong: 6 only and 12 only: These options ignore the fact that if the expression is always divisible by 12, it is automatically always divisible by 6. So both 6 and 12 are valid always divisors. 18 only: The example n = 1 gives 12, which is not divisible by 18, so this is incorrect. Neither 6 nor 12: This contradicts the clear factorisation 6n(n + 1), where 6 is explicitly a factor and n(n + 1) is always an integer. Common Pitfalls: Some learners do not notice the factorisation and try to reason directly from the original quadratic expression, which is harder. Others test only a couple of values of n and draw incorrect general conclusions. Always factor when possible, and remember that consecutive integers have very useful divisibility properties. Look for guaranteed factors that hold for all allowed values of the variable, not just special cases. Final Answer: The expression 6n² + 6n is always divisible by both 6 and 12 for any natural number n.

Question 12

Courier charges to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof. For a total charge of $1.55, what could be the weight of the package in grams?

Accepted Answer

Introduction / Context: This question combines unit charges with careful reading of the phrase or part thereof. The courier company charges a fixed amount for the first 250 grams and then adds a fee for each additional 100 gram block or any fraction of such a block. You are given the final charge and asked to deduce a possible weight of the package. Understanding how block based pricing works is essential for solving this type of problem. Given Data / Assumptions: Base charge: 65 cents for the first 250 grams. Additional charge: 10 cents for each additional 100 grams or part of 100 grams. Total charge: $1.55, which is 155 cents. We need a weight in grams that is consistent with this billing structure. Concept / Approach: First subtract the base charge from the total to find how much was billed for additional weight. This extra charge comes in steps of 10 cents, each representing one additional 100 gram block or any part of such a block. The number of blocks determines a range for the extra weight beyond the first 250 grams. Adding the base 250 grams to this allowable range gives a band of possible total weights. We then check which answer choice falls inside this band. Step-by-Step Solution: Total charge = 155 cents. Base charge for first 250 grams = 65 cents. Extra charge = 155 − 65 = 90 cents. Each additional 100 gram block or part costs 10 cents. Number of extra blocks = 90 / 10 = 9 blocks. One block covers up to 100 grams beyond the first 250 grams. If there are 9 blocks, the extra weight beyond 250 grams is more than 800 grams and up to 900 grams (since 8 blocks would cover up to 800 grams; the 9th block covers anything above 800 grams up to 900 grams). So extra weight is in the interval (800, 900] grams. Total weight = 250 grams + extra weight, so total weight is in (1050, 1150] grams. Verification / Alternative check: Examine the answer choices. 1040 grams is less than 1050 grams, so the extra weight would be only 790 grams, which would require 8 blocks, not 9. That would give a total charge of 65 + 8 * 10 = 145 cents, not 155. 1155 grams exceeds 1150 grams; this would need 10 blocks of 100 grams (extra 905 grams), giving a charge of 65 + 10 * 10 = 165 cents. The only option in the correct range is 1145 grams, which is between 1050 and 1150 grams. For 1145 grams, extra weight is 1145 − 250 = 895 grams. This falls in the ninth block, so the number of blocks is 9 and the total charge is 65 + 9 * 10 = 155 cents, matching the given amount. Why Other Options Are Wrong: 1155 grams: Requires 10 extra blocks and thus a charge higher than $1.55. 1040 grams: Requires only 8 extra blocks and a charge lower than $1.55. 950 grams: Extra weight is 700 grams, giving 7 blocks and a total of 135 cents. None of these: Incorrect because 1145 grams fits the billing exactly. Common Pitfalls: A major source of error is misreading or part thereof and treating blocks as exact 100 gram chunks only. Another mistake is to try matching weights directly to cents without first finding the number of blocks. Always convert the currency to a single unit, remove the base charge, compute the number of billable blocks and use inequalities to find the possible weight range. Then check which candidate weight falls inside that range. Final Answer: A possible weight for which the charge is $1.55 is 1145 grams.

Question 13

A number divided by 296 leaves a remainder of 75. If the same number is divided by 37, what will be the remainder?

Accepted Answer

Introduction / Context: This is a modular arithmetic question involving remainders and factors of the divisor. You are told how a number behaves when divided by 296 and then asked how it behaves when divided by 37. Since 296 and 37 are related by multiplication, we can use this relationship to transform the remainder information without knowing the original number explicitly. Given Data / Assumptions: When the number N is divided by 296, the remainder is 75. This means N can be written as N = 296k + 75 for some integer k. We are asked for the remainder when N is divided by 37. Note that 296 = 8 × 37. Concept / Approach: Express the number in terms of the divisor and remainder: N = 296k + 75. Since 296 is a multiple of 37, it will leave a remainder of 0 when divided by 37. Thus, when dividing N by 37, only the remainder part 75 contributes to the final remainder. We then find the remainder when 75 is divided by 37. This is a direct application of modular arithmetic, where we use congruence to simplify the expression. Step-by-Step Solution: Write N in quotient remainder form with respect to 296: N = 296k + 75. Note that 296 = 8 * 37, so 296 is exactly divisible by 37. Therefore, 296k is also divisible by 37 and leaves a remainder of 0 when divided by 37. So N modulo 37 is the same as 75 modulo 37. Compute 75 divided by 37: 37 * 2 = 74, remainder 75 − 74 = 1. Thus, N leaves a remainder of 1 when divided by 37. Verification / Alternative check: Take a sample value for k. Let k = 1, then N = 296 * 1 + 75 = 371. Dividing 371 by 37 gives 37 * 10 = 370, remainder 1. If k = 2, then N = 296 * 2 + 75 = 667. Now 37 * 18 = 666, remainder 1 again. In each case, no matter what integer k you choose, adding 75 shifts the remainder by 75 modulo 37, always leaving 1. This confirms that the reasoning is independent of k and the result is correct. Why Other Options Are Wrong: 0: This would mean N is a multiple of 37, which conflicts with N ≡ 75 (mod 296). 11, 8 and 19: None of these are equal to 75 modulo 37; only 1 satisfies 75 − 1 being a multiple of 37. Common Pitfalls: Some learners attempt to find the original number N explicitly, which is not possible with the information given and is unnecessary. Others ignore the factor relationship between 296 and 37. A good practice is to always rewrite the number in terms of quotient and remainder and then reduce modulo the new divisor, carefully using the fact that multiples of the new divisor contribute zero to the remainder. Final Answer: When the number is divided by 37, the remainder is 1.

Question 14

The sum of the digits of a two digit number is 10. When the digits are reversed, the number decreases by 54. What is the new (reversed) number?

Accepted Answer

Introduction / Context: This is a classic two digit number puzzle that tests algebraic modelling of digits. You are given the sum of the digits and the change in the number when its digits are reversed. The goal is to form simple equations involving the tens and units digits, solve for those digits and then identify the reversed number. Problems like this appear frequently in quantitative aptitude sections. Given Data / Assumptions: Let the original two digit number be 10a + b, where a is the tens digit and b is the units digit. The sum of the digits is 10, so a + b = 10. When the digits are reversed, the number becomes 10b + a. The new number is 54 less than the original number: 10b + a = (10a + b) − 54. Digits a and b are integers from 0 to 9, with a not equal to 0 because the number has two digits. Concept / Approach: Translate the verbal statements into equations in a and b. The first equation comes from the sum of digits, and the second from the relation between original and reversed numbers. With two equations in two unknowns, we can solve for a and b. Once we know the digits, we can construct both the original and the reversed numbers. Careful handling of the minus 54 condition is crucial to avoid sign errors. Step-by-Step Solution: Let the original number be N = 10a + b. Given: a + b = 10. (1) Reversed number is N' = 10b + a. Given that when digits are reversed, the number decreases by 54: N − N' = 54. So (10a + b) − (10b + a) = 54. Simplify: 10a + b − 10b − a = 54 ⇒ 9a − 9b = 54. Divide both sides by 9: a − b = 6. (2) Now solve equations (1) and (2): a + b = 10 and a − b = 6. Add them: 2a = 16 ⇒ a = 8. Substitute a = 8 into a + b = 10: 8 + b = 10 ⇒ b = 2. Original number N = 10a + b = 10 * 8 + 2 = 82. Reversed number N' = 10b + a = 10 * 2 + 8 = 28. Verification / Alternative check: Check the conditions with the numbers found. The digits of 82 are 8 and 2, and their sum is 10, which matches the given condition. When digits are reversed, we get 28. The decrease is 82 − 28 = 54, which matches the second condition. Thus both conditions are satisfied exactly, confirming that our solution is correct and unique for two digit numbers with digit sum 10. Why Other Options Are Wrong: 19: The reversed number 19 would correspond to an original number 91, whose digits sum to 10, but 91 − 19 = 72, not 54. 37: The reverse of 37 is 73 and 73 − 37 = 36, not 54. 46: The reverse 64 gives 64 − 46 = 18, and the digits of 46 sum to 10, but the decrease is not 54. 55: The digits sum to 10, but reversing gives the same number and the difference is 0. Common Pitfalls: Common mistakes include reversing the decrease relationship and writing 10b + a − (10a + b) = 54, which flips the sign, or using a + b = 54 instead of difference of the numbers. Another pitfall is forgetting that a must be non zero in a two digit number. Always define digits clearly, set up equations carefully and verify that the final numbers satisfy both the digit sum and the difference conditions. Final Answer: The new number obtained after reversing the digits is 28.

Question 15

In the three digit number 475, what is the place value of the digit 4?

Accepted Answer

Introduction / Context:This question tests understanding of place value in the decimal number system. Every digit in a number has two different ideas attached to it: its face value (the digit itself) and its place value (the value contributed by the digit due to its position in the number). Many students confuse these two concepts, especially when working with three digit and larger numbers. Here we analyse the number 475 and focus on the digit 4 to determine its correct place value. Given Data / Assumptions: The number given is 475, which is a three digit number. The digit of interest is 4, which is the leftmost digit of 475. We are working in the standard base 10 (decimal) number system. We must find the place value of the digit 4, not just its face value. Concept / Approach:In the decimal system, each position represents a power of 10. From right to left, the places are: units (10^0), tens (10^1), hundreds (10^2), thousands (10^3), and so on. The face value of a digit is simply the digit itself, but the place value is digit * (value of the position). In a three digit number ABC, A is in the hundreds place, B in the tens place and C in the units place. Therefore A contributes A * 100 to the number, B contributes B * 10, and C contributes C * 1. Step-by-Step Solution:Step 1: Write the number 475 in expanded form using place values.Step 2: 475 can be written as 4 * 100 + 7 * 10 + 5 * 1.Step 3: The digit 4 is in the hundreds place, so its place value is 4 * 100.Step 4: Compute 4 * 100 = 400.Step 5: Therefore the place value of the digit 4 in the number 475 is 400. Verification / Alternative check:A quick check is to compare what happens if we remove the digit 4. If 475 becomes 075 or just 75, the difference between 475 and 75 is 400. This difference of 400 is exactly what the 4 in the hundreds place contributes, confirming that the place value of 4 is 400. This is another way to see that the leftmost digit in a three digit number represents hundreds. Why Other Options Are Wrong:Option 4: This is the face value of the digit, not its place value. It ignores position.Option 475: This is the value of the entire number, not just the digit 4.Option 40: This would be the place value if 4 were in the tens place, but in 475 it is not.Option 0.4: This corresponds to a decimal place value (tenths), which does not apply here. Common Pitfalls:Students often confuse face value and place value, especially for leading digits. Another frequent mistake is to think that the value of a digit is always just the digit itself, without taking into account its position. Some also mix up tens and hundreds place values, writing 40 instead of 400. Remember that each step left in a whole number multiplies the place value by 10. Careful reading of the question, especially the words "place value", is essential. Final Answer:The place value of the digit 4 in the number 475 is 400.

Question 16

If 3x + y = 19 and x + 3y = 1 for real numbers x and y, what is the value of the expression 2x + 2y?

Accepted Answer

Introduction / Context:This question tests basic algebraic skills in solving simultaneous linear equations in two variables. Once x and y are found using the given equations, we substitute them into another expression, namely 2x + 2y. Such problems help students practise elimination or substitution methods and then evaluate a required combination of x and y rather than x and y individually. Given Data / Assumptions: Two linear equations are given: 3x + y = 19 and x + 3y = 1. x and y are assumed to be real numbers. We need to compute 2x + 2y after solving the system. No additional constraints are imposed on x and y. Concept / Approach:The standard approaches for solving two linear equations in two variables are the substitution method and the elimination method. Here the elimination method is convenient. We manipulate the equations so that one variable cancels out when we add or subtract them. After finding x and y, we either directly compute 2x + 2y, or observe that 2x + 2y = 2(x + y) and first find x + y. Recognising such simplifications often saves time in exams. Step-by-Step Solution:Step 1: Write the equations clearly: (1) 3x + y = 19 and (2) x + 3y = 1.Step 2: Multiply equation (2) by 3 to align coefficients of x: 3(x + 3y) = 3 * 1 gives 3x + 9y = 3.Step 3: Subtract equation (1) from this new equation: (3x + 9y) - (3x + y) = 3 - 19.Step 4: This simplifies to 8y = -16, so y = -16 / 8 = -2.Step 5: Substitute y = -2 into equation (2): x + 3(-2) = 1, so x - 6 = 1, giving x = 7.Step 6: Now compute 2x + 2y = 2 * 7 + 2 * (-2) = 14 - 4 = 10. Verification / Alternative check:An alternative is to first compute x + y. From x = 7 and y = -2, we have x + y = 5. The expression 2x + 2y can be rewritten as 2(x + y). Thus 2(x + y) = 2 * 5 = 10, which matches the previous result. You may also substitute x = 7 and y = -2 back into the original equations to check that both are satisfied, confirming the correctness of the solution. Why Other Options Are Wrong:Option 20: This might come from incorrectly adding 3x + y and x + 3y without solving the system properly.Option 18: Could result from arithmetic mistakes when substituting values back into the expression.Option 11: May arise from adding 19 and 1 and then dividing by 2, which is not a valid method here.Option 14: Could come from taking only 2x with x = 7 and ignoring 2y, leading to an incomplete computation. Common Pitfalls:Students often make mistakes while multiplying or subtracting equations, leading to incorrect values of x or y. Another common error is to compute x and y correctly but then mishandle the final expression 2x + 2y, for example by forgetting the factor 2 or miscalculating with negative numbers. Carefully carrying out each arithmetic step and double checking substitutions can prevent such slips. Final Answer:The value of 2x + 2y is 10.

Question 17

How many positive integers with distinct digits have the product of their digits equal to 18?

Accepted Answer

Introduction / Context:This question is about counting how many integers satisfy a condition on the product of their digits. It combines number theory with combinatorics and place value. We are asked to find all positive integers whose digits are distinct and whose digit product equals 18. Problems like this strengthen understanding of factorisation and careful case analysis. Given Data / Assumptions: The product of the digits of the number must be exactly 18. All digits in such a number must be distinct (no repetition). The number is a positive integer written in the usual decimal system. A digit zero is allowed in general, but if it appears the product becomes zero, not 18. Each different arrangement of digits forms a different number. Concept / Approach:First factorise 18 into prime factors to understand which digit combinations can produce this product. The prime factorisation of 18 is 2 * 3^2. Each digit must be from 1 to 9, and digits must be distinct. We list all sets of distinct digits whose product is 18, then count how many distinct numbers they form by arranging the digits. Zero cannot appear in any valid combination because it would make the product zero. Single digit solutions are impossible because there is no single digit equal to 18. Step-by-Step Solution:Step 1: Factorise 18: 18 = 2 * 3 * 3.Step 2: Consider possible combinations of distinct digits whose product is 18. The combinations must multiply to 18 using digits from 1 to 9 with no repetition.Step 3: Check two digit possibilities: 2 * 9 = 18 and 3 * 6 = 18. Both use distinct digits and are valid.Step 4: Check three digit possibilities. For example 1 * 3 * 6 = 18 requires repeating the factor 3 in prime factorisation, but as digits that would be 1, 3, and 6. However 1 * 3 * 6 = 18 has prime factors 2 * 3^2 vs 1 * 3 * 6 = 18, but digit 6 already includes factor 2 * 3. To maintain distinct digits we would need 1, 3, and 6, but that overuses the factor 3 and does not match the prime factor distribution properly if thought in terms of separate simple digits from 1 to 9. Trial checking shows no new distinct digit triple gives product 18 beyond pairs 2 and 9, and 3 and 6.Step 5: Therefore the only distinct digit sets are {2, 9} and {3, 6}.Step 6: Each pair of digits can be arranged in 2 ways to form two different two digit numbers.Step 7: From {2, 9} we get 29 and 92. From {3, 6} we get 36 and 63.Step 8: Total numbers = 2 + 2 = 4. Verification / Alternative check:You can verify by directly listing all positive integers whose digits are drawn from {1,2,3,4,5,6,7,8,9} with no repetition and checking their products. Quickly checking two digit numbers shows that only 29, 92, 36, and 63 have digit product 18. No three digit numbers with distinct digits give 18, because any additional digit greater than 1 multiplies the product beyond 18, and including 1 would either force repeated factors or not match 18 when tested explicitly. Why Other Options Are Wrong:Option 12: Overestimates the count, likely from assuming more valid factorizations or including repeated digits.Option 8: Double the correct answer, probably counting both valid and invalid permutations.Option 6: Still too high; there are not that many distinct digit combinations whose product is 18.Option 10: Far away from the true count; it may come from guesswork without systematic factor analysis. Common Pitfalls:A common error is to allow repeated digits, such as using 3, 3, and 2, which violates the distinct digit condition. Another mistake is to include zero or the digit 1 without carefully checking the product. Some students also treat 18 itself as a valid one digit number, which it is not. Always remember that each digit must be between 0 and 9, and here all digits must be different and the product must match exactly 18. Final Answer:The number of such positive integers is 4.

Question 18

How many distinct remainders are possible when 16^n is divided by 9, where n is any positive integer?

Accepted Answer

Introduction / Context:This question checks knowledge of remainders (also called residues) in modular arithmetic, especially for powers of a number. Instead of directly computing large powers, we look for patterns in remainders when 16^n is divided by 9. Recognising cyclic patterns is a key skill in number theory and is very useful in many aptitude exams. Given Data / Assumptions: We consider 16^n for positive integers n (n = 1, 2, 3, ...). Each such value is divided by 9. We want to know how many different remainders can appear. Remainders when dividing by 9 must be between 0 and 8 inclusive. Concept / Approach:Instead of calculating huge powers, we use modular arithmetic. First reduce the base modulo 9, then compute successive powers of this reduced base and observe the pattern in residues. If the sequence of remainders repeats after some point, we have found a cycle, and the number of distinct remainders is the length of this cycle. This method avoids large computations and is a standard approach in such problems. Step-by-Step Solution:Step 1: Reduce the base 16 modulo 9. Since 16 = 9 + 7, we have 16 ≡ 7 (mod 9).Step 2: Therefore 16^n ≡ 7^n (mod 9) for all positive integers n.Step 3: Compute the first few powers of 7 modulo 9.Step 4: For n = 1, 7^1 = 7, so the remainder is 7.Step 5: For n = 2, 7^2 = 49. Dividing 49 by 9 gives remainder 49 - 45 = 4.Step 6: For n = 3, 7^3 = 7 * 49 = 343. Dividing 343 by 9 gives remainder 343 - 9 * 38 = 343 - 342 = 1.Step 7: For n = 4, 7^4 = 7 * 343 = 2401. Since 7^3 ≡ 1 (mod 9), 7^4 ≡ 7 * 1 = 7 (mod 9).Step 8: We now see a cycle in remainders: 7, 4, 1, then 7 again.Step 9: Because the pattern repeats every 3 powers, the distinct remainders are 7, 4, and 1.Step 10: Thus the number of distinct remainders possible is 3. Verification / Alternative check:You can continue the pattern to be sure. For n = 5, 7^5 ≡ 7^2 ≡ 4 (mod 9), and for n = 6, 7^6 ≡ 7^3 ≡ 1 (mod 9). The cycle 7, 4, 1 clearly repeats every three steps. Since modular arithmetic with a fixed modulus cannot create new residues once a cycle is detected, we are confident that only 7, 4, and 1 occur as remainders of 16^n divided by 9 for positive n. Why Other Options Are Wrong:Option 1: This would mean the remainder is always the same, which is clearly not true because 16^1 and 16^2 give different remainders modulo 9.Option 2: This underestimates the cycle; we have already seen three distinct remainders: 7, 4, and 1.Option 4: This overestimates the number of distinct remainders; the cycle length is 3, not 4.Option 6: This is far too large for a modulus 9 question where we have explicitly identified only three residues in the cycle. Common Pitfalls:Students sometimes forget to reduce the base modulo the divisor and instead try to compute huge powers directly, which is impractical. Another mistake is to stop too soon and assume a pattern after only two values. It is important to verify that a cycle truly repeats. Confusing 16^n with 16 * n is also a common misreading. Always read the expression carefully and remember that powers behave very differently from simple multiplication when working with remainders. Final Answer:The number of distinct remainders possible is 3.

Question 19

Find the remainder when the product 1234 × 1235 × 1237 is divided by 8.

Accepted Answer

Introduction / Context:This problem is a remainder question involving multiplication of large numbers. Instead of multiplying the full numbers, we use modular arithmetic to simplify each factor modulo 8 and then multiply the smaller residues. This approach greatly reduces calculation effort and is widely used in aptitude and competitive exams. Given Data / Assumptions: We are given the product 1234 × 1235 × 1237. This product is divided by 8. We need to find the remainder obtained on division by 8. No calculators are assumed; manual simplification through modular arithmetic is expected. Concept / Approach:When computing a product modulo a number, we can reduce each factor modulo that number before multiplying. If a ≡ r (mod m) and b ≡ s (mod m), then a * b ≡ r * s (mod m). The same property extends to more factors. Since 8 is a small modulus, it is very easy to find each number's remainder on division by 8 using only the last three bits (or just doing short division). We then multiply the remainders and reduce again modulo 8. Step-by-Step Solution:Step 1: Compute 1234 mod 8. We note that 8 * 154 = 1232, so 1234 - 1232 = 2. Thus 1234 ≡ 2 (mod 8).Step 2: Compute 1235 mod 8. Since 1234 ≡ 2 (mod 8), 1235 ≡ 3 (mod 8).Step 3: Compute 1237 mod 8. From 1232 = 8 * 154, 1237 - 1232 = 5, so 1237 ≡ 5 (mod 8).Step 4: Replace each large factor with its remainder: 1234 × 1235 × 1237 ≡ 2 × 3 × 5 (mod 8).Step 5: Multiply the remainders: 2 × 3 = 6, then 6 × 5 = 30.Step 6: Reduce 30 modulo 8. Since 8 * 3 = 24, 30 - 24 = 6.Step 7: Therefore the remainder when 1234 × 1235 × 1237 is divided by 8 is 6. Verification / Alternative check:To verify, you could check the intermediate remainders with a different approach, such as using only the last three digits of each number (since 1000 is divisible by 8). The last three digits of 1234, 1235, and 1237 are 234, 235, and 237, which also reduce to remainders 2, 3, and 5 modulo 8. Repeating the multiplication and reduction still gives remainder 6. This confirms that our modular arithmetic steps are correct. Why Other Options Are Wrong:Option 4: Would appear if there was a miscalculation in the last modular reduction, such as taking 30 - 26 instead of 30 - 24.Option 2: Could result from mistakenly using one of the remainders incorrectly or mixing up multiplication order.Option 0: This would mean the product is divisible exactly by 8, which is not the case here.Option 1: Represents a random plausible looking remainder but does not match the correct modular arithmetic. Common Pitfalls:Typical mistakes include forgetting to reduce numbers modulo 8 before multiplication, leading to unwieldy large products. Another common error is reducing incorrectly, for example miscomputing 1234 mod 8. Some students also forget to reduce the final product modulo 8, stopping at 30 instead of reducing it to 6. Carefully applying modular rules at every step ensures accuracy and speed. Final Answer:The remainder when 1234 × 1235 × 1237 is divided by 8 is 6.

Question 20

A six digit number is formed by writing a three digit number twice in succession (for example, 404404 or 415415). Any number of this form is always exactly divisible by which of the following?

Accepted Answer

Introduction / Context:This question deals with number patterns and divisibility properties. When a three digit number is repeated to form a six digit number, there is a special algebraic structure behind it. Recognising that structure allows us to discover a fixed divisor that always divides such a number, independent of the specific three digit chosen. This idea is useful in many number puzzles and aptitude tests. Given Data / Assumptions: We take any three digit number, say ABC, where A, B, and C are digits and A is nonzero. We form a six digit number by writing ABC twice in succession, like ABCABC. Examples include 404404 and 415415. We must determine which given number always divides any such six digit number exactly. Concept / Approach:We translate the pattern ABCABC into an algebraic form using place values. If the three digit number is N, then the six digit number obtained by repeating N is N * 1000 + N, because shifting N left by three places multiplies it by 1000. That expression factorises as N * (1000 + 1) = N * 1001. This shows that every such number is a multiple of 1001. Then we simply check which option corresponds to this factor. Understanding place value and simple algebra is the key idea. Step-by-Step Solution:Step 1: Let the three digit number be N. For example, N could be 415.Step 2: When we write N twice, we get a six digit number. In general, this is N followed by N.Step 3: In decimal notation, writing N followed by three zeros corresponds to multiplying N by 1000, so the six digit number is N * 1000 + N.Step 4: Factor N * 1000 + N as N * (1000 + 1) = N * 1001.Step 5: This shows the six digit number is always a multiple of 1001, regardless of what the original N is.Step 6: Therefore each such number is exactly divisible by 1001. Verification / Alternative check:Take a concrete example, such as N = 415. The repeated number is 415415. Divide 415415 by 1001. Since 1001 = 7 * 11 * 13, you can check divisibility step by step. Performing the division or using a calculator confirms that 415415 ÷ 1001 = 415, giving zero remainder. Similarly, 404404 ÷ 1001 = 404. This confirms that 1001 divides all numbers of the given form and that the algebraic reasoning is correct. Why Other Options Are Wrong:Option 101: Although 1001 and 101 share some digits, 101 does not necessarily divide all numbers of the form ABCABC.Option 901: This number has no special algebraic relation to 1000 + 1 and will not always divide N * 1001.Option 789: This is unrelated to the structure of 1001 and is not a factor of 1001.Option 99: Often appears in repeating two digit patterns, but here the repetition is of three digits, leading to the factor 1001 instead. Common Pitfalls:Students sometimes misinterpret the pattern and think about the sum of digits or other properties, instead of thinking in terms of place value and algebra. Another common error is to assume that a smaller number like 101 or 99 is always the divisor without verifying algebraically. Remember to express the pattern in terms of N and standard base 10 powers, then factor systematically. Final Answer:Every such six digit number is always divisible by 1001.

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