Question 1

A teacher multiplies 987 by a certain integer and gets 556781 as the product. It is known that in this answer only the digits 6 and 7 are wrong, and all the other digits are correct. What is the correct product?

Accepted Answer

Introduction / Context:This problem combines multiplication with error detection. We know that the teacher tried to multiply 987 by some integer and obtained a six digit result, but exactly two digits in the middle of the product are wrong. Our task is to identify the correct product using the information that the positions of the incorrect digits are fixed and the rest of the digits are correct. This type of question tests logical reasoning and understanding of divisibility. Given Data / Assumptions: The intended multiplication is 987 times some integer k. The incorrect reported product is 556781. Only the digits 6 and 7 in this six digit result are wrong. The digits 5, 5, 8, and 1 in their positions are correct. Therefore, the correct product must have the form 55XY81, where X and Y are digits replacing 6 and 7. The correct product must be exactly divisible by 987. Concept / Approach:We know the correct answer must match the pattern 55XY81 and must be a multiple of 987. Instead of guessing the unknown integer k directly, we can test the given answer options, all of which fit the general pattern of keeping the first two and last two digits similar. We simply divide each option by 987 and see which one gives an integer with no remainder. The one that divides exactly will be the correct product. Because the options are close together, this approach is efficient. Step-by-Step Solution:Step 1: Note that the wrong answer is 556781. Only the third and fourth digits (6 and 7) may change.Step 2: The correct answer must be of the form 55XY81, matching the first two and last two digits.Step 3: Examine each option: 553681, 555181, 556581, and 555681 (plus any extra distractor) to see which is divisible by 987.Step 4: Divide 553681 by 987; it does not give an integer, so it is not correct.Step 5: Divide 555181 by 987; again it does not divide exactly.Step 6: Divide 556581 by 987; still not a whole number.Step 7: Divide 555681 by 987; this division gives an exact integer quotient (555681 ÷ 987 = 563), with remainder zero.Step 8: Therefore the correct product must be 555681. Verification / Alternative check:To cross check, compute 987 × 563 explicitly. First 987 × 500 = 493500, then 987 × 60 = 59220, and 987 × 3 = 2961. Add these: 493500 + 59220 = 552720, then 552720 + 2961 = 555681. This matches the candidate option exactly, confirming that 555681 is the true product when 987 is multiplied by 563. Since this product has the digits 5, 5, 5, 6, 8, 1 and only the third and fourth digits differ from 556781, the error description is satisfied. Why Other Options Are Wrong:Option 553681: Even though it looks similar to the wrong answer, it is not divisible by 987 exactly, so it cannot be the correct product.Option 555181: This also fails to be a multiple of 987; division leaves a nonzero remainder.Option 556581: Dividing by 987 again results in a non integer; it does not correspond to 987 multiplied by an integer k.Option 556351: This option neither fits the specific error pattern nor divides exactly by 987. Common Pitfalls:Students may misread the statement and assume that any two digits may be wrong, not specifically the digits 6 and 7 in the middle. Others might try to guess the integer multiplier without using divisibility, leading to unnecessary effort. Some may also forget to test divisibility completely and rely only on visual similarity. Always use the key condition that the correct answer must be a multiple of 987 and must differ only in the specified digit positions. Final Answer:The correct product of 987 and the unknown integer is 555681.

Question 2

What is the rightmost digit (units place) of the sum 65776759 + 54697467?

Accepted Answer

Introduction / Context:This question focuses on finding the units digit of the sum of two large numbers. In many competitive exams, you do not need the entire sum; only the last digit is required. Recognising that only the units digits affect the units digit of the sum is a quick and efficient trick that saves time and avoids big calculations. Given Data / Assumptions: Two large integers are given: 65776759 and 54697467. We need only the rightmost digit (units digit) of their sum. All standard addition rules of decimal arithmetic apply. We do not need to compute the full sum explicitly. Concept / Approach:The units digit of a sum depends only on the units digits of the addends. When we add two numbers, carry from the tens place does not change which pair of units digits we initially add; instead it influences higher places. So to get the final units digit, we just add the units digits of each number and then take the units digit of that small sum. This method is very fast and works for any size of integers. Step-by-Step Solution:Step 1: Identify the units digit of each number.Step 2: For 65776759, the units digit is 9.Step 3: For 54697467, the units digit is 7.Step 4: Add the units digits: 9 + 7 = 16.Step 5: The units digit of 16 is 6.Step 6: Therefore, the units digit of the sum 65776759 + 54697467 is 6. Verification / Alternative check:If you wish, you can perform the full addition to verify. 65776759 + 54697467 = 120474226. The last digit of 120474226 is 6, which matches the units digit we found using the shortcut. However, for exam speed, computing the entire sum is unnecessary. The simple method using only units digits is more efficient and less error prone. Why Other Options Are Wrong:Option 4: This might arise from incorrect addition of 9 and 7, for example mistaking 9 + 7 as 14 and then misreading the units digit.Option 9: May come from mistakenly copying one of the original units digits instead of summing them.Option 0: Would require the units digits of the addends to sum to a multiple of 10, which is not the case here (9 + 7 = 16, not 10 or 20).Option 2: Could result from a miscalculation during mental arithmetic. Common Pitfalls:Students sometimes try to add the full numbers and may make mistakes in the higher place values, which can then incorrectly influence their confidence in the units digit. Others might misread the question and compute something different, like the tens digit. Always read carefully and remember that for the units digit of a sum, only the units digits of the addends matter. Simple mental addition is sufficient. Final Answer:The rightmost digit (units place) of the sum is 6.

Question 3

A box contains 90 screws, each weighing 100 grams, and 100 bolts, each weighing 150 grams. The total weight of the box plus its contents is 35.5 kilograms. What is the weight of the empty box (in kilograms)?

Accepted Answer

Introduction / Context:This is a straightforward arithmetic and unit conversion problem that tests understanding of mass, addition, and subtraction. We are given the total weight of a box and its contents, as well as the individual weight and count of screws and bolts inside. The goal is to find the weight of the empty box itself by subtracting the weight of all the items from the total weight. Given Data / Assumptions: Number of screws = 90. Weight of each screw = 100 grams. Number of bolts = 100. Weight of each bolt = 150 grams. Total weight of box plus contents = 35.5 kilograms. 1 kilogram = 1000 grams. Concept / Approach:The total weight of the contents equals the sum of the weight of all screws and the weight of all bolts. Once we find this in kilograms, we subtract it from the total weight (box plus contents) to get the weight of the empty box. Unit conversion is required because the item weights are given in grams while the total is in kilograms. Converting everything to kilograms avoids confusion and keeps calculations consistent. Step-by-Step Solution:Step 1: Compute total weight of screws. Each screw weighs 100 grams, and there are 90 screws.Step 2: Total screw weight = 90 * 100 grams = 9000 grams.Step 3: Convert 9000 grams to kilograms: 9000 / 1000 = 9 kilograms.Step 4: Compute total weight of bolts. Each bolt weighs 150 grams, and there are 100 bolts.Step 5: Total bolt weight = 100 * 150 grams = 15000 grams.Step 6: Convert 15000 grams to kilograms: 15000 / 1000 = 15 kilograms.Step 7: Total weight of contents = 9 kilograms + 15 kilograms = 24 kilograms.Step 8: Total weight of box plus contents = 35.5 kilograms.Step 9: Weight of empty box = 35.5 kilograms - 24 kilograms = 11.5 kilograms. Verification / Alternative check:You can convert the final answer back to grams to double check. Weight of empty box = 11.5 kilograms = 11500 grams. Adding back the contents: 11500 grams (box) + 9000 grams (screws) + 15000 grams (bolts) = 35500 grams. Converting 35500 grams to kilograms gives 35500 / 1000 = 35.5 kilograms, which matches the given total weight. This confirms that the calculation is consistent. Why Other Options Are Wrong:Option 10.5 kgs: This would imply contents weigh 25 kilograms, which does not match the given numbers.Option 12 kgs: Would lead to total contents weight of 23.5 kilograms, contradicting the actual content weights.Option 9.6 kgs: Comes from an incorrect combination or unit conversion error.Option 8.5 kgs: Underestimates the box weight significantly and does not satisfy the total of 35.5 kilograms. Common Pitfalls:Students may forget to convert grams to kilograms, leading to incorrect subtraction. Another frequent mistake is miscalculating the total number of grams by mixing up the counts or weights of screws and bolts. Carefully separating each part of the calculation and checking the arithmetic at every step helps avoid these errors. Final Answer:The weight of the empty box is 11.5 kgs.

Question 4

When 85.44 is multiplied by 0.08455, how many digits will there be to the right of the decimal point in the product?

Accepted Answer

Introduction / Context:This problem checks your understanding of how decimal places behave when two decimal numbers are multiplied. Instead of calculating the full product of 85.44 and 0.08455, we only need to determine how many digits will appear to the right of the decimal point in the final result. This is a standard concept in working with decimals and helps in quick estimation. Given Data / Assumptions: First decimal number: 85.44. Second decimal number: 0.08455. We need the total number of digits to the right of the decimal point in their product. No rounding is mentioned; we count all decimal digits resulting from exact multiplication. Concept / Approach:When multiplying two decimals, the total number of digits to the right of the decimal point in the product is equal to the sum of the numbers of digits to the right of the decimal points in each factor. This rule comes from the fact that moving the decimal point is equivalent to multiplying or dividing by powers of 10. We use this rule without computing the actual product value, which saves time and avoids arithmetic mistakes. Step-by-Step Solution:Step 1: Count the number of digits to the right of the decimal point in 85.44.Step 2: In 85.44, there are 2 digits after the decimal point (4 and 4).Step 3: Count the number of digits to the right of the decimal point in 0.08455.Step 4: In 0.08455, there are 5 digits after the decimal point (0, 8, 4, 5, and 5).Step 5: Add these counts: 2 + 5 = 7.Step 6: Therefore the product 85.44 × 0.08455 will have 7 digits to the right of the decimal point. Verification / Alternative check:To verify, imagine removing the decimal points and treating the numbers as integers, then adjusting the decimal afterwards. 85.44 can be seen as 8544 / 100, and 0.08455 can be seen as 8455 / 100000. Their product is (8544 * 8455) / (100 * 100000) = (8544 * 8455) / 10^7. Since the denominator is 10^7, the decimal representation of the product will have 7 digits after the decimal point, which matches our earlier count. Why Other Options Are Wrong:Option 5: Ignores the decimal places of one of the factors or miscounts them.Option 4: May come from counting only digits after the decimal in one number.Option 6: Represents adding incorrectly, for example 2 + 5 and then subtracting 1 by mistake.Option 8: Overestimates the total number of decimal places; there is no basis for this count from the given numbers. Common Pitfalls:A common mistake is to miscount the digits after the decimal point, especially when there are leading zeros, as in 0.08455. Students sometimes count 4 digits instead of 5. Another error is to think that multiplication somehow shortens the decimal part, which is not true in general. Always remember: the number of decimal places in the product is the sum of the decimal places in the factors. Final Answer:The product has 7 digits to the right of the decimal point.

Question 5

Consider the increasing sequence of natural numbers in which no term is a perfect square and no term is a perfect cube. The first few terms are 2, 3, 5, 6, 7, 10, ... What is the 300th term of this sequence?

Accepted Answer

Introduction / Context:This problem is about constructing and analysing a special sequence of natural numbers. The sequence contains all positive integers that are neither perfect squares nor perfect cubes. We are asked to find the 300th term in this sequence. This is a classic inclusion excluding type counting problem that requires careful reasoning rather than listing all terms one by one. Given Data / Assumptions: The sequence contains natural numbers greater than or equal to 1, but numbers that are perfect squares or perfect cubes are excluded. The first few valid terms are 2, 3, 5, 6, 7, 10, and so on, confirming that 1, 4, 8, 9, 16, 25, 27, etc. are omitted. We need to determine the 300th term of this sequence. We assume standard definitions: perfect square means n^2 for an integer n, and perfect cube means n^3 for an integer n. Concept / Approach:Instead of listing all numbers, we use counting. For a given positive integer N, let us count how many integers from 1 to N are excluded because they are either perfect squares or perfect cubes. The remaining numbers (from 1 to N) are exactly those that belong to the sequence. Using inclusion exclusion, we subtract the count of squares and cubes and then add back the count of sixth powers, which have been subtracted twice. Then we choose N such that the count of allowed numbers is 300 and identify the smallest such N. That N will be the 300th term of the sequence. Step-by-Step Solution:Step 1: For a given N, define A(N) as the number of natural numbers from 1 to N that are perfect squares.Step 2: A(N) = floor(sqrt(N)), because the squares are 1^2, 2^2, 3^2, ..., up to floor(sqrt(N))^2.Step 3: Define B(N) as the number of natural numbers from 1 to N that are perfect cubes.Step 4: B(N) = floor(N^(1/3)), because the cubes are 1^3, 2^3, 3^3, ..., up to floor(N^(1/3))^3.Step 5: Some numbers are both perfect squares and perfect cubes; these are perfect sixth powers (because lcm(2, 3) = 6). The count of such numbers from 1 to N is C(N) = floor(N^(1/6)).Step 6: The number of integers from 1 to N that are either squares or cubes is A(N) + B(N) - C(N) by inclusion exclusion.Step 7: Therefore, the number of integers from 1 to N that are neither perfect squares nor perfect cubes is S(N) = N - (A(N) + B(N) - C(N)).Step 8: We need N such that S(N) = 300.Step 9: By testing values around 320, we find that for N = 320, S(320) = 299, and for N = 321, S(321) = 300.Step 10: Hence the 300th number in the sequence is N = 321. Verification / Alternative check:To verify, compute A(321) = floor(sqrt(321)) = 17 (since 17^2 = 289 and 18^2 = 324 > 321). Compute B(321) = floor(321^(1/3)) = 6 (since 6^3 = 216 and 7^3 = 343 > 321). Compute C(321) = floor(321^(1/6)) = 2 (since 2^6 = 64 and 3^6 = 729 > 321). Then A(321) + B(321) - C(321) = 17 + 6 - 2 = 21. So S(321) = 321 - 21 = 300. For N = 320, the counts of squares, cubes, and sixth powers are the same, giving S(320) = 320 - 21 = 299. Thus 321 is the smallest N with 300 acceptable terms, confirming that it is the 300th term of the sequence. Why Other Options Are Wrong:Option 322: For N = 322, one more allowed number is added, so S(322) = 301, which would be the 301st term.Option 319: For N = 319, the count of valid numbers is less than 300, so 319 cannot be the 300th term.Option 320: As computed, S(320) = 299, so 320 corresponds to the 299th term.Option 323: This is larger than 321 and corresponds to a later term in the sequence, not the 300th. Common Pitfalls:Students may try to list terms manually, which is impractical for the 300th term and very error prone. Another frequent mistake is to subtract only squares or only cubes, forgetting the inclusion exclusion principle. Some also forget about sixth powers, which leads to double subtraction of numbers that are both squares and cubes. Carefully using the formula S(N) = N - (floor(sqrt(N)) + floor(N^(1/3)) - floor(N^(1/6))) avoids these mistakes. Final Answer:The 300th term of the sequence is 321.

Question 6

Which of the following numbers is not a prime number?

Accepted Answer

Introduction / Context:This question tests the basic concept of prime numbers. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. Any number with more than two positive divisors is called a composite number. Here we must identify which option is not prime, that is, which number has additional divisors besides 1 and itself. Given Data / Assumptions: The options given are 5, 11, 21, 37, and 41. We assume the standard definition of prime numbers. We check each candidate by trying to factor it into smaller positive integers. We only need to find the one that is clearly composite. Concept / Approach:To test whether a number n is prime, we try to divide it by all prime numbers less than or equal to sqrt(n). If none of these primes divides n exactly, then n is prime. For relatively small numbers like those given, we can check divisibility by small primes such as 2, 3, 5, and 7. A composite number will factor into a product of smaller integers greater than 1. Step-by-Step Solution:Step 1: Check 5. It is not divisible by 2, 3, or any other smaller prime except 1 and 5, so 5 is prime.Step 2: Check 11. It is not divisible by 2, 3, or 5. The only positive divisors are 1 and 11, so 11 is prime.Step 3: Check 21. Test divisibility by 2: 21 is odd, so not divisible by 2. Test divisibility by 3: 2 + 1 = 3, and 3 divides 21, so 21 = 3 * 7.Step 4: Since 21 has divisors 1, 3, 7, and 21, it is not prime; it is composite.Step 5: Check 37. It is not divisible by 2, 3, or 5. The only positive divisors are 1 and 37, so 37 is prime.Step 6: Check 41 similarly. 41 is not divisible by 2, 3, or 5. It has only 1 and 41 as divisors and is therefore prime.Step 7: The only non prime (composite) number among the options is 21. Verification / Alternative check:You can confirm 21 is composite using divisibility rules. The sum of its digits is 2 + 1 = 3, which is divisible by 3, so 21 is divisible by 3. Dividing 21 by 3 gives 7, so 21 = 3 * 7. Since it clearly has factors other than 1 and itself, it cannot be prime. For the other numbers, quick divisibility checks by small primes show no exact divisors apart from 1 and the number itself. Why Other Options Are Wrong:Option 5: 5 has only divisors 1 and 5, so it is prime and does not satisfy the condition of being non prime.Option 11: 11 is a standard example of a prime number greater than 10.Option 37: 37 is not divisible by any prime less than or equal to 6, making it a prime number.Option 41: 41 is also a prime number; it has no divisors other than 1 and 41. Common Pitfalls:Students sometimes confuse prime and composite numbers or assume that any odd number is prime, which is not true. For example, 21 is odd but composite. Another mistake is not checking divisibility by all relevant small primes, such as forgetting to test divisibility by 3 or 7. Always recall that a prime number greater than 2 is odd, but not every odd number is prime. Factorisation or divisibility tests are essential to be sure. Final Answer:The number that is not prime is 21.

Question 7

For an integer n, which of the following expressions can never be odd (that is, is always even)?

Accepted Answer

Introduction / Context:This question tests understanding of parity, that is, whether expressions are odd or even. We are given several algebraic expressions in terms of an integer n and asked to identify which expression can never be odd. In other words, we must find the expression that is always even regardless of the value of n. This is a fundamental idea in number theory and algebra. Given Data / Assumptions: n is an integer (it may be even or odd). We consider the expressions n + 3, 2n, n + 4, 3n, and n - 1. We need to find the expression that is always even and therefore never odd. We use standard definitions: an even integer is divisible by 2, and an odd integer is of the form 2k + 1 for some integer k. Concept / Approach:We analyse the parity of each expression in terms of whether n is even or odd. An even number can be written as 2k, and an odd number as 2k + 1. By substituting these forms into each expression, we determine whether the result is always even, always odd, or can be both depending on n. The expression that is always a multiple of 2, no matter what n is, is the one that can never be odd. Step-by-Step Solution:Step 1: Consider the expression 2n. If n is any integer, then 2n = 2 * n, which is always divisible by 2.Step 2: Therefore 2n is always even and can never be odd.Step 3: Now check n + 3. If n = 2k (even), then n + 3 = 2k + 3, which is odd. If n = 2k + 1 (odd), then n + 3 = 2k + 4 = 2(k + 2), which is even. So n + 3 can be odd for some n and even for others.Step 4: Check n + 4. If n = 2k (even), then n + 4 = 2k + 4 = 2(k + 2), which is even. If n = 2k + 1 (odd), then n + 4 = 2k + 5, which is odd. So n + 4 also can be odd for some n.Step 5: Check 3n. If n is even (2k), then 3n = 3 * 2k = 6k, which is even. If n is odd (2k + 1), then 3n = 3(2k + 1) = 6k + 3, which is odd. So 3n can be odd or even depending on n.Step 6: Check n - 1. If n = 2k (even), then n - 1 = 2k - 1, which is odd. If n = 2k + 1 (odd), then n - 1 = 2k, which is even. So n - 1 also can be odd.Step 7: Among all expressions, only 2n is guaranteed to be even for every integer n and can never be odd. Verification / Alternative check:You can test specific values of n to verify your conclusions. For example, let n = 2 (even). Then n + 3 = 5 (odd), 2n = 4 (even), n + 4 = 6 (even), 3n = 6 (even), and n - 1 = 1 (odd). Now let n = 3 (odd). Then n + 3 = 6 (even), 2n = 6 (even), n + 4 = 7 (odd), 3n = 9 (odd), and n - 1 = 2 (even). We see that every expression except 2n can switch between odd and even, while 2n always stays even. This confirms our reasoning. Why Other Options Are Wrong:Option n + 3: This expression is odd for even n and even for odd n, so it can be odd.Option n + 4: Similarly, it is even for even n and odd for odd n, so it can be odd.Option 3n: When n is odd, 3n is odd; when n is even, 3n is even, so it can be odd.Option n - 1: This expression is odd for even n and even for odd n, so it can be odd as well. Common Pitfalls:Students sometimes misinterpret the phrase "cannot be odd" as meaning "is odd", which reverses the logic of the question. Another frequent error is to assume that if an expression is even for some values, then it is always even. Always check both cases, n even and n odd, to understand full parity behaviour. Remember that a coefficient of 2 guarantees that an expression is even, but only if every term has a factor of 2. Final Answer:The expression that can never be odd is 2n.

Question 8

A perfect cube is an integer whose cube root is an integer (for example, 27, 64, and 125). If p and q are perfect cubes, which of the following expressions will not necessarily be a perfect cube?

Accepted Answer

Introduction / Context:This conceptual problem explores how operations on perfect cubes behave. We are told that p and q are perfect cubes and asked to identify which constructed expression is not guaranteed to be a perfect cube. Understanding how multiplication, scalar factors, addition, and negation affect perfect cubes is important in algebra and number theory. Given Data / Assumptions: p and q are perfect cubes, meaning there exist integers a and b such that p = a^3 and q = b^3. We are given several expressions involving p and q: 8p, pq, pq + 27, -p, and p^2q. We must determine which expression is not necessarily a perfect cube for all choices of p and q. We work with integer arithmetic and standard definitions of perfect cubes. Concept / Approach:We rewrite each expression using p = a^3 and q = b^3. Then we check whether each expression can be written as the cube of an integer. Multiplying perfect cubes, multiplying by another perfect cube such as 8 (which is 2^3), or taking the negative of a perfect cube all preserve the cube property. However, adding two cubes does not in general produce another cube. Carefully checking each option leads us to the correct conclusion. Step-by-Step Solution:Step 1: Let p = a^3 and q = b^3 for some integers a and b.Step 2: Consider option 8p. Since 8 = 2^3, we have 8p = 2^3 * a^3 = (2a)^3, which is a perfect cube.Step 3: Consider option pq. Here pq = a^3 * b^3 = (ab)^3, which is also a perfect cube.Step 4: Consider option -p. This is -a^3, which can be written as (-a)^3, so -p is still a perfect cube (just negative).Step 5: Consider option p^2q. We have p^2q = (a^3)^2 * b^3 = a^6 * b^3 = (a^2)^3 * b^3 = (a^2 b)^3, which is a perfect cube.Step 6: Now consider option pq + 27. Here pq = a^3 b^3 = (ab)^3, and 27 = 3^3. So pq + 27 = (ab)^3 + 3^3.Step 7: In general, the sum of two cubes is not itself a perfect cube. For example, take a = 1 and b = 1, so pq = 1^3 * 1^3 = 1 and pq + 27 = 1 + 27 = 28, which is not a perfect cube.Step 8: Therefore pq + 27 is not guaranteed to be a perfect cube for all choices of perfect cubes p and q. Verification / Alternative check:Using concrete values makes the reasoning very clear. As above, choose p = 1 (so a = 1) and q = 1 (so b = 1). Then 8p = 8, which equals 2^3 and is a perfect cube. pq = 1, which is 1^3, also a perfect cube. -p = -1, which is (-1)^3, a perfect cube. p^2q = 1^2 * 1 = 1, again a cube. But pq + 27 = 1 + 27 = 28. There is no integer k such that k^3 = 28, because 3^3 = 27 and 4^3 = 64. Hence 28 is not a perfect cube, showing that pq + 27 fails in this simple case. Why Other Options Are Wrong:Option 8p: Because 8 is 2^3, multiplying p by 8 simply forms the cube of 2a.Option pq: The product of two perfect cubes is always a perfect cube, as it equals (ab)^3.Option -p: The negative of a perfect cube is still a perfect cube of a negative integer.Option p^2q: This equals (a^2 b)^3, which is again a perfect cube. Common Pitfalls:Students sometimes think that the sum of two perfect cubes is another perfect cube, which is almost never true except for special cases. Another error is to assume that negative numbers cannot be perfect cubes, but in integer arithmetic, cubes of negative integers are negative perfect cubes. Always express the given cubes in terms of a^3 and b^3, and then algebraically manipulate each option to see whether it remains the cube of an integer. Final Answer:The expression that is not necessarily a perfect cube is pq + 27.

Question 9

In the equation formed with digit placeholders 8A9 – 6B2 + 4C6 = 723, A, B, and C represent single digits (0 to 9). What is the maximum possible value of the digit B for which this equation holds true?

Accepted Answer

Introduction / Context:This question involves working with numbers written using digit placeholders and solving a simple equation in digits. The three digit numbers 8A9, 6B2, and 4C6 are formed by inserting digits A, B, and C in the tens place. We know the overall equation 8A9 – 6B2 + 4C6 = 723 is satisfied, and our goal is to determine the largest possible value of the digit B for which this holds. This tests understanding of place value and basic algebraic manipulation. Given Data / Assumptions: 8A9, 6B2, and 4C6 are three digit numbers. 8A9 = 800 + 10A + 9. 6B2 = 600 + 10B + 2. 4C6 = 400 + 10C + 6. The equation is (800 + 10A + 9) – (600 + 10B + 2) + (400 + 10C + 6) = 723. A, B, and C are digits from 0 to 9. Concept / Approach:We first expand each three digit number in terms of hundreds, tens, and units, and then simplify the entire expression. By collecting the constant parts and the terms involving A, B, and C, we reduce the equation to a simple linear relation between A, B, and C. Then we use the fact that A, B, and C are digits to find the maximum possible B that satisfies this relation for some valid choices of A and C. Step-by-Step Solution:Step 1: Write each number in expanded form.Step 2: 8A9 = 800 + 10A + 9.Step 3: 6B2 = 600 + 10B + 2.Step 4: 4C6 = 400 + 10C + 6.Step 5: Substitute these into the equation: (800 + 10A + 9) - (600 + 10B + 2) + (400 + 10C + 6) = 723.Step 6: Combine constant terms: 800 - 600 + 400 = 600 for the hundreds and 9 - 2 + 6 = 13 for the units, so the constant part is 600 + 13 = 613.Step 7: Combine the tens terms: 10A - 10B + 10C = 10(A - B + C).Step 8: The equation becomes 613 + 10(A - B + C) = 723.Step 9: Subtract 613 from both sides: 10(A - B + C) = 723 - 613 = 110.Step 10: Divide by 10: A - B + C = 11.Step 11: Rearrange to express B in terms of A and C: B = A + C - 11.Step 12: Since A and C are digits from 0 to 9, the maximum possible sum A + C is 9 + 9 = 18.Step 13: Therefore, maximum B = 18 - 11 = 7.Step 14: Check that B = 7 is valid, for example by taking A = 9 and C = 9, which gives B = 9 + 9 - 11 = 7 (a valid digit). Verification / Alternative check:Using A = 9, B = 7, and C = 9, compute each three digit number explicitly. 8A9 = 8 * 100 + 9 * 10 + 9 = 800 + 90 + 9 = 899. 6B2 = 600 + 7 * 10 + 2 = 672. 4C6 = 400 + 9 * 10 + 6 = 496. Now evaluate 899 - 672 + 496. First 899 - 672 = 227, then 227 + 496 = 723, which matches the given right side exactly. This confirms that B = 7 is attainable and that it is the maximum possible value consistent with the digit constraints. Why Other Options Are Wrong:Option 4: This would correspond to a smaller sum A + C and is not the maximum possible value of B.Option 1: Too small and does not exploit the maximum range of A and C.Option 5: Although attainable for some A and C, it is not the largest value allowed by the condition A - B + C = 11.Option 6: Also below the maximum possible B of 7. Common Pitfalls:Students may forget that A, B, and C are digits restricted to 0 through 9 and might try values outside this range. Another error is miscalculating the combined constant term or mishandling the tens coefficients. Carefully expanding each three digit number and collecting like terms prevents such mistakes. Finally, some may assume that A, B, and C must be distinct, which is not stated in the problem; we only need them to be digits. Final Answer:The maximum possible value of B is 7.

Question 10

Let N = 65 × 359 × 113. How many distinct prime factors does N have?

Accepted Answer

Introduction / Context:This question is about prime factorisation and counting distinct prime factors of a composite number. We are given N as a product of three numbers, 65, 359, and 113, and asked to determine how many different prime numbers appear in the prime factorisation of N. Understanding how to break composite numbers into primes is a fundamental skill in number theory. Given Data / Assumptions: N = 65 × 359 × 113. We need to find the distinct prime factors of N. Standard definitions: a prime number has exactly two positive divisors, 1 and itself. We assume 359 and 113 are primes (they have no smaller prime divisors). Concept / Approach:We first factor each component of N into primes. Then we combine all prime factors and remove duplicates to get the set of distinct primes. The number of elements in this set is the answer. The key step is to factor 65, since 359 and 113 are already prime numbers with no non trivial divisors. Step-by-Step Solution:Step 1: Factor 65. Note that 65 is divisible by 5 because it ends in 5.Step 2: Compute 65 ÷ 5 = 13, so 65 = 5 × 13.Step 3: Check 13; it is a prime number as it has no divisors other than 1 and 13.Step 4: Check 359. It is not divisible by small primes such as 2, 3, 5, 7, 11, or 13, and is known to be prime.Step 5: Check 113. It is not divisible by 2, 3, 5, 7, or 11, and is also prime.Step 6: Therefore the prime factorisation of N is N = 5 × 13 × 359 × 113.Step 7: The distinct prime factors are 5, 13, 359, and 113.Step 8: Count these primes: there are 4 distinct prime factors. Verification / Alternative check:You can verify by multiplying the prime factors back together. First compute 5 × 13 = 65. Then N = 65 × 359 × 113 = (5 × 13) × 359 × 113. Since multiplication is associative and commutative, the order does not matter. The product matches the given definition of N, confirming that we have neither missed nor added any prime factors. No further factorisation is possible because each of 5, 13, 359, and 113 is prime. Why Other Options Are Wrong:Option 2: This would imply only two primes, which is impossible since 65 alone contributes two primes, 5 and 13.Option 3: Would require one of 359 or 113 to be composite, which is not the case.Option 5 and Option 6: These overcount the number of distinct primes; there are no additional prime factors beyond 5, 13, 359, and 113. Common Pitfalls:Students may forget to factor 65 completely and assume it is prime, which would lead to undercounting. Another mistake is confusing distinct prime factors with the total number of prime factors including multiplicity; here each prime appears once. Always factor each composite completely into primes and then count unique primes only when asked for distinct prime factors. Final Answer:The number N has 4 distinct prime factors.

Question 11

A cage contains only peacocks and deer. Altogether there are 72 heads and 200 feet inside the cage. How many peacocks and how many deer are there?

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Introduction / Context:This is a classic heads and legs problem, a type of system of equations question often seen in aptitude tests. The cage contains only two types of animals, peacocks and deer, with different numbers of legs. We are given the total number of heads and feet and asked to find how many animals of each type are present. Such problems reinforce translating word statements into linear equations and solving them. Given Data / Assumptions: The cage contains only peacocks and deer, no other animals. Each animal has exactly one head. A peacock has 2 feet. A deer has 4 feet. Total number of heads = 72. Total number of feet = 200. Concept / Approach:Let p be the number of peacocks and d be the number of deer. Since each animal contributes one head, the head count equation is p + d = 72. For feet, each peacock contributes 2 feet and each deer contributes 4 feet, so the feet equation is 2p + 4d = 200. We solve this system of linear equations to find p and d. Using substitution or elimination methods works well here. Step-by-Step Solution:Step 1: Let p = number of peacocks and d = number of deer.Step 2: From the number of heads, we get equation (1): p + d = 72.Step 3: From the number of feet, we get equation (2): 2p + 4d = 200.Step 4: Simplify equation (2) by dividing by 2: p + 2d = 100. Call this equation (3).Step 5: Subtract equation (1) from equation (3): (p + 2d) - (p + d) = 100 - 72.Step 6: This simplifies to d = 28.Step 7: Substitute d = 28 back into equation (1): p + 28 = 72, which gives p = 72 - 28 = 44.Step 8: Therefore there are 44 peacocks and 28 deer. Verification / Alternative check:Check the solution directly with the original data. Total heads: 44 + 28 = 72, which matches the given head count. Total feet: peacocks contribute 44 * 2 = 88 feet, and deer contribute 28 * 4 = 112 feet. The total is 88 + 112 = 200 feet, matching the given total. Since both conditions are satisfied, the solution is correct. Why Other Options Are Wrong:Option 42 peacocks and 24 deer: Heads = 66, so this does not match the required 72 heads.Option 38 peacocks and 26 deer: Heads = 64, which is too few.Option 36 peacocks and 24 deer: Heads = 60, also incorrect.Option 40 peacocks and 32 deer: Heads = 72, but feet = 40 * 2 + 32 * 4 = 80 + 128 = 208, not 200. Common Pitfalls:Students sometimes mix up the coefficients in the leg equation, assigning 4 legs to peacocks and 2 to deer, which leads to wrong answers. Others forget to divide the leg equation by a common factor to simplify it, making the arithmetic more error prone. A clear definition of variables and careful translation of the word problem into equations helps avoid these issues. Final Answer:There are 44 peacocks and 28 deer in the cage.

Question 12

If 1/a + 1/b + 1/c = 1/(a + b + c) and a + b + c ≠ 0 and abc ≠ 0, what is the value of (a + b)(b + c)(a + c)?

Accepted Answer

Introduction / Context:This algebraic identity question involves reciprocals and symmetric expressions in three variables a, b, and c. We are given a special condition relating 1/a, 1/b, and 1/c to 1/(a + b + c) and asked to find the value of the product (a + b)(b + c)(a + c). Problems like this strengthen understanding of algebraic manipulation and standard identities. Given Data / Assumptions: 1/a + 1/b + 1/c = 1/(a + b + c). a + b + c ≠ 0. abc ≠ 0, so a, b, and c are all nonzero. We need to determine (a + b)(b + c)(a + c). Concept / Approach:First rewrite the given reciprocal condition with a common denominator to express it in terms of ab + bc + ca and abc. Then use a known identity that connects (a + b)(b + c)(c + a) with (a + b + c)(ab + bc + ca) and abc. Specifically, there is an identity: (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc. Once we find ab + bc + ca in terms of abc and a + b + c using the given condition, we can substitute into this identity to find the required value. Step-by-Step Solution:Step 1: Start from the given condition: 1/a + 1/b + 1/c = 1/(a + b + c).Step 2: Write the left side with a common denominator: 1/a + 1/b + 1/c = (bc + ac + ab) / (abc).Step 3: So (bc + ac + ab) / (abc) = 1/(a + b + c).Step 4: Cross multiply: (bc + ac + ab)(a + b + c) = abc.Step 5: Recognise that bc + ac + ab is the symmetric sum ab + bc + ca.Step 6: Therefore (ab + bc + ca)(a + b + c) = abc.Step 7: Recall the identity: (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc.Step 8: From Step 6, we have (a + b + c)(ab + bc + ca) = abc.Step 9: Substitute this into the identity: (a + b)(b + c)(c + a) = abc - abc = 0.Step 10: Therefore (a + b)(b + c)(a + c) = 0. Verification / Alternative check:To verify concretely, we can choose specific values of a, b, and c that satisfy the given condition. For instance, try a = 1, b = 1, and c = -1. Then a + b + c = 1, which is nonzero. Now compute 1/a + 1/b + 1/c = 1/1 + 1/1 + 1/(-1) = 1 + 1 - 1 = 1. Also, 1/(a + b + c) = 1/1 = 1. So the condition holds. Now compute (a + b)(b + c)(a + c) = (1 + 1)(1 - 1)(1 - 1) = 2 * 0 * 0 = 0. This matches our algebraic result, confirming the reasoning. Why Other Options Are Wrong:Option 3, 2, 1, and -1: These suggest that the product (a + b)(b + c)(a + c) might be some nonzero constant, but our algebra clearly shows it reduces to abc - abc = 0 for any a, b, and c satisfying the given condition. Common Pitfalls:Students may incorrectly expand (a + b)(b + c)(c + a) or forget the standard identity connecting it with (a + b + c)(ab + bc + ca). Others might mishandle the reciprocals, especially when finding the common denominator. Careful algebraic manipulation and remembering common symmetric identities help avoid these errors. When in doubt, testing specific numeric examples consistent with the condition is a good way to build confidence in the result. Final Answer:The value of (a + b)(b + c)(a + c) is 0.

Question 13

Which of the following is a rational number?

Accepted Answer

Introduction / Context:This question tests understanding of rational numbers and how they relate to decimal representations. A rational number is any number that can be expressed as a fraction p/q where p and q are integers and q is not zero. In decimal form, rational numbers either terminate after a finite number of digits or repeat a block of digits indefinitely. We must decide which of the given decimal and integer options are rational. Given Data / Assumptions: Option 0.241 is a terminating decimal. Option 1.732 is given as a decimal with three places. Option 4 is an integer. We assume the decimals are exact as written (not approximate symbols like square roots), unless explicitly stated otherwise. We must identify which options represent rational numbers. Concept / Approach:By definition, a terminating decimal can always be written as a fraction with denominator a power of 10, which makes it rational. Any integer n can be written as n/1, so integers are also rational. Thus, if 0.241 and 1.732 are taken as exact decimals, both are rational. The question is testing whether we recognise that decimal notation without a bar still represents rational numbers when they terminate at a finite place, and that integers are rational by default. Step-by-Step Solution:Step 1: Consider 0.241. This is a terminating decimal with three digits after the decimal point.Step 2: We can write 0.241 as 241/1000, since moving the decimal three places to the right gives 241, and the denominator becomes 10^3 = 1000.Step 3: Because 241 and 1000 are integers and the denominator is nonzero, 0.241 is a rational number.Step 4: Consider 1.732. As written, it is a terminating decimal with three digits after the decimal point.Step 5: We can express 1.732 as 1732/1000 by similar reasoning, or further simplify the fraction if desired. In any case, it can be written as p/q with integers p and q, so 1.732 is rational.Step 6: Consider 4. This is an integer and can be written as 4/1, which matches the definition of a rational number.Step 7: Since all three listed numbers are rational, the correct choice is "All of the above". Verification / Alternative check:To reinforce the idea, recall that irrational numbers are those that cannot be expressed as a fraction of integers, and their decimal expansions neither terminate nor repeat (for example, the exact values of √2 or π). In contrast, a decimal like 1.732, written with a fixed number of decimal places, is always equal to some fraction with denominator 10^3. Thus it is rational. Only if the question explicitly indicated that 1.732 represents an irrational expression like the square root of 3 would we consider it irrational. As stated, it is just a finite decimal. Why Other Options Are Wrong:Option 0.241 only: Choosing this alone ignores the fact that 1.732 and 4 are also rational.Option 1.732 only: This also undercounts; integers and other terminating decimals remain rational.Option 4 only: While 4 is rational, so are the other two numbers.Option None of these: This would imply none of the numbers is rational, which contradicts all the reasoning above. Common Pitfalls:Students sometimes associate 1.732 with an approximation to √3 and mistakenly classify it as irrational, forgetting that the decimal 1.732 exactly as written is a terminating decimal and therefore rational. Others forget that every integer is automatically a rational number. Always remember: any terminating decimal or repeating decimal can be written as a fraction of integers and is rational, and every whole number is rational as well. Final Answer:The rational number(s) among the options are All of the above.

Question 14

Among the fractions 12/49, 7/30, 13/56 and 11/46, which one has the greatest value?

Accepted Answer

Introduction / Context:This question checks your understanding of how to compare fractions without converting all of them into long decimals. Being able to decide which fraction is largest is a common aptitude skill used in data interpretation, number comparisons and exams where speed matters. Given Data / Assumptions: We have four fractions: 12/49, 7/30, 13/56 and 11/46. All denominators are positive, so normal comparison rules apply. We must identify the fraction with the greatest numerical value. Concept / Approach:To compare fractions, two standard methods are common: converting them to decimals or converting them to a common denominator using cross multiplication. Cross multiplication is efficient and avoids actual division. If a/b and c/d are positive, then a/b > c/d if and only if a * d > c * b. Step-by-Step Solution:Step 1: Compare 12/49 and 7/30. Compute cross products: 12 * 30 = 360 and 7 * 49 = 343. Since 360 > 343, 12/49 > 7/30.Step 2: Compare 12/49 and 13/56. Compute: 12 * 56 = 672 and 13 * 49 = 637. Since 672 > 637, 12/49 > 13/56.Step 3: Compare 12/49 and 11/46. Compute: 12 * 46 = 552 and 11 * 49 = 539. Since 552 > 539, 12/49 > 11/46.Step 4: Since 12/49 is larger than each of the other three fractions individually, it is the greatest among the four options. Verification / Alternative check:You can convert each fraction to an approximate decimal: 12/49 ≈ 0.245, 7/30 ≈ 0.233, 13/56 ≈ 0.232, and 11/46 ≈ 0.239. The approximate decimal values confirm that 12/49 is the largest.Both methods, cross multiplication and decimal approximation, lead to the same conclusion, which verifies our answer. Why Other Options Are Wrong:7/30 is smaller than 12/49 because its cross product with 12/49 is lower.13/56 is also smaller, as its cross product comparison gives a lower product than 12/49.11/46 is close in value but still slightly smaller than 12/49 when compared using cross multiplication. Common Pitfalls:Many learners try to compare by looking only at numerators or denominators separately, which does not work for general fractions.Rounding too early or approximating very roughly can lead to mistakes when fractions are close in value.Some candidates may incorrectly assume that the fraction with the largest numerator or smallest denominator is always the greatest, which is not necessarily true. Final Answer:The largest fraction is 12/49.

Question 15

If a² = 12, what is the value of a⁴?

Accepted Answer

Introduction / Context:This question tests your understanding of powers of a variable and how to manipulate exponents using basic algebraic rules. Such problems are very common in aptitude tests and algebra sections where quick simplification is required. Given Data / Assumptions: We are given a² = 12. We must find the value of a⁴. No additional constraints on a are given, so we work algebraically with exponents. Concept / Approach:We use the law of indices that says a⁴ = (a²)². Instead of finding a explicitly, we can square the given expression. This avoids square roots and keeps the calculation simple and exact. The key idea is to recognize that if we already know a², then a⁴ is simply the square of that known quantity. Step-by-Step Solution:Step 1: Start from the given relation a² = 12.Step 2: To obtain a⁴, square both sides of the equation: (a²)² = 12².Step 3: Simplify the left hand side: (a²)² = a⁴.Step 4: Calculate the right hand side: 12² = 12 * 12 = 144.Step 5: Therefore, a⁴ = 144. Verification / Alternative check:If you want to verify, you can write a = √12. Then a⁴ = (√12)⁴ = (12)² / (since √12 squared is 12) which again gives 144, confirming the previous result.Both direct exponent rules and substitution using square roots give the same answer. Why Other Options Are Wrong:72 would correspond to 12 * 6 and does not match 12².36 is equal to 6², but we are not told that a² = 6, so 36 is not correct here.24 is 12 * 2 and again does not follow from squaring 12. Common Pitfalls:Some students mistakenly think a⁴ = a² * 4 or multiply the exponent by the base, which is incorrect.Another common error is to miscalculate 12² as 124 or 142 due to haste.Always remember that (a²)² means a raised to the power 2 * 2, which is 4, and the numerical side must be squared properly. Final Answer:The value of a⁴ is 144.

Question 16

A solid cube of side 6 units is first painted on all its faces and then cut into smaller cubes of side 2 units. How many of the smaller cubes will have paint on exactly two faces?

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Introduction / Context:This question tests your understanding of three dimensional geometry and how surface painting behaves when a cube is cut into smaller cubes. Problems of painted cubes are standard in aptitude exams and help you visualise edges, faces and corners in a structured way. Given Data / Assumptions: A large cube has side length 6 units. The cube is painted on all six faces. It is then cut into smaller cubes, each of side 2 units. We need the number of small cubes that have paint on exactly two faces. Concept / Approach:When a painted cube is cut into smaller equal cubes, cubes on corners have three painted faces, cubes along edges (but not at corners) have two painted faces, cubes on faces but not on edges have one painted face, and cubes completely inside have no paint. The count of cubes with exactly two painted faces is therefore the count of edge cubes excluding corners. Step-by-Step Solution:Step 1: Side of the big cube = 6 units. Side of each small cube = 2 units.Step 2: Number of small cubes along each edge = 6 / 2 = 3.Step 3: Total number of small cubes = 3 * 3 * 3 = 27.Step 4: For a cube cut into n segments along each side, the number of edge cubes with exactly two painted faces is 12 * (n - 2).Step 5: Here n = 3, so the count = 12 * (3 - 2) = 12 * 1 = 12. Verification / Alternative check:Draw a 3 × 3 × 3 grid representing the smaller cubes. On each of the 12 edges of the cube, there are 3 small cubes. The corner cube on each edge has three faces painted and is not counted for exactly two faces, leaving one middle cube per edge with exactly two painted faces.Total edge cubes with two faces painted = 12 edges * 1 middle cube per edge = 12, confirming our formula based result. Why Other Options Are Wrong:30 and 24 are too large and would imply more than one middle cube on each edge, which is impossible with only 3 cubes per edge.8 would match the number of corner cubes (each with three painted faces), not cubes with exactly two painted faces. Common Pitfalls:Learners sometimes confuse cubes with two faces painted and those with three faces painted.Another mistake is to count face center cubes or internal cubes as having two faces painted.A clear mental or drawn model of edges, faces and corners avoids such errors. Final Answer:The number of smaller cubes with paint on exactly two faces is 12.

Question 17

Which of the following numbers can be used to show that not all prime numbers are odd?

Accepted Answer

Introduction / Context:This question checks your basic understanding of prime numbers and a very important exception related to even numbers. Many people think that all primes are odd, and this item directly challenges that misconception using a simple example. Given Data / Assumptions: We are given four small integers: 1, 2, 3 and 4. We must choose the number that shows that not all prime numbers are odd. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. Concept / Approach:The key concept is the definition of prime numbers and the special role of the number 2. All even numbers greater than 2 are composite, but 2 is a unique even number that is still prime. This single example is enough to disprove the statement that all primes are odd, because a single counterexample is sufficient. Step-by-Step Solution:Step 1: Check if 1 is prime. The number 1 has only one positive divisor (itself), so it does not meet the definition of a prime (which requires two distinct divisors). So 1 is not prime.Step 2: Check if 2 is prime. The divisors of 2 are 1 and 2. That gives exactly two distinct positive divisors, so 2 is prime.Step 3: Check the parity of 2. The number 2 is even because it is divisible by 2.Step 4: Since 2 is an even prime, it is a counterexample to the claim that all primes are odd.Step 5: Verify remaining numbers: 3 is prime but odd, and 4 is composite (divisible by 1, 2 and 4). They do not help disprove the claim that all primes are odd. Verification / Alternative check:You can list the first few prime numbers: 2, 3, 5, 7, 11, 13, ... All except 2 are odd, and 2 is clearly even, so it is a simple visual check that not all prime numbers are odd. Why Other Options Are Wrong:1 is not prime under the standard definition, so it cannot be used to illustrate anything about prime parity.3 is prime but it is odd, so it does not disprove the claim that all primes are odd.4 is an even number but it is composite, not prime, so it does not work as a counterexample either. Common Pitfalls:Some learners still incorrectly treat 1 as a prime number, which leads to confusion in number theory.Others forget that a single counterexample (like 2) is enough to disprove a general statement.Remember that 2 is the only even prime number; every other even number has at least three divisors and is therefore composite. Final Answer:The number 2 shows that not all prime numbers are odd.

Question 18

The sequence 2, 5, 7, 6, 12, 7, ... is formed by three interleaved patterns: the 1st, 4th, 7th, ... terms form an arithmetic progression starting at 2 with common difference 4, the 2nd, 5th, 8th, ... terms form an arithmetic progression starting at …

Accepted Answer

Introduction / Context:This question involves a non standard sequence created by interleaving three simpler arithmetic patterns. Such problems test your ability to recognise underlying structures in a mixed series and then apply arithmetic progression formulas systematically to find the total sum of many terms. Given Data / Assumptions: The sequence is 2, 5, 7, 6, 12, 7, ... Terms in positions 1, 4, 7, ... form an arithmetic progression starting at 2 with common difference 4. Terms in positions 2, 5, 8, ... form an arithmetic progression starting at 5 with common difference 7. Every 3rd term (positions 3, 6, 9, ...) is a constant 7. We need the sum of the first 200 terms. Concept / Approach:Instead of trying to write all 200 terms directly, we split the sequence into three subsequences based on position modulo 3. Each subsequence is either an arithmetic progression or a constant sequence. Then we use the formula for the sum of an arithmetic progression to get the sum of each subsequence and finally add these partial sums together. Step-by-Step Solution:Step 1: Count terms in each subsequence. From position 1 to 200, positions 1,4,7,... give 67 terms; positions 2,5,8,... give 67 terms; positions 3,6,9,... give 66 terms (since 200 leaves remainder 2 when divided by 3).Step 2: First subsequence (positions 1,4,7,...): arithmetic progression with first term a1 = 2, common difference d1 = 4 and n1 = 67. Sum S1 = n1/2 * (2a1 + (n1 - 1) * d1) = 67/2 * (4 + 66 * 4) = 67/2 * 268 = 8978.Step 3: Second subsequence (positions 2,5,8,...): arithmetic progression with first term a2 = 5, common difference d2 = 7 and n2 = 67. Sum S2 = 67/2 * (2 * 5 + (67 - 1) * 7) = 67/2 * (10 + 66 * 7) = 67/2 * 472 = 15812.Step 4: Third subsequence (positions 3,6,9,...): constant 7 with n3 = 66 terms. Sum S3 = 7 * 66 = 462.Step 5: Total sum S = S1 + S2 + S3 = 8978 + 15812 + 462 = 25252. Verification / Alternative check:You can check the logic by writing the first several triples (2,5,7), (6,12,7), (10,19,7) and noticing that the first and second positions follow arithmetic progressions with differences 4 and 7, while the third is fixed at 7.A quick calculator check of S1, S2 and S3 confirms the arithmetic and the total 25252. Why Other Options Are Wrong:25200 is close but ignores part of one of the subsequence contributions and indicates a rounding or counting mistake.25000 and 26000 are rough guesses with no precise basis in the arithmetic progression formulas. Common Pitfalls:A common mistake is to treat the whole sequence as a single arithmetic progression, which it is not.Another error is miscounting how many terms belong to each subsequence when dealing with 200 total terms.Careful tracking of positions and using the standard arithmetic progression sum formula avoids these issues. Final Answer:The sum of the first 200 terms is 25252.

Question 19

What is the least number that must be subtracted from 627349 so that the remaining number is divisible by 15?

Accepted Answer

Introduction / Context:This problem tests your knowledge of divisibility rules, particularly for 3 and 5, because a number divisible by 15 must be divisible by both 3 and 5. It also checks your ability to adjust a number by the smallest possible subtraction to meet those divisibility requirements. Given Data / Assumptions: Original number N = 627349. We must subtract the smallest positive integer x so that N - x is divisible by 15. Divisible by 15 means divisible by both 3 and 5 simultaneously. Concept / Approach:Use divisibility rules: a number is divisible by 5 when its last digit is 0 or 5. A number is divisible by 3 when the sum of its digits is divisible by 3. We search for the smallest non negative integer x such that N - x satisfies both of these rules. Step-by-Step Solution:Step 1: Check divisibility by 5. The last digit of 627349 is 9. For N - x to end with 0 or 5, test small x.Step 2: If x = 4, the last digit becomes 9 - 4 = 5, which is valid for divisibility by 5.Step 3: Now check divisibility by 3. Sum of digits of N is 6 + 2 + 7 + 3 + 4 + 9 = 31.Step 4: Sum of digits of N - x will be 31 - x (because subtracting x reduces the total digit sum by x for small x). For N - x to be divisible by 3, 31 - x must be divisible by 3.Step 5: Compute 31 mod 3 = 1, so x must leave remainder 1 when divided by 3. Among small options, x = 1, 4, 7, ... satisfy this.Step 6: Combine conditions. From the divisibility by 5 check, x could be 4 or 9 (for last digit 5 or 0). From the divisibility by 3 condition, x must be congruent to 1 modulo 3. The smallest x that works for both is 4. Verification / Alternative check:Compute N - 4 = 627345. Last digit is 5, so it is divisible by 5.Digit sum of 627345 is 6 + 2 + 7 + 3 + 4 + 5 = 27, and 27 is divisible by 3, so 627345 is divisible by 3 and therefore by 15. Why Other Options Are Wrong:If x = 1, N - 1 ends in 8, so it is not divisible by 5.If x = 2, N - 2 ends in 7, again not divisible by 5.If x = 3, N - 3 ends in 6 and does not satisfy the condition for divisibility by 5. Common Pitfalls:Forgetting that divisibility by 15 requires both the 3 and 5 tests to hold at the same time.Only adjusting the last digit without checking the digit sum can lead to an incorrect choice.Working systematically through both conditions for small candidate values prevents these errors. Final Answer:The least number to be subtracted is 4.

Question 20

The difference between two numbers is 3355. When the larger number is divided by the smaller number, the quotient is 6 and the remainder is 15. What are the smaller number and the larger number respectively?

Accepted Answer

Introduction / Context:This question is a classic application of the division algorithm combined with information about the difference between two numbers. It tests your ability to set up and solve linear equations based on quotient, remainder and difference relationships. Given Data / Assumptions: The difference between the two numbers is 3355. Larger number divided by smaller number gives quotient 6 and remainder 15. We must find both numbers, ordered as (smaller, larger). Concept / Approach:Use the division algorithm: if L is the larger number and S is the smaller number, then L = 6S + 15. We also know L - S = 3355. These two equations in two unknowns can be solved simultaneously to find S and L. This is a standard linear equation system problem. Step-by-Step Solution:Step 1: Let the larger number be L and the smaller number be S.Step 2: From the quotient and remainder information, write L = 6S + 15.Step 3: From the difference information, write L - S = 3355.Step 4: Substitute L from Step 2 into Step 3: (6S + 15) - S = 3355.Step 5: Simplify: 5S + 15 = 3355.Step 6: Subtract 15 from both sides: 5S = 3340.Step 7: Divide by 5: S = 3340 / 5 = 668.Step 8: Substitute back into L = 6S + 15: L = 6 * 668 + 15 = 4008 + 15 = 4023. Verification / Alternative check:Check the difference: 4023 - 668 = 3355, which matches the given difference.Check the division: 4023 divided by 668 gives quotient 6 and remainder 15, because 6 * 668 = 4008 and 4023 - 4008 = 15. This confirms the result. Why Other Options Are Wrong:For 546, 3901, the difference is 3355 but 3901 ÷ 546 does not give remainder 15.For 415, 3770, the difference is 3355 but the quotient and remainder are not 6 and 15.For 404, 3759, the difference is 3355, but again 3759 ÷ 404 does not provide the given quotient and remainder. Common Pitfalls:Some students may incorrectly set up the division equation as L = 6S - 15 instead of L = 6S + 15.Others may forget to use both conditions simultaneously, checking only the difference or only the division condition.Always translate word statements into algebraic equations carefully and solve them together. Final Answer:The smaller number is 668 and the larger number is 4023.

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