Question 1

Compute the missing divisor: 62976 divided by N equals 123 (that is, 62976 ÷ N = 123). What is the value of N?

Accepted Answer

Introduction / Context:This is a direct division problem presented as a missing factor question. Solving requires rearranging to N = 62976 / 123 and computing the quotient accurately. Recognizing powers of two may speed up checking the candidate options. Given Data / Assumptions: 62976 ÷ N = 123. N is a positive integer. We must find N. Concept / Approach:Rearrange the relation to isolate N. Then compute the division or check by multiplication with candidate options. Since 512 is a familiar power of two (2^9), quickly testing 123 * 512 is efficient. Step-by-Step Solution: Rearrange: N = 62976 / 123.Test 512: 123 * 512 = 123 * (500 + 12) = 61500 + 1476 = 62976.Therefore, N = 512 exactly. Verification / Alternative check:Perform long division 62976 ÷ 123 to confirm the quotient 512 with no remainder. This validates the selection. Why Other Options Are Wrong:412, 502, and 522 do not produce 62976 when multiplied by 123. 123 would imply 123 * 123 = 15129, which is far too small. Common Pitfalls:Arithmetic slips in multiplication or assuming that the divisor must be near 123. Using estimation helps but should be confirmed by exact multiplication. Final Answer:512

Question 2

Given x / y = 3 / 4 with x and y positive, evaluate the expression: 6/7 + (y − x) / (y + x). Choose the exact value.

Accepted Answer

Introduction / Context:This problem blends a fixed ratio with a rational expression. By parameterizing x and y with a common factor, the complex-looking fraction simplifies neatly. This technique is standard in ratio problems and avoids cumbersome algebra. Given Data / Assumptions: x / y = 3 / 4 with x, y > 0. We must evaluate 6/7 + (y − x) / (y + x). Arithmetic should be exact and simplified. Concept / Approach:Let x = 3k and y = 4k for some positive k to satisfy the ratio x / y = 3 / 4. Then substitute these into the expression to simplify. The common factor k cancels cleanly in both numerator and denominator for the second term. Step-by-Step Solution: Set x = 3k and y = 4k.Compute y − x = 4k − 3k = k.Compute y + x = 4k + 3k = 7k.Thus (y − x) / (y + x) = k / (7k) = 1/7.Therefore, the whole expression is 6/7 + 1/7 = 1. Verification / Alternative check:Choose a specific k, for example k = 1, giving x = 3 and y = 4. Substitute: 6/7 + (4 − 3) / (4 + 3) = 6/7 + 1/7 = 1. This confirms the general result. Why Other Options Are Wrong:5/7 and 6/7 omit the contribution from the second term. 11/7 and 2 overcount the total by miscomputing (y − x) / (y + x). Common Pitfalls:Plugging numbers without maintaining the 3:4 ratio, or mistakenly computing (y − x) / (y + x) as (y / x) or (x / y). Keeping the ratio structure avoids such errors. Final Answer:1

Question 3

Fraction Removal from a Number — Solve for the Original Value If three-fourths (3/4) of a number is subtracted from the number itself, the value obtained is 163. What is that original number?

Accepted Answer

Introduction / Context:This aptitude question translates a sentence about fractions into a simple linear equation. The expression “three-fourths of a number subtracted from the number” is a common prompt to test algebraic modeling and careful reading. Solving it requires only basic manipulation once the equation is set up correctly. Given Data / Assumptions: The unknown number is positive and denoted by N. Subtracting 3/4 of N from N gives 163. We are to find the exact value of N. Concept / Approach:The wording maps directly to the equation: N − (3/4)N = 163. Combine like terms on the left to isolate a single multiple of N, then solve by division. The structure highlights how fractional parts of a whole relate back to the whole number. Step-by-Step Solution:Write the relation: N − (3/4)N = 163.Combine: (1 − 3/4)N = (1/4)N = 163.Solve for N: N = 163 * 4 = 652.Therefore, the original number is 652. Verification / Alternative check:Compute 3/4 of 652 = 489. Then N − 3/4 N = 652 − 489 = 163, matching the given condition exactly. Why Other Options Are Wrong:625, 562, 632, and 640 produce N − 3/4 N values of 156.25, 140.5, 158, and 160 respectively, none equal to 163. Common Pitfalls:Misreading “three-fourths” as “three-four” or interpreting the phrase as (3/4N − N). Also, some candidates multiply 163 by 3 inadvertently instead of by 4. Final Answer:652

Question 4

Number and Its Reciprocal — Translate to a Quadratic A positive number, when increased by 10, equals 200 times its reciprocal. What is the number?

Accepted Answer

Introduction / Context:Problems linking a number to its reciprocal typically lead to a quadratic equation. This question tests your ability to convert a verbal condition into algebra and then solve efficiently while keeping track of the positivity constraint. Given Data / Assumptions: N is a positive real number. N + 10 = 200 * (1/N). We must determine N. Concept / Approach:Multiplying through by N removes the reciprocal and yields a standard quadratic in N. Solve via the quadratic formula or factoring (after computing the discriminant). Discard any negative root because the problem specifies a positive number. Step-by-Step Solution:Start with N + 10 = 200/N.Multiply both sides by N: N^2 + 10N = 200.Rearrange: N^2 + 10N − 200 = 0.Discriminant: 10^2 + 4*200 = 100 + 800 = 900; sqrt = 30.Roots: N = (−10 ± 30)/2 → N = 10 or N = −20. Take the positive root N = 10. Verification / Alternative check:Check N = 10: left side 10 + 10 = 20; right side 200 * (1/10) = 20. The equality holds. Why Other Options Are Wrong:100, 20, and 200 do not satisfy the equation; 5 also fails since 5 + 10 = 15 while 200*(1/5) = 40. Common Pitfalls:Forgetting to multiply through by N; mishandling signs when applying the quadratic formula; accidentally keeping the negative root even though the number must be positive. Final Answer:10

Question 5

Sharing an Estate — Sum of Fractions Given Away A man gives 1/4 of his property to his daughter, 1/2 to his sons, and 1/5 to charity. What fraction of his property has he given away in total?

Accepted Answer

Introduction / Context:This problem evaluates fluency with adding fractions that have different denominators. The setup mimics practical distributions (family and charity), making it a good test of fraction arithmetic in everyday contexts. Given Data / Assumptions: Daughter's share: 1/4 of the property. Sons' share: 1/2 of the property. Charity: 1/5 of the property. Goal: total fraction given away. Concept / Approach:Add the fractions using a common denominator. The least common multiple of 4, 2, and 5 is 20. Convert each fraction to twentieths and sum. The result can be simplified if needed, but here it will already be in simplest terms. Step-by-Step Solution:Convert to twentieths: 1/4 = 5/20; 1/2 = 10/20; 1/5 = 4/20.Add: 5/20 + 10/20 + 4/20 = 19/20.Therefore, 19/20 of the property has been given away.Remaining portion = 1 − 19/20 = 1/20 (for completeness). Verification / Alternative check:Use decimal equivalents: 0.25 + 0.5 + 0.2 = 0.95 = 19/20. This matches the fraction sum. Why Other Options Are Wrong:1/20 is the remainder, not the given amount; 1/10 and 9/10 miscompute the sum; 17/20 omits part of a share. Common Pitfalls:Adding numerators and denominators directly (invalid); forgetting to use a common denominator; mixing up “given away” with “leftover.” Final Answer:19/20

Question 6

Using Sum and Product — Sum of Reciprocals Two numbers have sum 10 and product 20. What is the sum of their reciprocals?

Accepted Answer

Introduction / Context:This classic identity-based question rewards recognizing a known expression. When you know the sum and product of two numbers, you can compute the sum of their reciprocals without finding the numbers individually. Given Data / Assumptions: a + b = 10. a * b = 20. We seek (1/a) + (1/b). Concept / Approach:Use the identity: (1/a) + (1/b) = (a + b) / (a * b). Substituting the given sum and product yields the result directly. This avoids solving a quadratic to find a and b separately. Step-by-Step Solution:Apply the identity: (1/a) + (1/b) = (a + b) / (ab).Plug in values: = 10 / 20.Simplify: 10 / 20 = 1/2.Hence, the sum of reciprocals is 1/2. Verification / Alternative check:As a check, solve a and b as roots of t^2 − 10t + 20 = 0. Compute reciprocals and confirm their sum is 1/2; the identity guarantees consistency. Why Other Options Are Wrong:1, 3/5, 11/6, and 2/5 arise from misapplying formulas or arithmetic slips in simplification. Common Pitfalls:Attempting to find a and b explicitly (extra work); inverting the identity incorrectly as (ab)/(a + b). Final Answer:1/2

Question 7

Sharing a Chocolate — Fractional Distribution to Friends A chocolate bar has 12 equal pieces. Manju gives 1/4 of it to Anju, 1/3 of it to Raj, and 1/6 of it to Fiza. How many pieces remain with Manju afterward?

Accepted Answer

Introduction / Context:This problem applies fraction addition to a concrete sharing scenario. You translate fractional portions of a whole into actual piece counts, ensuring the fractions sum correctly before subtracting from the total pieces. Given Data / Assumptions: Total pieces = 12. Given to Anju: 1/4 of the bar. Given to Raj: 1/3 of the bar. Given to Fiza: 1/6 of the bar. Concept / Approach:Compute the total fraction given away by adding 1/4 + 1/3 + 1/6. Use the common denominator 12. Multiply the resulting fraction by 12 to find the number of pieces given. Subtract from 12 to get the remainder with Manju. Step-by-Step Solution:Common denominator 12: 1/4 = 3/12; 1/3 = 4/12; 1/6 = 2/12.Sum of given fractions: 3/12 + 4/12 + 2/12 = 9/12 = 3/4.Pieces given: (3/4)*12 = 9 pieces.Pieces remaining: 12 − 9 = 3 pieces. Verification / Alternative check:Compute piece counts separately: 1/4 of 12 = 3; 1/3 of 12 = 4; 1/6 of 12 = 2; total given = 3 + 4 + 2 = 9; remaining = 3. Why Other Options Are Wrong:1, 2, 4, and 5 are not consistent with the total given pieces equaling 9. Common Pitfalls:Adding denominators directly; failing to convert all fractions to the same denominator; arithmetic slips when converting fractions to piece counts. Final Answer:3

Question 8

Counting Divisors — Total Factors of 120 How many positive factors (divisors) does 120 have in total?

Accepted Answer

Introduction / Context:Counting divisors relies on prime factorization and a standard formula. This is a staple of number system questions and tests whether you can decompose a number quickly and apply the multiplicative rule for divisor counts. Given Data / Assumptions: Target number: 120. We seek the count of all positive divisors, including 1 and the number itself. Concept / Approach:Factor 120 into primes. If N = p1^a * p2^b * p3^c, then the total number of divisors d(N) = (a+1)(b+1)(c+1). This counts choices for each prime power exponent from 0 up to its maximum. Step-by-Step Solution:Prime factorization: 120 = 2^3 * 3^1 * 5^1.Apply formula: d(120) = (3+1)(1+1)(1+1) = 4 * 2 * 2.Compute: 4 * 2 * 2 = 16.Therefore, 120 has 16 positive divisors. Verification / Alternative check:List quickly in pairs (1×120, 2×60, 3×40, 4×30, 5×24, 6×20, 8×15, 10×12) to see eight pairs → 16 divisors. Why Other Options Are Wrong:3 and 10 are far below the correct count; 12 misses several divisors; 18 exceeds the true total. Common Pitfalls:Incorrect prime factorization (e.g., missing a factor 3); forgetting to add 1 to each exponent when applying the formula. Final Answer:16

Question 9

Divisibility of n(n^2 − 1) for Odd n > 1 If n is any odd integer greater than 1, what is n(n^2 − 1) = n(n − 1)(n + 1) always divisible by?

Accepted Answer

Introduction / Context:This question examines guaranteed divisibility of the product of three consecutive integers centered at an odd n. Recognizing structural factors in consecutive triples is a core technique in number theory problems on aptitude tests. Given Data / Assumptions: n is odd and greater than 1. Expression: n(n − 1)(n + 1). We must find a fixed integer that always divides this product. Concept / Approach:Among three consecutive integers, one is a multiple of 3, so the product is divisible by 3. Because n is odd, both neighbors (n − 1) and (n + 1) are even; they are consecutive even numbers, so one is a multiple of 2 and the other is a multiple of 4. Hence the product has at least a factor of 2 * 4 = 8 from the even terms, and a factor of 3 from the “multiple-of-3” property, totaling 8 * 3 = 24. Step-by-Step Solution:Since n is odd, n − 1 and n + 1 are even.One of these consecutive even numbers is divisible by 4, and the other by 2 → combined factor 8.Among n − 1, n, n + 1, one is a multiple of 3 → additional factor 3.Therefore, n(n − 1)(n + 1) is always divisible by 8 * 3 = 24. Verification / Alternative check:Try n = 3: product = 3 * 2 * 4 = 24 (divisible by 24). Try n = 5: 5 * 4 * 6 = 120 (24 * 5). The pattern persists for all odd n. Why Other Options Are Wrong:Divisibility by 12 is too weak (24 is guaranteed). 48 and 96 require an extra factor of 2 or 2^2 that is not ensured for every odd n. “None of these” is false because 24 always divides the product. Common Pitfalls:Assuming both neighbors contribute a 4 each (not guaranteed); overlooking the “multiple-of-3” property of any three consecutive integers. Final Answer:divisible by 24 always

Question 10

Constant Divisors of 7^(6n) − 6^(6n) For n a positive integer, the expression 7^(6n) − 6^(6n) is divisible by which of the following?

Accepted Answer

Introduction / Context:Expressions of the form a^m − b^m have well-known divisibility properties. If d divides m, then a^m − b^m is divisible by a^d − b^d. This problem asks you to recognize constant divisors that work for every positive integer n when the exponent is 6n. Given Data / Assumptions: Expression: 7^(6n) − 6^(6n) with n ≥ 1. We are checking divisibility by 13, 127, and 559. 6 divides 6n, so 7^(6n) − 6^(6n) is divisible by 7^6 − 6^6. Concept / Approach:Since 6 | 6n, a standard result guarantees 7^(6n) − 6^(6n) is divisible by 7^6 − 6^6. Therefore, any factor of 7^6 − 6^6 will also divide 7^(6n) − 6^(6n) for all n. Compute 7^6 − 6^6 and factor (or test) it against the candidate divisors. Step-by-Step Solution:Compute 7^6 = 117,649 and 6^6 = 46,656.Difference: 117,649 − 46,656 = 70,993.Check factors: 70,993 ÷ 13 = 5,461 (integer) → divisible by 13.Further factor: 70,993 = 127 × 559 (since 127×500 = 63,500; 127×59 = 7,493; sum = 70,993).Thus 127 and 559 also divide 7^6 − 6^6, and therefore divide 7^(6n) − 6^(6n) for all n. Verification / Alternative check:Note also that 7 ≡ −6 (mod 13), so for any even exponent 6n, 7^(6n) − 6^(6n) ≡ (−6)^(6n) − 6^(6n) = 0 (mod 13). The factorization of 70,993 confirms divisibility by 127 and 559 as well. Why Other Options Are Wrong:“None of these” is wrong because all three listed numbers divide the expression for every n ≥ 1. Selecting only one of 13, 127, or 559 would be incomplete; hence “All of these” is the only correct choice. Common Pitfalls:Attempting to test only n = 1 and missing the general argument; assuming divisibility holds for one modulus but not recognizing all listed moduli are factors of 7^6 − 6^6. Final Answer:All of these

Question 11

Sum of the First 25 Natural Numbers — Apply the Standard Formula Compute the value of 1 + 2 + 3 + … + 25.

Accepted Answer

Introduction / Context:Summation formulas save time in arithmetic progressions. The series of the first n natural numbers has a well-known closed form, allowing you to compute large sums instantly without manual addition. Given Data / Assumptions: We sum integers from 1 through 25 inclusive. Series is an arithmetic progression with first term 1, last term 25, common difference 1. Concept / Approach:Use the formula S_n = n(n + 1)/2. This formula arises from pairing terms equidistant from the ends (1 with 25, 2 with 24, etc.), each pair summing to 26, with 12 full pairs and one middle term 13—equivalently handled by the formula. Step-by-Step Solution:Identify n = 25.Apply S_n = n(n + 1)/2 = 25 * 26 / 2.Compute 26 / 2 = 13 → S_25 = 25 * 13.Multiply: 25 * 13 = 325. Verification / Alternative check:Quick mental check: The average of numbers 1 to 25 is (1 + 25)/2 = 13. With 25 terms, total = 13 * 25 = 325. Matches the formula result. Why Other Options Are Wrong:432, 315, 335, and 300 are off due to arithmetic mistakes or misuse of the formula. Common Pitfalls:Forgetting to divide by 2; summing a subset (e.g., up to 24) by mistake; miscomputing 25 * 13 under time pressure. Final Answer:325

Question 12

Sum of Squares — First 35 Natural Numbers Find the value of 1^2 + 2^2 + 3^2 + … + 35^2 using the standard formula.

Accepted Answer

Introduction / Context:Summing squares is a frequent task in quantitative aptitude and statistics. Rather than computing each square, rely on the closed-form formula for the sum of the first n squares to obtain the result quickly and accurately. Given Data / Assumptions: We need 1^2 + 2^2 + … + 35^2. n = 35 is a positive integer. Concept / Approach:The standard formula is S = n(n + 1)(2n + 1)/6. Substitute n = 35 and compute carefully by simplifying before multiplying to reduce arithmetic effort and minimize errors. Step-by-Step Solution:Identify n = 35 → n + 1 = 36, 2n + 1 = 71.Apply S = n(n + 1)(2n + 1)/6 = 35 * 36 * 71 / 6.Simplify 36 / 6 = 6 → S = 35 * 6 * 71.Compute 35 * 6 = 210; then 210 * 71 = 210 * (70 + 1) = 14,700 + 210 = 14,910. Verification / Alternative check:Estimate: average square near (35/2)^2 ≈ 17.5^2 ≈ 306; 35 terms × 306 ≈ 10,710 (a rough underestimate). Precise formula gives 14,910—consistent with exact calculation, not the rough estimate. Why Other Options Are Wrong:15510, 14510, 16510, and 15210 arise from slips such as not reducing by 6 first, mis-multiplying 210 * 71, or transposing digits. Common Pitfalls:Multiplying 35 * 36 * 71 first (risking large-number errors) instead of simplifying by 6; arithmetic mistakes when distributing 210 across 71. Final Answer:14910

Question 13

Series summation — cubes of natural numbers Find the exact sum of the cubes of the first 15 natural numbers (i.e., 1^3 + 2^3 + … + 15^3).

Accepted Answer

Introduction / Context: Summations of powers appear frequently in number system and algebra questions. A classic identity states that the sum of the first n cubes equals the square of the nth triangular number. Applying the identity correctly saves time and avoids long arithmetic. Given Data / Assumptions: We need S = 1^3 + 2^3 + … + 15^3. Use the standard closed-form: sum of first n cubes. Basic integer arithmetic is sufficient; no calculators required if steps are organized. Concept / Approach: The identity is: sum_{k=1 to n} k^3 = (n*(n+1)/2)^2. That means we first compute the triangular sum T = 1 + 2 + … + n = n*(n+1)/2, and then square T to obtain the sum of cubes. This both reduces effort and limits error risk. Step-by-Step Solution:Let n = 15.Compute T = n*(n+1)/2 = 15*16/2 = 120.Square T: S = T^2 = 120^2 = 14400.Therefore, the required sum is 14400. Verification / Alternative check: Spot-check by partial sums (e.g., up to 5: 1+8+27+64+125 = 225, and (5*6/2)^2 = 15^2 = 225) to reinforce confidence in the identity, then rely on the same logic for n = 15. Why Other Options Are Wrong:15400, 16800, 13300 result from squaring an incorrect triangular sum or arithmetic slips in 120^2. Common Pitfalls: Forgetting to square the triangular sum; computing 15*16 incorrectly; mixing the square of the sum with the sum of squares by mistake. Always remember: sum of cubes = (sum of first n integers)^2. Final Answer: 14400

Question 14

Series summation — first 84 even natural numbers Compute the exact sum: 2 + 4 + 6 + … up to 84 terms.

Accepted Answer

Introduction / Context: Sums of arithmetic progressions (AP) are common in aptitude exams. The even numbers form an AP with first term 2 and common difference 2. Knowing and applying the AP sum formula makes the calculation quick and reliable. Given Data / Assumptions: Total terms n = 84. First term a = 2, common difference d = 2. We need S = 2 + 4 + … (84 terms). Concept / Approach: The sum of the first n even numbers equals 2*(1 + 2 + … + n). Hence S = 2 * (n*(n+1)/2) = n*(n+1). This compact relation eliminates the need to list terms individually. Step-by-Step Solution:Use S = n*(n+1).For n = 84, S = 84*85.Compute 84*85 = (84*80) + (84*5) = 6720 + 420 = 7140.Therefore, the sum is 7140. Verification / Alternative check: AP formula: S = n/2 * (2a + (n-1)d) = 84/2 * (4 + 83*2) = 42 * (4 + 166) = 42 * 170 = 7140, confirming the result. Why Other Options Are Wrong:7540, 6720, 8832 arise from off-by-one term counts or multiplying 84 by 84 instead of 85. Common Pitfalls: Confusing the number of terms with the largest even value; forgetting that the nth even number is 2n; and mishandling 84*85 mental multiplication. Final Answer: 7140

Question 15

Series summation — first 37 odd natural numbers Find the exact sum: 1 + 3 + 5 + … (37 terms).

Accepted Answer

Introduction / Context: A well-known identity states that the sum of the first n odd numbers equals n^2. This elegant result bypasses lengthy addition and is widely used in number theory and aptitude problems. Given Data / Assumptions: We seek S = 1 + 3 + 5 + … (37 terms). Each term increases by 2 (an AP of odds). Use the identity for the sum of the first n odd numbers. Concept / Approach: Identity: sum of first n odd numbers = n^2. For n = 37, compute 37^2 directly. This avoids the AP sum formula steps and leads to an exact integer quickly. Step-by-Step Solution:Let n = 37.Compute n^2: 37^2 = (40 - 3)^2 = 1600 - 240 + 9 = 1369.So S = 1369. Verification / Alternative check: Using AP sum: S = n/2 * (first + last). The 37th odd is 2*37 - 1 = 73. Then S = 37/2 * (1 + 73) = 37/2 * 74 = 37 * 37 = 1369, same result. Why Other Options Are Wrong:1295, 1388, 1875 do not equal 37^2 and come from miscounting terms or miscomputing 37^2. Common Pitfalls: Using the largest term (73) as n; mixing odds with evens; arithmetic slips in the square computation. Remember the identity: n odds sum to n^2. Final Answer: 1369

Question 16

Arithmetic progression — first 15 multiples of 8 Compute the sum of 8, 16, 24, … up to the 15th multiple of 8.

Accepted Answer

Introduction / Context: Sums of multiples are AP sums. Recognizing the structure allows using a compact formula to avoid enumerating every term. Here the common difference is 8 and we want the first 15 terms. Given Data / Assumptions: First term a = 8. Common difference d = 8. Number of terms n = 15. Concept / Approach: The n-th multiple of 8 is 8n, and the sum of the first n multiples of 8 equals 8*(1 + 2 + … + n) = 8 * (n*(n+1)/2). Using this relation is faster and less error-prone than listing terms. Step-by-Step Solution:Compute T = 1 + 2 + … + 15 = 15*16/2 = 120.Multiply by 8: S = 8 * 120 = 960.Hence, the sum is 960. Verification / Alternative check: AP formula: S = n/2 * (2a + (n-1)d) = 15/2 * (16 + 14*8) = 15/2 * (16 + 112) = 15/2 * 128 = 15 * 64 = 960, confirming the value. Why Other Options Are Wrong:660, 1060, 1200 reflect arithmetic errors such as wrong n, misapplied formula, or skipping the /2 step. Common Pitfalls: Confusing the number of terms with the last term value; forgetting the factor 1/2 in the AP formula; or miscomputing 15*16. Final Answer: 960

Question 17

Units digit evaluation — large exponent Determine the units digit (ones place) of (6817)^754.

Accepted Answer

Introduction / Context: For huge exponents, only the base’s last digit affects the final units digit. Recognizing the repeating cycle of units digits for powers of a digit greatly simplifies computation. Here, 6817 ends with 7, so we analyze powers of 7 modulo 10. Given Data / Assumptions: We need the units digit of 6817^754. Only the last digit 7 matters for units digit analysis. Cycles of powers modulo 10 repeat with small periods. Concept / Approach: Powers of 7 mod 10 cycle as: 7, 9, 3, 1, and then repeat every 4. Therefore, we compute the exponent modulo 4 to select the correct position in the cycle. Step-by-Step Solution:Cycle for 7^k mod 10: [7, 9, 3, 1] with period 4.Compute 754 mod 4: since 4*188 = 752, remainder = 2.Remainder 2 corresponds to the second element in the cycle, which is 9.Thus, the units digit is 9. Verification / Alternative check: Confirm smaller examples: 7^1 → 7, 7^2 → 49 (ends 9), 7^3 → 343 (ends 3), 7^4 → 2401 (ends 1), and repeats. The mapping of remainders 1–4 to [7,9,3,1] is consistent. Why Other Options Are Wrong:8, 4, 2 correspond to other digit cycles, not the 7^k cycle at remainder 2. Common Pitfalls: Using the exponent itself instead of its remainder modulo the cycle length; mis-ordering the cycle; or assuming every digit has the same period length. Final Answer: 9

Question 18

Units digit of a sum with large exponents Find the units digit of 3^65 + 6^59 + 7^71.

Accepted Answer

Introduction / Context: To find the units digit of a sum involving large exponents, we compute each term’s units digit independently using known cycles, then add and reduce modulo 10. This approach avoids dealing with massive numbers. Given Data / Assumptions: Expression: 3^65 + 6^59 + 7^71. Only last digits matter: 3, 6, 7. Each base has a small repeating cycle modulo 10. Concept / Approach: Cycles: 3^k mod 10 → [3, 9, 7, 1] (period 4). 6^k mod 10 → always 6. 7^k mod 10 → [7, 9, 3, 1] (period 4). Compute exponents modulo 4 where applicable, then add resulting units digits and reduce modulo 10. Step-by-Step Solution:For 3^65: 65 mod 4 = 1 → units digit 3.For 6^59: always 6 → units digit 6.For 7^71: 71 mod 4 = 3 → units digit 3.Sum of units digits: 3 + 6 + 3 = 12 → 12 mod 10 = 2. Verification / Alternative check: Quick sanity: 3-cycle position 1 gives 3; 7-cycle position 3 gives 3; 6 is constant. The final 2 is consistent with common results for such mixed-base problems. Why Other Options Are Wrong:1, 4, 6 arise from selecting the wrong cycle position or forgetting to reduce the final sum modulo 10. Common Pitfalls: Forgetting that 6^n ends in 6 for all n ≥ 1; computing 65 mod 4 incorrectly; adding the units digits but not reducing modulo 10 at the end. Final Answer: 2

Question 19

Remainder with a smaller divisor from a larger modulus condition A number leaves remainder 29 when divided by 56. Find the remainder when the same number is divided by 8.

Accepted Answer

Introduction / Context: When you know a number's remainder modulo a larger divisor, you can find its remainder modulo any factor of that divisor by reducing the given remainder accordingly. This leverages congruence compatibility. Given Data / Assumptions: N ≡ 29 (mod 56). Need N mod 8. 8 divides 56, so reduction is direct. Concept / Approach: If N ≡ r (mod m) and k divides m, then N ≡ r (mod k) but r must be reduced modulo k to get the canonical remainder between 0 and k − 1. Compute r mod k directly. Step-by-Step Solution:Given N ≡ 29 (mod 56).Compute 29 mod 8: 8*3 = 24; remainder = 29 − 24 = 5.Therefore, N ≡ 5 (mod 8). Verification / Alternative check: Any number of the form N = 56q + 29 will be 8*(7q) + 5, so the remainder on division by 8 is always 5. Why Other Options Are Wrong:4, 6, 7 are simply other residues modulo 8; only 5 matches 29 reduced modulo 8. Common Pitfalls: Dividing 29 by 8 incorrectly; trying to recompute N explicitly instead of using modular reduction; forgetting to bring the remainder into the 0–7 range. Final Answer: 5

Question 20

Remainder transformation between related moduli A number leaves remainder 39 when divided by 357. What remainder does it leave when divided by 17?

Accepted Answer

Introduction / Context: Converting a known remainder modulo one divisor to another is simple if the second divisor divides the first. Because congruences are compatible under such inclusion, we can reduce the known remainder directly. Given Data / Assumptions: N ≡ 39 (mod 357). We want N mod 17. 357 is divisible by 17 (357 = 17 * 21). Concept / Approach: If N ≡ r (mod m) and k | m, then N ≡ r (mod k). Reduce r modulo k to get the canonical remainder between 0 and k − 1. No need to know the actual quotient. Step-by-Step Solution:Compute 39 mod 17: 17*2 = 34; remainder = 39 − 34 = 5.Therefore, N ≡ 5 (mod 17).Hence the remainder upon division by 17 is 5. Verification / Alternative check: Write N = 357q + 39. Modulo 17, 357q ≡ 0 (since 357 is a multiple of 17). So N ≡ 39 ≡ 5 (mod 17), confirming the result. Why Other Options Are Wrong:3, 6, 7 are not equal to 39 mod 17 and thus do not satisfy the congruence transformation. Common Pitfalls: Not recognizing that 357 is divisible by 17; mistakenly trying to compute N explicitly; arithmetic slips when reducing 39 modulo 17. Final Answer: 5

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