Question 1

Remainder of a square modulo 5 If a number leaves remainder 3 when divided by 5, what remainder will its square leave when divided by 5?

Accepted Answer

Introduction / Context: Squaring a residue modulo a small base is a standard modular arithmetic task. By replacing the number with its remainder and then squaring under the same modulus, we get the new remainder efficiently. Given Data / Assumptions: N ≡ 3 (mod 5). We require N^2 mod 5. Modular arithmetic properties apply: operations respect the modulus. Concept / Approach: If N ≡ r (mod m), then N^2 ≡ r^2 (mod m). Therefore compute 3^2 mod 5. No need to know N explicitly. Step-by-Step Solution: Given r = 3 under modulus 5.Compute r^2 = 3^2 = 9.Reduce modulo 5: 9 mod 5 = 4.Therefore, N^2 ≡ 4 (mod 5). Verification / Alternative check: Example: Let N = 8 (which is 3 mod 5). Then N^2 = 64; 64 mod 5 = 4. This confirms the general reasoning. Why Other Options Are Wrong: 3 is the original residue, not the square's residue; 5 is not a valid remainder; 0 would require N to be a multiple of 5. Common Pitfalls: Forgetting to reduce after squaring; mixing up N mod 5 with N^2 mod 5; assuming residues remain unchanged after exponentiation. Final Answer: 4

Question 2

Find the divisor from two remainder statements A number leaves remainder 24 when divided by a certain divisor. Twice the number leaves remainder 11 with the same divisor. Determine the divisor.

Accepted Answer

Introduction / Context: Problems that provide remainders for N and for 2N invite the use of modular equivalences. By translating each sentence into a congruence and comparing them, we can deduce the divisor without ever computing N itself. Given Data / Assumptions: N ≡ 24 (mod d). 2N ≡ 11 (mod d). d is a positive integer greater than the remainders where applicable. Concept / Approach: From N ≡ 24 (mod d), doubling gives 2N ≡ 48 (mod d). But we also have 2N ≡ 11 (mod d). Therefore, 48 ≡ 11 (mod d) which implies d divides (48 − 11) = 37. Since a positive divisor greater than the remainder 24 is required, the only valid d is 37. Step-by-Step Solution: Start: N ≡ 24 (mod d).Double: 2N ≡ 48 (mod d).Compare with given: 2N ≡ 11 (mod d) ⇒ 48 ≡ 11 (mod d).Hence d | (48 − 11) = 37 ⇒ d = 37 (since d > 24). Verification / Alternative check: Construct N = 37k + 24. Then 2N = 74k + 48. Mod 37, 2N ≡ 11 because 48 mod 37 = 11, consistent with the condition for any integer k. Why Other Options Are Wrong: 13, 59, 35 do not divide 37. Only 37 satisfies 48 ≡ 11 (mod d) and is larger than 24. Common Pitfalls: Forgetting to double the first congruence; assuming d equals the difference between remainders without checking divisibility; overlooking the constraint that remainders must be less than the divisor. Final Answer: 37

Question 3

Divisibility check — For the 14-digit integer 58129745812974, determine which divisibility claim is correct. Use standard tests for 9, 11, and 4 and select the single best statement.

Accepted Answer

Introduction / Context: This question asks you to apply quick divisibility tests to a large integer (58129745812974) without long division. The classic tests for 9 and 11 rely on digit sums and alternating sums, while the test for 4 depends on the last two digits only. Mastering these quick checks saves time on exams and interviews. Given Data / Assumptions: Integer under test: 58129745812974. Divisibility tests needed: by 9, by 11, and by 4. Choose exactly one correct statement about divisibility. Concept / Approach: Use well-known rules: a number is divisible by 9 if the sum of its digits is a multiple of 9; divisible by 11 if the alternating sum (sum of digits in odd positions minus sum of digits in even positions) is a multiple of 11 (including 0); divisible by 4 if the last two digits form a number divisible by 4. Step-by-Step Solution: 1) Sum of digits for 9: 5+8+1+2+9+7+4+5+8+1+2+9+7+4 = 72. Since 72/9 = 8, the number is divisible by 9.2) Alternating sum for 11 (from leftmost as position 1): (5−8+1−2+9−7+4−5+8−1+2−9+7−4) = 0. A result of 0 means divisibility by 11.3) Test for 4: last two digits are 74; 74/4 is not an integer, so it is not divisible by 4. Verification / Alternative check: Because tests for 9 and 11 are independent, being divisible by both is consistent. A quick calculator check would confirm divisibility by 99 (since 9×11 = 99), but that is not necessary once the tests pass. Why Other Options Are Wrong: “Divisible by 9 only” and “Divisible by 11 only” ignore that both tests pass; “Divisible by 4” fails because 74 is not a multiple of 4; “Not divisible by any of these” is contradicted by both successful tests. Common Pitfalls: Forgetting to alternate signs correctly for 11; mis-adding a long digit sum; using only the last digit for 4 (the rule requires the last two digits). Final Answer: Divisible by both 9 and 11

Question 4

Counting with inclusion–exclusion — How many integers between −11 and 11 (inclusive) are multiples of 2 or multiples of 3? Count 0 as a multiple of every integer.

Accepted Answer

Introduction / Context: This problem checks your comfort with inclusion–exclusion and with symmetric intervals around 0. We count integers from −11 to 11 inclusive that are multiples of 2 or 3. Because 0 is a multiple of every integer, ensure it is correctly included once, not multiple times. Given Data / Assumptions: Interval: from −11 to 11, inclusive (i.e., −11, −10, …, 10, 11). Criterion: divisible by 2 or by 3 (or by both). 0 qualifies (divisible by every integer). Concept / Approach: Use inclusion–exclusion: count multiples of 2, add multiples of 3, subtract multiples of 6 (since those were counted twice). Alternatively, list quickly near 0 by symmetry. Either approach should deliver the same number. Step-by-Step Solution: 1) Multiples of 2 in [−11, 11]: −10, −8, −6, −4, −2, 0, 2, 4, 6, 8, 10 → 11 numbers.2) Multiples of 3 in [−11, 11]: −9, −6, −3, 0, 3, 6, 9 → 7 numbers.3) Multiples of 6 (common to both sets): −6, 0, 6 → 3 numbers.4) Inclusion–exclusion total = 11 + 7 − 3 = 15. Verification / Alternative check: A direct, quick listing centered at 0 confirms there are exactly 15 such integers. Many students find symmetry helps avoid misses on the negative side. Why Other Options Are Wrong: 11 and 14 undercount by missing some 3-multiples or misusing inclusion–exclusion; 16 overcounts by double-counting common multiples; “None of these” is invalid because 15 is attainable. Common Pitfalls: Forgetting to include 0; forgetting to subtract multiples of 6; omitting − multiples on the negative side; treating the interval as exclusive of endpoints by mistake. Final Answer: 15

Question 5

Divisibility by 11 — Among the following eight-digit numbers, identify which one is divisible by 11 using the alternating-sum rule.

Accepted Answer

Introduction / Context: The divisibility rule for 11 avoids long division by converting the digits into an alternating sum. If the alternating sum is a multiple of 11 (including 0), the number is divisible by 11. We apply this rule to each option to find the only correct choice. Given Data / Assumptions: Candidates: 45678940, 54857266, 87524398, 93455120. Rule: sum of digits in odd positions minus sum in even positions must be 0 or a multiple of 11. Counting positions from left as position 1. Concept / Approach: Compute alternating sum quickly, preferably grouping digits to reduce errors. Because we only need one correct option, stop once you find a number with alternating sum 0 or ±11, ±22, etc., but verify carefully to avoid slips. Step-by-Step Solution: Example for 93455120: positions (from left) 1..8 → (9−3+4−5+5−1+2−0) = 11 − 6 − 1 + 2 = 6? Recheck cleanly: (9 − 3) + (4 − 5) + (5 − 1) + (2 − 0) = 6 − 1 + 4 + 2 = 11.Alternating sum = 11, which is a multiple of 11 → divisible by 11.The other options produce alternating sums not equal to a multiple of 11 on careful calculation. Verification / Alternative check: A second pass or quick calculator confirms only 93455120 satisfies the 11-rule. If in doubt, compute modulo 11 directly for each candidate. Why Other Options Are Wrong: 45678940, 54857266, and 87524398 each yield an alternating sum that is not a multiple of 11, so they fail the rule. Common Pitfalls: Mislabeling odd/even positions; dropping a digit in the alternating sum; stopping early without confirming the multiple-of-11 condition. Final Answer: 93455120

Question 6

Remainder problem — Find the remainder when 17^200 is divided by 18. Use modular patterns or parity arguments for efficiency.

Accepted Answer

Introduction / Context: Large exponents are best handled via modular arithmetic patterns. Here, 17 is close to 18, which offers a quick simplification. Recognizing small congruences prevents unnecessary exponentiation and speeds up remainder calculations dramatically. Given Data / Assumptions: Compute 17^200 mod 18. Use congruences and exponent rules. We only need the remainder (0–17). Concept / Approach: Observe that 17 ≡ −1 (mod 18). Then 17^200 ≡ (−1)^200 (mod 18). Since any even power of −1 equals 1, the remainder is immediate. No need for Euler’s theorem here, though it would also work. Step-by-Step Solution: 1) Reduce base: 17 ≡ −1 (mod 18).2) Raise to the 200th power: (−1)^200 = 1.3) Therefore, 17^200 ≡ 1 (mod 18).4) Remainder upon division by 18 is 1. Verification / Alternative check: Euler’s theorem: φ(18) = 6, so 17^6 ≡ 1 (mod 18). Then 17^200 = (17^6)^{33} * 17^2 ≡ 1^{33} * 17^2 ≡ (−1)^2 = 1, confirming the same result. Why Other Options Are Wrong: 3, 4, 5, and 7 do not match the clean parity conclusion (even power of −1). Any non-1 remainder would contradict the simple congruence. Common Pitfalls: Expanding 17^200 directly; forgetting that even powers of −1 yield +1; misapplying Euler's theorem without reducing correctly. Final Answer: 1

Question 7

Remainder problem — Determine the remainder when 4^1000 is divided by 7. Identify and use the exponent cycle modulo 7.

Accepted Answer

Introduction / Context: For remainders with large exponents, spotting cycles modulo a small base is the standard trick. The base 4 modulo 7 repeats with a short period due to Euler's theorem and orders of elements, allowing a quick reduction of the exponent 1000. Given Data / Assumptions: Compute 4^1000 mod 7. Know φ(7) = 6 and that powers of 4 modulo 7 cycle. We need only the final remainder. Concept / Approach: Since 7 is prime, 4^6 ≡ 1 (mod 7). Therefore reduce 1000 modulo 6 to find the equivalent smaller exponent. Then compute 4^r mod 7 for the remainder exponent r and read off the result. Step-by-Step Solution: 1) Because 4^6 ≡ 1 (mod 7), compute 1000 mod 6.2) 1000 = 6*166 + 4 → remainder exponent r = 4.3) Compute 4^4 mod 7: 4^2 = 16 ≡ 2 (mod 7), then 4^4 = (4^2)^2 ≡ 2^2 = 4 (mod 7).4) Hence, 4^1000 ≡ 4 (mod 7). Verification / Alternative check: Direct cycle: 4, 2, 1, 4, 2, 1, … every 3 steps within the 6-length residue pattern also shows the 4th term equals 4, confirming the result. Why Other Options Are Wrong: 1 or 2 would require 1000 to land on different positions in the cycle; 0 is impossible since 7 ∤ 4; “None of these” is invalid because 4 fits perfectly. Common Pitfalls: Reducing 1000 modulo 7 instead of modulo 6; arithmetic slips computing 4^4; assuming the remainder must be small like 1 by default. Final Answer: 4

Question 8

Number theory — Identify a common factor shared by (41^43 + 43^43) and (41^41 + 43^41). Use the gcd property for sums with odd exponents.

Accepted Answer

Introduction / Context: This problem probes a useful gcd identity for sums of like powers. When exponents are odd, the greatest common divisor of a^m + b^m and a^n + b^n is a^{gcd(m,n)} + b^{gcd(m,n)} (up to trivial factors). Recognizing this reduces what looks like a huge arithmetic task to a simple algebraic conclusion. Given Data / Assumptions: Two numbers: N1 = 41^43 + 43^43 and N2 = 41^41 + 43^41. Both exponents 43 and 41 are odd. We seek a nontrivial common factor from the options. Concept / Approach: For odd exponents m, n: gcd(a^m + b^m, a^n + b^n) = a^{g} + b^{g}, where g = gcd(m, n). Here, gcd(43, 41) = 1, hence the gcd simplifies to a^1 + b^1 = a + b = 41 + 43. Step-by-Step Solution: 1) Note that 43 and 41 are both odd primes, so gcd(43, 41) = 1.2) Apply the identity to a = 41, b = 43: gcd(41^43 + 43^43, 41^41 + 43^41) = 41^{1} + 43^{1}.3) Therefore a definitive common factor is 41 + 43 = 84.4) Among the options, this corresponds to “41 + 43.” Verification / Alternative check: Since both quantities are sums of odd powers, each is divisible by (41 + 43) individually (because a + b divides a^k + b^k for odd k). Hence (41 + 43) divides both, confirming it as a common factor. Why Other Options Are Wrong: “43 − 41” is 2, which need not divide sums of odd powers generally; “41^41 + 43^41” and “41^43 + 43^43” are the original numbers, not guaranteed as common factors; “41*43” is unrelated to the odd-power sum divisibility rule. Common Pitfalls: Misapplying the identity for even exponents; confusing a + b with a − b; assuming the gcd must be huge rather than a simple expression. Final Answer: 41 + 43

Question 9

Remainder problem — Compute the remainder when 9^19 + 6 is divided by 8. Use modular reduction of the base.

Accepted Answer

Introduction / Context: We are asked for a remainder modulo 8. Reducing bases modulo 8 and recognizing simple patterns allow quick answers for large exponents. This is a standard exercise in modular arithmetic fluency for aptitude tests. Given Data / Assumptions: Expression: 9^19 + 6. Modulus: 8. Basic rules: reduce the base first, then apply exponent laws. Concept / Approach: Since 9 ≡ 1 (mod 8), any power 9^k ≡ 1^k ≡ 1 (mod 8). Then just add the remaining constant 6 and reduce modulo 8. Step-by-Step Solution: 1) Reduce base: 9 ≡ 1 (mod 8).2) Raise to power: 9^19 ≡ 1^19 = 1 (mod 8).3) Add 6: 1 + 6 = 7.4) Therefore (9^19 + 6) mod 8 = 7. Verification / Alternative check: Because 9 ≡ 1 (mod 8) exactly, no cycle tracking is needed. Any exponent would still collapse to 1 before adding 6, confirming the same remainder. Why Other Options Are Wrong: 2, 3, 5, and 1 do not match the straightforward reduction result of 7. Common Pitfalls: Attempting to compute 9^19 explicitly; using 9 ≡ −7 (mod 8) and making sign errors; forgetting to add the constant term 6 at the end. Final Answer: 7

Question 10

Remainder problem — What is the remainder when 19^100 is divided by 20? Use the symmetry 19 ≡ −1 (mod 20).

Accepted Answer

Introduction / Context: Powers of numbers that are “one less than” the modulus often simplify neatly. Here 19 ≡ −1 (mod 20), making high powers easy to evaluate using parity. This technique is a must-know for number system questions involving large exponents and small moduli. Given Data / Assumptions: Compute 19^100 mod 20. Observation: 19 ≡ −1 modulo 20. Even powers of −1 equal +1. Concept / Approach: Replace 19 with −1 under modulo 20 and apply exponent rules. Because the exponent 100 is even, the result collapses to 1 immediately, without any need for cycles or Euler's theorem. Step-by-Step Solution: 1) 19 ≡ −1 (mod 20).2) 19^100 ≡ (−1)^100 (mod 20).3) Since 100 is even, (−1)^100 = 1.4) Therefore, the remainder is 1. Verification / Alternative check: Euler’s theorem with φ(20) = 8 also works but is unnecessary. The parity shortcut is the simplest and least error-prone approach here. Why Other Options Are Wrong: 19 would correspond to an odd power; 3 or 29 do not arise from the parity shortcut; 0 is impossible because 20 does not divide 19. Common Pitfalls: Mixing up odd/even exponents; misapplying Euler’s theorem; forgetting that the remainder must be in 0–19 inclusive. Final Answer: 1

Question 11

Divisibility inference — If (2^32 + 1) is divisible by some positive integer m, which of the following is guaranteed also to be divisible by the same m?

Accepted Answer

Introduction / Context: We are given that m | (2^32 + 1). Such numbers arise in the context of Fermat-type expressions. The key idea is to use modular arithmetic: if 2^32 ≡ −1 (mod m), then powers of 2 that are multiples of 32 behave predictably, allowing us to deduce divisibility of related expressions. Given Data / Assumptions: m divides 2^32 + 1 → 2^32 ≡ −1 (mod m). We must test which option is also ≡ 0 (mod m). Basic exponent rules in modular arithmetic apply. Concept / Approach: If 2^32 ≡ −1 (mod m), then raise both sides to powers. In particular, (2^32)^3 = 2^96 ≡ (−1)^3 = −1 (mod m). Adding 1 gives 2^96 + 1 ≡ 0 (mod m), guaranteeing divisibility by m. Other options generally do not follow from the given congruence. Step-by-Step Solution: 1) From m | (2^32 + 1), get 2^32 ≡ −1 (mod m).2) Cube both sides: (2^32)^3 = 2^96 ≡ (−1)^3 = −1 (mod m).3) Add 1: 2^96 + 1 ≡ 0 (mod m).4) Hence m | (2^96 + 1), which matches option “2^96 + 1.” Verification / Alternative check: Similarly, (2^32)^2 = 2^64 ≡ 1 (mod m), implying m | (2^64 − 1), not (2^64 + 1). This further confirms why option e is not guaranteed. Why Other Options Are Wrong: 2^16 ± 1 and 7*2^13 have no direct necessity from 2^32 ≡ −1; 2^64 + 1 is not forced because 2^64 ≡ 1, making 2^64 + 1 ≡ 2 (mod m), not 0. Common Pitfalls: Assuming any multiple power automatically works; confusing (a ≡ −1) with (a ≡ 1); overlooking that squaring gives +1 while cubing keeps −1. Final Answer: 2^96 + 1

Question 12

Exact divisibility — Determine which number divides 7^12 − 4^12. Use parity, small prime tests, and algebraic factorization reasoning.

Accepted Answer

Introduction / Context: We must decide which candidate divides 7^12 − 4^12. Observations about parity (odd/even), simple modular reductions (mod 3, 5, 7, 11), and known factorization patterns for a^n − b^n lead quickly to the correct choice without computing the huge powers explicitly. Given Data / Assumptions: Expression: 7^12 − 4^12. Options: 34, 33, 35, 36, 45. We check divisibility via modular arithmetic and reasoning. Concept / Approach: First, note 7^12 is odd and 4^12 is even, so the result is odd; thus any even option (34, 36) is eliminated. Next check divisibility by 3 and 11 (since 33 = 3×11). Also test 35 = 5×7 and 45 = 5×9 for completeness. Use small-base reductions: 7 ≡ 1 (mod 3), 4 ≡ 1 (mod 3); 7 ≡ −4 (mod 11) helps too. Step-by-Step Solution: 1) Mod 2: 7^12 − 4^12 is odd − even = odd → not divisible by 2 ⇒ eliminate 34 and 36.2) Mod 3: 7 ≡ 1, 4 ≡ 1 (mod 3) → 7^12 ≡ 1, 4^12 ≡ 1 → difference 0 → divisible by 3.3) Mod 11: Use 7 ≡ −4 (mod 11). Then 7^12 − 4^12 ≡ (−4)^12 − 4^12 = 4^12 − 4^12 = 0 → divisible by 11.4) Therefore divisible by 3 and 11 simultaneously → divisible by 33.5) Check 35: mod 5, 7^12 ≡ (2)^12, 4^12 ≡ (−1)^12 = 1, 2^12 = 4096 ≡ 1 (mod 5) → difference 0 (so divisible by 5) but mod 7: 4 ≡ 4 so 4^12 ≡ 4^{12}, 7^12 ≡ 0? No, mod 7 gives 7^12 ≡ 0 only if the base were 7, but we test 7 dividing the expression; 4^12 mod 7 = (4^3)^4 = 64^4 ≡ (1)^4 = 1, while 7^12 ≡ 0 (mod 7) actually means 7 divides 7^12; hence expression ≡ 0 − 1 = −1 (mod 7) not divisible → thus 35 fails. Verification / Alternative check: Because it is divisible by both 3 and 11 and is odd, 33 is the strongest candidate. A quick computation confirms 7^12 − 4^12 ≡ 0 (mod 33). Why Other Options Are Wrong: 34 and 36 are even; 35 fails modulo 7 as shown; 45 requires divisibility by 5 and 9, but modulo 9 the value is not guaranteed 0. Common Pitfalls: Assuming divisibility by 35 because it works modulo 5; overlooking the parity test; not leveraging the identity a^n − b^n and small-mod checks. Final Answer: 33

Question 13

Number theory — For any natural number a, determine the largest integer that always divides the expression a^3 - a. Provide reasoning based on factor patterns.

Accepted Answer

Introduction / Context: This problem asks for a divisor that works for all natural numbers a when applied to the expression a^3 - a. Such questions test your grasp of algebraic factorization and guaranteed divisibility results that hold universally (independent of the specific value of a). Recognizing standard identities quickly leads to a clean, general proof without case-by-case calculation. Given Data / Assumptions: a is a natural number (1, 2, 3, …). Target expression: a^3 - a. We must find the largest fixed integer that always divides a^3 - a for every a. Concept / Approach: First, factor the expression using a common algebraic identity. Then analyze the factors for universal properties: in particular, parity (evenness) and multiples of 3. A classic result is that any product of three consecutive integers is divisible by 3 and by 2. We will connect a^3 - a to that pattern. Step-by-Step Solution: Factor the expression: a^3 - a = a(a^2 - 1) = a(a - 1)(a + 1).The three factors are consecutive integers: (a - 1), a, (a + 1).Among any three consecutive integers, one is divisible by 3. Hence a(a - 1)(a + 1) is divisible by 3.Among any two consecutive integers, one is even. Here, we have three consecutive integers, so at least one is even. In fact, among three consecutive integers, at least one is divisible by 2. Therefore the product is divisible by 2.Combining these: the product is divisible by 2 * 3 = 6 for every natural a. Verification / Alternative check: Test a few values: a=1 → 1^3 - 1 = 0 (divisible by any integer); a=2 → 8 - 2 = 6 (divisible by 6); a=3 → 27 - 3 = 24 (divisible by 6). These examples support the general proof from factorization. Why Other Options Are Wrong: 4 / 5 / 7 / 8: None of these always divide a^3 - a for all natural a. For example, a=2 gives 6, which is not divisible by 4, 5, 7, or 8. The guaranteed universal divisor is 6. Common Pitfalls: Forgetting to factor fully; assuming higher fixed powers of 2 divide the product (they do not always); relying only on small tests instead of recognizing the general consecutive-integers structure. Final Answer: 6

Question 14

Divisibility check — Evaluate 19^5 + 21^5 and state whether it is divisible by 10, by 20, by both, or by neither. Use modular reasoning (units and mod 4).

Accepted Answer

Introduction / Context: Instead of computing huge powers, divisibility questions can be solved using modular arithmetic. Here you assess 19^5 + 21^5 with respect to 10 and 20. Divisibility by 10 requires a last digit 0 (i.e., divisibility by 2 and 5), while divisibility by 20 requires divisibility by both 4 and 5. We will use units-digit patterns and modular reductions to conclude quickly. Given Data / Assumptions: Expression: 19^5 + 21^5. We test divisibility by 10 and by 20. We use modular arithmetic (mod 10 for last digit, mod 4 for the factor 4, and mod 5 for the factor 5 if needed). Concept / Approach: Use periodicity of last digits for powers to handle mod 10. For mod 4, reduce bases to their remainders modulo 4. Finally, confirm the factor 5 via mod 5 or simple observation from the last digit. If the number is divisible by both 4 and 5, it is divisible by 20. Step-by-Step Solution: Units digits: 19 → 9, 21 → 1. For odd exponents, 9^odd ends in 9 and 1^odd ends in 1.Hence last digit of 19^5 + 21^5 is 9 + 1 = 10 → ends with 0 → divisible by 10.Check mod 4: 19 ≡ 3 (mod 4), 3^2 ≡ 1, so 3^5 ≡ 3 (mod 4). Also, 21 ≡ 1 (mod 4), 1^5 ≡ 1. Sum ≡ 3 + 1 = 0 (mod 4).Since the sum is divisible by 4 and ends with 0 (divisible by 5), it is divisible by 20. Verification / Alternative check: Observe mod 5: 19 ≡ -1 (mod 5) so 19^5 ≡ -1; 21 ≡ 1 (mod 5) so 21^5 ≡ 1; sum ≡ 0 (mod 5). Combined with the mod 4 result, divisibility by 20 is confirmed via the lcm approach. Why Other Options Are Wrong: Only 10 / Only 20 / Neither: The sum satisfies divisibility by both 10 and 20 as shown, so these are incorrect. Only 5: Too weak; the sum is divisible by 5 but also by 4, hence by 20 and 10. Common Pitfalls: Trying to compute the full powers; ignoring mod 4; assuming “ends with 0” implies exactly 10 but not checking for 20; overlooking periodicity of units digits. Final Answer: Both 10 and 20

Question 15

Always divisible? — For natural number x, analyze the expression 6x^2 + 6x and determine which fixed divisibility statement is always true.

Accepted Answer

Introduction / Context: Divisibility questions often become straightforward after factoring. Recognizing common factors and parity properties lets us determine strong, universal divisibility claims for expressions that depend on natural numbers. Here we analyze 6x^2 + 6x and decide which divisors always occur. Given Data / Assumptions: x is a natural number. Expression: 6x^2 + 6x. We must pick the statement that is always true for all x. Concept / Approach: Factor the expression first, then use the property of consecutive integers. The product x(x + 1) involves two consecutive integers, which guarantees evenness (and hence at least one factor 2). Combine this with the 6 outside to determine the minimal universal power-of-2 and power-of-3 factors, and thus the largest guaranteed composite divisor. Step-by-Step Solution: Factor: 6x^2 + 6x = 6x(x + 1).Among consecutive integers x and x+1, one is even → x(x+1) is even → contributes at least one factor 2.Therefore 6x(x+1) has factors 6 * 2 = 12 at minimum, for any x.Hence the expression is always divisible by 12. Divisible by 12 implies divisible by 6 as well. Verification / Alternative check: Test x=1: 6*1^2 + 6*1 = 12 (divisible by 12). Test x=2: 24 + 12 = 36 (divisible by 12). Test x=3: 54 + 18 = 72 (divisible by 12). These checks match the general proof. Why Other Options Are Wrong: 12 only: Misleading wording—while the expression is always divisible by 12, it is also divisible by 6; “only” suggests otherwise. 6 only / 3 only: Too weak; a stronger always-true statement is that it is divisible by 12. 24: Not guaranteed (counterexample x=1 gives 12). Common Pitfalls: Stopping after extracting factor 6; forgetting the evenness of x(x+1); over-claiming divisibility by 24 without verifying small x. Final Answer: 6 and 12

Question 16

Prime triples — If N, N + 2, and N + 4 are all prime numbers, how many possible values of N exist? Explain using divisibility by 3.

Accepted Answer

Introduction / Context: This is a classic number theory puzzle about “prime triplets.” Most sets of three odd numbers spaced by 2 cannot all be prime because one is divisible by 3. The task is to determine how many such N exist that make N, N+2, and N+4 all prime simultaneously. Given Data / Assumptions: N is an integer (we are seeking prime values for the three terms). N, N+2, N+4 are all required to be prime. Primes are integers greater than 1 with exactly two positive divisors. Concept / Approach: Consider the three numbers modulo 3. Any three consecutive odd numbers (spaced by 2) will cover all residues mod 3. That means at least one of them must be divisible by 3. The only way to avoid a composite member is if that divisible-by-3 term is actually 3 itself, which is prime. This restriction pins down a unique feasible value for N. Step-by-Step Solution: Among N, N+2, N+4, one is ≡ 0 (mod 3).If that term is greater than 3, it becomes composite and the triple fails.Thus, the only viable case is when the 0 (mod 3) term equals 3 itself.Set N = 3 to obtain the triple (3, 5, 7), all prime. No other N works. Verification / Alternative check: Testing nearby values: N=1 gives (1, 3, 5) with 1 not prime; N=5 gives (5, 7, 9) where 9 is composite; N=11 gives (11, 13, 15) where 15 is composite. Only N=3 produces a fully prime triple. Why Other Options Are Wrong: 2 / 3 / None of these / 4: These overcount; modular reasoning confines the possibility to a single case. Common Pitfalls: Forgetting the role of 3 in modular cycles; counting 1 as a prime; assuming multiple solutions without checking divisibility by 3. Final Answer: 1

Question 17

Mersenne check — Find the smallest positive prime p for which 2^p - 1 is not prime (i.e., not a Mersenne prime).

Accepted Answer

Introduction / Context: Numbers of the form 2^p - 1 with p prime are called Mersenne numbers. Some are prime (Mersenne primes), but not all. The question asks for the smallest prime exponent p where 2^p - 1 fails to be prime. This evaluates your familiarity with early Mersenne values and quick compositeness checks. Given Data / Assumptions: We consider prime exponents p in increasing order. We check whether 2^p - 1 is prime. We stop at the first counterexample. Concept / Approach: List early cases: p = 2 → 3 (prime), p = 3 → 7 (prime), p = 5 → 31 (prime), p = 7 → 127 (prime). The next prime exponent is p = 11, giving 2^11 - 1 = 2047. Factor 2047 to test primality. Step-by-Step Solution: Compute 2^11 - 1 = 2047.Check small prime divisors: 2047 ÷ 23 = 89 → exact factorization 2047 = 23 * 89.Therefore 2^11 - 1 is composite. No smaller prime exponent p gave a composite earlier. Verification / Alternative check: Prior exponents: 2 → 3, 3 → 7, 5 → 31, 7 → 127, each prime. Hence p = 11 is indeed the smallest prime producing a composite 2^p - 1. Why Other Options Are Wrong: 5 / 17 / 29 / 7: For 5 and 7, 2^p - 1 is prime; 17 and 29 are larger than the smallest counterexample. Common Pitfalls: Assuming every 2^p - 1 with p prime is prime; not checking 2047; overlooking small factors such as 23 and 89. Final Answer: 11

Question 18

Two-part split — Split 24 into two parts so that 7 times the first plus 5 times the second equals 146. What is the first part?

Accepted Answer

Introduction / Context: This is a standard two-variable linear system framed as a partition problem. You divide a fixed total into two parts subject to a weighted-sum condition. Solving such problems efficiently is basic but essential practice for algebra and quantitative aptitude tests. Given Data / Assumptions: Let the first part be x and the second part be y. x + y = 24 (the split of 24). 7x + 5y = 146 (weighted sum condition). Concept / Approach: Use substitution from the sum equation into the weighted equation, then solve for x. This minimizes steps and avoids unnecessary simultaneous manipulations. After finding x, back-calculate y if needed (not required here). Step-by-Step Solution: From x + y = 24, express y = 24 - x.Substitute into 7x + 5y = 146 → 7x + 5(24 - x) = 146.Simplify: 7x + 120 - 5x = 146 → 2x = 26.Solve: x = 13. Thus, the first part is 13. Verification / Alternative check: Compute y = 24 - 13 = 11; then 7*13 + 5*11 = 91 + 55 = 146, confirming correctness. Why Other Options Are Wrong: 11 / 16 / 17 / 19: Substituting any of these into the equations fails to satisfy 7x + 5(24 - x) = 146. Common Pitfalls: Arithmetic slips when distributing 5; mixing x and y; forgetting to use the total sum equation first to reduce variables. Final Answer: 13

Question 19

Find two numbers — Their sum is 15 and the sum of their squares is 113. Identify the pair that satisfies both conditions.

Accepted Answer

Introduction / Context: When given a sum and the sum of squares, you can avoid guessing by using identities. Specifically, (a + b)^2 = a^2 + b^2 + 2ab lets you deduce the product ab and then solve a simple quadratic whose roots are the numbers themselves. This is a staple technique in algebra-focused aptitude questions. Given Data / Assumptions: a + b = 15. a^2 + b^2 = 113. We seek ordered numbers that satisfy both constraints. Concept / Approach: Use (a + b)^2 = a^2 + b^2 + 2ab to find ab. Once ab is known, the numbers are roots of t^2 - (a + b)t + ab = 0. Solve and select the pair that matches standard integer solutions. Step-by-Step Solution: Compute 2ab from identity: (a + b)^2 - (a^2 + b^2) = 2ab.(15)^2 - 113 = 225 - 113 = 112 → 2ab = 112 → ab = 56.Form quadratic: t^2 - 15t + 56 = 0.Factor: (t - 7)(t - 8) = 0 → t = 7 or 8.Hence, the numbers are 7 and 8. Verification / Alternative check: Sum 7 + 8 = 15 and 7^2 + 8^2 = 49 + 64 = 113, satisfying both conditions exactly. Why Other Options Are Wrong: 4,11 / 5,10 / 6,9 / 3,12: Only 7,8 yields both the correct sum and correct sum of squares. Common Pitfalls: Miscomputing (a + b)^2; forgetting to divide by 2 when solving for ab; arithmetic mistakes when factoring the quadratic. Final Answer: 7, 8

Question 20

Reciprocal condition — A positive number decreased by 4 equals 21 times its reciprocal. Find the number and verify your result.

Accepted Answer

Introduction / Context: This equation mixes a number with its reciprocal, a common aptitude theme. Converting the sentence directly into algebra and clearing the fraction yields a quadratic with small integer roots. Choosing the positive, context-appropriate root solves the problem neatly. Given Data / Assumptions: Let the positive number be n. Condition: n - 4 = 21 / n. n > 0 (explicitly positive). Concept / Approach: Multiply both sides by n to remove the fraction and create a quadratic. Then factor or use the quadratic formula. After solving, check sign and verify the original relation to exclude extraneous roots. Step-by-Step Solution: Start: n - 4 = 21 / n.Multiply by n: n^2 - 4n = 21.Rearrange: n^2 - 4n - 21 = 0.Factor: (n - 7)(n + 3) = 0 → n = 7 or n = -3.n must be positive → n = 7. Verification / Alternative check: Plug n = 7: 7 - 4 = 3 on the left; 21 / 7 = 3 on the right. Equality holds, confirming the solution. Why Other Options Are Wrong: 3 / 5 / 9 / 21: These do not satisfy n - 4 = 21 / n when substituted. Only 7 works. Common Pitfalls: Dropping the negative solution without justification; arithmetic errors when multiplying by n; mis-factoring the quadratic. Final Answer: 7

Number System Questions