Q1. What is the least positive integer k such that 840k is a perfect square?Correct Option B … Explanation: Prime factorising, 840 = 2³ × 3 × 5 × 7. For a perfect square every prime must appear to an even power, so the deficient primes are 2 (one more needed), 3, 5 and 7. The smallest multiplier is therefore 2 × 3 × 5 × 7 = 210, giving 840 × 210 = 176400 = 420². Option A (70 = 2 × 5 × 7) leaves 3 with an odd power. Option C (21) ignores the extra 2 and the 5. Option D (35) misses both 2 and 3. Option E (140) contains an unnecessary extra 2 while still omitting 3. Hence, option B.Q2. The HCF of two positive integers is 34, and their LCM is 11730. If one integer is 510, what is the other integer?Correct Option A … Explanation: For any two positive integers, HCF × LCM = product of the numbers. So the other integer = (34 × 11730) ÷ 510 = 398820 ÷ 510 = 782. The check holds: 510 = 2 × 3 × 5 × 17 and 782 = 2 × 17 × 23, giving HCF 34 and LCM 11730. Options B (672), C (534) and E (918) are not even multiples of 34, so they cannot share an HCF of 34 with 510. Option D (884 = 4 × 13 × 17) is a multiple of 34 but yields an HCF of 34 with an LCM of 13260, not 11730. Hence, option A.Q3. The product of six consecutive positive integers is always divisible by which of the following? I. 12 II. 15 III. 20Correct Option E … Explanation: The product of any n consecutive integers is divisible by n!, so a product of six consecutive integers is always divisible by 6! = 720. Since 720 = 12 × 60 = 15 × 48 = 20 × 36, all three of 12, 15 and 20 divide it without exception. Structurally, six consecutive numbers must contain at least three even terms (supplying 2² and more), at least two multiples of 3, and at least one multiple of 5, which covers 12 = 2²×3, 15 = 3×5 and 20 = 2²×5. Options A, B, C and D each reject at least one divisor that the 6! argument guarantees. Hence, option E.Q4. What is the units digit of 7²⁰²⁴ × 3²⁰²⁵ × 9²⁰²⁶?Correct Option B … Explanation: Units digits repeat in cycles. For 7 the cycle is 7, 9, 3, 1 with period 4; since 2024 is a multiple of 4, 7²⁰²⁴ ends in 1. For 3 the cycle is 3, 9, 7, 1; since 2025 leaves remainder 1 on division by 4, 3²⁰²⁵ ends in 3. For 9 the cycle is 9, 1 with period 2; since 2026 is even, 9²⁰²⁶ ends in 1. The units digit of the product is that of 1 × 3 × 1 = 3. Option A comes from wrongly taking 3²⁰²⁵ as ending in 1, option C from misreading the 7-cycle, option D from taking an odd exponent for 9, and option E cannot arise since no factor ends in 5 or 0. Hence, option B.Q5. How many trailing zeros does the product 1 × 2 × 3 × ⋯ × 125 have?Correct Option D … Explanation: Trailing zeros are governed by the number of 5s in the prime factorisation of 125!, since 2s are far more plentiful. Applying Legendre's formula, ⌊125/5⌋ + ⌊125/25⌋ + ⌊125/125⌋ = 25 + 5 + 1 = 31. Option A stops at the first term and ignores the extra 5s contributed by 25, 50, 75, 100 and 125. Option C stops after two terms and forgets that 125 = 5³ contributes a third five. Options B and E do not correspond to any stage of the calculation. Hence, option D.Q6. Two positive integers have a sum of 861 and an HCF of 123. How many unordered pairs of such integers are possible?Correct Option C … Explanation: Write the numbers as 123m and 123n with m and n coprime. Then 123(m + n) = 861, so m + n = 7. The coprime unordered pairs summing to 7 are (1, 6), (2, 5) and (3, 4) — all automatically coprime because consecutive-sum pairs with a prime total share no factor with 7. These give the pairs (123, 738), (246, 615) and (369, 492), so exactly 3 pairs exist. Option B undercounts by dropping one valid pair, while options A, D and E arise from counting ordered pairs or from including non-coprime splits such as (0, 7). Hence, option C.Q7. What is the greatest integer n such that 14ⁿ divides 321!?Correct Option A … Explanation: Since 14 = 2 × 7, the limiting prime is the scarcer one, 7. The exponent of 7 in 321! is ⌊321/7⌋ + ⌊321/49⌋ + ⌊321/343⌋ = 45 + 6 + 0 = 51, whereas the exponent of 2 is 160 + 80 + 40 + 20 + 10 + 5 + 2 + 1 = 318. As 318 far exceeds 51, the answer is 51. Option C (45) stops at the first term and misses the six extra 7s from multiples of 49. Option D (160) is the count of 2s alone at the first step, which is not the binding constraint. Options B and E do not follow from the computation. Hence, option A.Q8. Positive integers a and b have HCF 12 and LCM 420. How many values can (a + b) take?Correct Option D … Explanation: Put a = 12m and b = 12n with m and n coprime. Then LCM = 12mn = 420, so mn = 35 = 5 × 7. The coprime factor pairs of 35 are (1, 35) and (5, 7), giving (a, b) = (12, 420) or (60, 84). The corresponding sums are 432 and 144, so a + b takes exactly 2 distinct values. Option A wrongly assumes the pair is unique. Option B counts ordered pairs (12, 420), (420, 12), (60, 84), (84, 60) as four different sums even though they duplicate. Options C and E overcount by admitting non-coprime splits such as m = 35, n = 1 duplicated or mn factorisations that violate the HCF condition. Hence, option D.Q9. If x is a positive integer such that x/4 and x/6 are both integers, and x/10 is not an integer, which of the following cannot be the remainder when x is divided by 30?Correct Option E … Explanation: If 4 and 6 both divide x, then LCM(4, 6) = 12 divides x, so x is a multiple of 12 and every remainder modulo 30 must itself be a multiple of 6 — the cycle of multiples of 12 modulo 30 is 12, 24, 6, 18, 0. The remainder 0 is ruled out because x ≡ 0 (mod 30) together with 12 | x forces 60 | x, which would make x/10 an integer. That leaves 6, 12, 18 and 24 as attainable, all of which appear as options A, C, D and B respectively (for example x = 36, 12, 48 and 24). Since 16 is not even a multiple of 6, no multiple of 12 can leave it as a remainder on division by 30. Hence, option E.Q10. A number, when divided by 8, leaves a remainder of 5, and when divided by 12, leaves a remainder of 9. What is the remainder when the number is divided by 24?Correct Option D … Explanation: Both conditions can be rewritten as a shortfall of 3: the number is 3 less than a multiple of 8 and 3 less than a multiple of 12. So the number is 3 less than a common multiple of 8 and 12, i.e. 3 less than a multiple of LCM(8, 12) = 24, giving a remainder of 24 − 3 = 21. Checking, 21 leaves 5 on division by 8 and 9 on division by 12. Options A and B simply repeat the given remainders without combining the moduli. Option C satisfies neither condition (15 leaves 7 with 8). Option E fails the second condition (17 leaves 5 with 12). Hence, option D.Q11. What is the remainder when 7²⁰²⁶ is divided by 25?Correct Option C … Explanation: Note that 7² = 49 ≡ −1 (mod 25). Writing the exponent as 7²⁰²⁶ = (7²)¹⁰¹³, the expression becomes (−1)¹⁰¹³ ≡ −1 (mod 25), and −1 is equivalent to 24 modulo 25. Option D would require an even power of 7², i.e. an exponent that is a multiple of 4, whereas 2026 is only a multiple of 2. Option B is impossible since any remainder modulo 25 must be less than 25. Options A and E do not arise from the cycle 7, 24, 18, 1 of powers of 7 modulo 25 at the position 2026 mod 4 = 2. Hence, option C.Q12. If a and b are positive integers such that aᵇ = 4096, and b is even, what is the sum of all possible values of a?Correct Option E … Explanation: Since 4096 = 2¹², any solution has a = 2ᵈ with d × b = 12. Restricting b to even divisors of 12 gives b = 2, 4, 6, 12, and correspondingly a = 64, 8, 4, 2. The sum of these values is 64 + 8 + 4 + 2 = 78. Option A adds only 2 + 4 + 8 and drops the largest case b = 2. Option B keeps just 2 and 64 while dropping the middle cases. Option D omits 64 and adds spuriously. Option C matches no valid subset; note that a = 4096 with b = 1 is excluded because b must be even. Hence, option E.Q13. What is the remainder when 2¹⁰⁰ + 3¹⁰⁰ is divided by 5?Correct Option A … Explanation: Powers of 2 modulo 5 cycle as 2, 4, 3, 1 with period 4, and 100 is a multiple of 4, so 2¹⁰⁰ ≡ 1. Powers of 3 modulo 5 cycle as 3, 4, 2, 1, also with period 4, so 3¹⁰⁰ ≡ 1 as well. The sum is 1 + 1 = 2 modulo 5. Option C results from evaluating only one of the two terms. Option D would require the two residues to be additive inverses, which they are not. Options B and E come from misplacing the position in the cyclicity, for instance treating 100 as leaving remainder 2 or 3 on division by 4. Hence, option A.Q14. What is the smallest positive integer that leaves a remainder of 3 when divided by 5, a remainder of 4 when divided by 7, and a remainder of 5 when divided by 9?Correct Option E … Explanation: Combining the first two congruences, numbers that are 3 more than a multiple of 5 and 4 more than a multiple of 7 are of the form 18 + 35k. Imposing the third condition, 18 + 35k ≡ 5 (mod 9) reduces to 8k ≡ 5 (mod 9), i.e. k ≡ 4 (mod 9), and the smallest such k = 4 gives 18 + 140 = 158. Verification: 158 = 5(31) + 3 = 7(22) + 4 = 9(17) + 5. Option A fails the third condition (193 leaves 4 with 9), option B fails the first (238 leaves 3 with 5 but 0 with 7), option C fails the second, and option D fails the first (203 leaves 3 with 5 but 0 with 7). Hence, option E.Q15. Two bells ring simultaneously at 8:00 am. One bell rings every 45 seconds, and the other every 60 seconds. How many times will they ring together between 8 and 9 am?Correct Option C … Explanation: The bells coincide at intervals of LCM(45, 60) = 180 seconds, that is, once every 3 minutes. Starting from the 8:00 am coincidence, the joint rings occur at 8:03, 8:06, …, 8:57 and 9:00. Strictly between 8 and 9 am — excluding both the 8:00 am ring given in the stem and the 9:00 am ring at the closing boundary — there are 57/3 = 19 coincidences. Option E counts every coincidence from 8:00 to 9:00 inclusive, and option D counts 9:00 but not 8:00; both read "between" as inclusive of an endpoint. Options A and B come from using an interval of 4 minutes or from dividing 60 by 45-second and 60-second gaps separately. Hence, option C.Q16. If a positive integer n is divisible by 9, 12, and 15, and has exactly 30 positive divisors, what is the smallest possible value of n?Correct Option D … Explanation: Divisibility by 9, 12 and 15 forces n to be a multiple of LCM(9, 12, 15) = 180 = 2² × 3² × 5. Testing multiples of 180 in order: 180 has (2+1)(2+1)(1+1) = 18 divisors; 360 = 2³ × 3² × 5 has 4 × 3 × 2 = 24; 540 = 2² × 3³ × 5 has 3 × 4 × 2 = 24; 720 = 2⁴ × 3² × 5 has 5 × 3 × 2 = 30, which meets the requirement. Option E (900 = 2² × 3² × 5²) has 27 divisors and is larger besides. Options A, B and C all fall short of 30 divisors as computed. Hence, option D.Q17. What is the remainder when 10! is divided by 13?Correct Option B … Explanation: By Wilson's theorem, since 13 is prime, 12! ≡ −1 (mod 13). Writing 12! = 12 × 11 × 10! and reducing, 12 ≡ −1 and 11 ≡ −2, so (−1)(−2) × 10! ≡ −1, i.e. 2 × 10! ≡ 12 (mod 13), which gives 10! ≡ 6. Option C is the value of −1 modulo 13, which applies to 12! and not 10!. Option E would follow from wrongly assuming Wilson's theorem yields a remainder of 1. Options A and D come from arithmetic slips while halving 12 modulo 13. Hence, option B.Q18. An event planner is assembling gift bags using 120 scented candles, 168 luxury soaps, and 264 bath bombs. Each bag must contain the same combination of items, and no items can be left over. What can be the minimum number of items in a bag?Correct Option C … Explanation: Minimising the items per bag means maximising the number of bags, which is the HCF of 120, 168 and 264. Since 120 = 2³ × 3 × 5, 168 = 2³ × 3 × 7 and 264 = 2³ × 3 × 11, the HCF is 24 bags. Each bag then holds 120/24 = 5 candles, 168/24 = 7 soaps and 264/24 = 11 bath bombs, totalling 5 + 7 + 11 = 23 items. Option D mistakes the number of bags (24) for the number of items per bag. Option B corresponds to using 12 bags as the divisor rather than the full HCF, which does not minimise the bag size. Options A and E do not correspond to any valid equal division of all three quantities. Hence, option C.Q19. How many positive integers n less than 100 satisfy the condition that n(n+1) is divisible by 6?Correct Option D … Explanation: The product of two consecutive integers is always even, so divisibility by 2 is automatic and only divisibility by 3 needs checking. Either n or n + 1 is a multiple of 3 precisely when n ≡ 0 or 2 (mod 3); the condition fails only when n ≡ 1 (mod 3). Among n = 1 to 99 there are 33 integers of the form 3k + 1 (namely 1, 4, …, 97), so the count of successes is 99 − 33 = 66. Option C counts exactly the failing cases instead of the successful ones. Option E assumes every n works, ignoring the multiple-of-3 requirement. Options A and B come from applying a 3-in-4 or 1-in-2 density that does not match the modulo-3 pattern. Hence, option D.Q20. A school has between 200 and 250 students. When arranged in rows of 7, 2 students are left over. When arranged in rows of 11, 3 students are left over. How many students will be left if they are arranged in rows of 15?Correct Option A … Explanation: Numbers leaving 3 on division by 11 take the form 11k + 3; requiring 11k + 3 ≡ 2 (mod 7) gives 4k ≡ 6 (mod 7), so k ≡ 5 (mod 7) and the solutions are of the form 58 + 77t. The values are 58, 135, 212, 289, and only 212 lies between 200 and 250. Dividing, 212 = 15 × 14 + 2, so 2 students remain. Options B, C, D and E would require the strength to be 213, 214, 216 or 218, none of which satisfies both original conditions (for instance 213 leaves 3 with 7, and 218 leaves 9 with 11). Hence, option A.
