Q1. If n is a positive integer, what is the largest integer that always divides n(n² + 5)?Correct Option 3 … Explanation: Rewrite the expression as n³ + 5n = (n³ − n) + 6n = (n − 1)n(n + 1) + 6n. The product of three consecutive integers is always divisible by 6, and 6n is clearly divisible by 6, so the whole expression is always divisible by 6. To check that nothing larger always works, put n = 1: 1 × (1 + 5) = 6. So no integer greater than 6 can divide the expression for every n. Option 1 (3) always divides, but it is not the largest. Options 2, 4 and 5 (12, 18, 30) all fail at n = 1. Hence, option 3.Q2. For how many integer values of n is (n + 17)/(n − 3) an integer?Correct Option 5 … Explanation: Write (n + 17)/(n − 3) = [(n − 3) + 20]/(n − 3) = 1 + 20/(n − 3). This is an integer exactly when (n − 3) divides 20. 20 has 6 positive divisors (1, 2, 4, 5, 10, 20), and each has a negative counterpart, giving 12 integer divisors in total. Each one gives a distinct value of n. Option 3 (6) counts only the positive divisors. Option 4 (20) mistakes the number itself for its count of divisors. Hence, option 5.Q3. Integers p, q and r satisfy the following conditions: pq is even, q + r is odd, and p + r is odd. Which of the following must be even?Correct Option 2 … Explanation: Since q + r and p + r are both odd, p and q each have the opposite parity to r, so p and q have the same parity. If both were odd, pq would be odd, which contradicts the given condition. So p and q are both even, and r is odd. Then r + 1 is even, so (r + 1)(p + q + 1) is even regardless of the second factor. Option 1: pq + r = even + odd = odd. Option 3: q + r is given as odd. Option 4: p + q + r = even + even + odd = odd. Option 5: p + r² = even + odd = odd. Hence, option 2.Q4. How many positive integers less than 200 have exactly three positive divisors?Correct Option 3 … Explanation: A number has exactly three positive divisors only if it is the square of a prime, p², whose divisors are 1, p and p². We need p² < 200, which means p ≤ 14. The primes in that range are 2, 3, 5, 7, 11 and 13, giving 4, 9, 25, 49, 121 and 169. That is 6 numbers. Option 2 (5) misses one of these squares, typically 169. Option 4 (7) wrongly includes 17² = 289, which exceeds 200. Hence, option 3.Q5. Which of the following is the greatest?Correct Option 4 … Explanation: Compare the logarithms (base 10) of each number. 5⁹⁰ gives 90 × 0.699 ≈ 62.9. 3¹³⁰ gives 130 × 0.477 ≈ 62.0. 2²¹⁰ gives 210 × 0.301 ≈ 63.2. 7⁷⁵ gives 75 × 0.845 ≈ 63.4. 11⁶⁰ gives 60 × 1.041 ≈ 62.5. The largest logarithm belongs to 7⁷⁵. You can confirm the closest contest by reducing exponents by their common factor 15: compare 7⁵ = 16807 with 2¹⁴ = 16384. Since 7⁵ is larger, 7⁷⁵ > 2²¹⁰. Option 3 (2²¹⁰) is the close runner-up. The others are clearly smaller. Hence, option 4.Q6. For how many positive integers n ≤ 100 is n/420 a terminating decimal?Correct Option 2 … Explanation: A fraction in lowest terms terminates only when its denominator has no prime factors other than 2 and 5. Since 420 = 2² × 3 × 5 × 7, the factors 3 and 7 must cancel. That means n must be a multiple of 21. The multiples of 21 up to 100 are 21, 42, 63 and 84, which is 4 values. Option 1 (5) wrongly includes 105, which exceeds 100. Option 3 (20) counts multiples of 5, which has nothing to do with cancelling 3 and 7. Hence, option 2.Q7. When a positive integer N is divided by a positive integer D, the remainder is 12. When 2N is divided by the same D, the remainder is 10. What is D?Correct Option 1 … Explanation: Let N = Dq + 12, so 2N = 2Dq + 24. The remainder of 2N on division by D therefore equals the remainder of 24 on division by D, which is given as 10. So 24 − 10 = 14 must be a multiple of D. Since the remainder 12 must be less than D, we need D > 12, which leaves only D = 14. Check: N = 26 gives remainder 12, and 2N = 52 = 3 × 14 + 10. Option 3 (7) is ruled out because a remainder of 12 is impossible with divisor 7. Option 5 is wrong because D is uniquely fixed. Hence, option 1.Q8. How many trailing zeros does the product 24 × 25 × 26 × … × 60 have?Correct Option 5 … Explanation: Trailing zeros come from pairs of 2 and 5, and factors of 2 are far more plentiful, so we only need to count factors of 5. The multiples of 5 from 24 to 60 are 25, 30, 35, 40, 45, 50, 55 and 60, which contribute 8 fives. Two of them, 25 and 50, are multiples of 25 and each contribute one extra 5, adding 2 more. The total is 10 trailing zeros. Option 1 (8) forgets the extra factors from 25 and 50. Hence, option 5.Q9. If p is a prime number greater than 3, what is the largest integer that must divide p² − 1?Correct Option 3 … Explanation: Write p² − 1 = (p − 1)(p + 1). Since p is odd, p − 1 and p + 1 are consecutive even numbers, so one of them is a multiple of 4, and their product is divisible by 8. Of the three consecutive integers p − 1, p and p + 1, one is a multiple of 3, and it cannot be the prime p. So the product is also divisible by 3, and therefore by 24. For p = 5, p² − 1 = 24, so nothing larger always works. Option 1 (12) always divides, but it is not the largest. Options 2, 4 and 5 fail at p = 5. Hence, option 3.Q10. How many positive divisors of 3600 are multiples of 15?Correct Option 3 … Explanation: First factorise: 3600 = 2⁴ × 3² × 5². A divisor that is a multiple of 15 must contain at least one 3 and at least one 5. The power of 2 can be 0 to 4 (5 choices), the power of 3 can be 1 or 2 (2 choices), and the power of 5 can be 1 or 2 (2 choices). That gives 5 × 2 × 2 = 20 divisors. Equivalently, these are the divisors of 3600/15 = 240 = 2⁴ × 3 × 5, which has 5 × 2 × 2 = 20 divisors. Hence, option 3.Q11. The repeating decimal 0.4121212… (the digits 12 repeat forever) equals p/q, where p/q is in lowest terms. What is p + q?Correct Option 1 … Explanation: Let x = 0.41212… Then 1000x = 412.1212… and 10x = 4.1212… Subtracting the second from the first gives 990x = 408, so x = 408/990. Dividing numerator and denominator by 6 gives 68/165, which is in lowest terms. So p + q = 68 + 165 = 233. Option 2 (1411) comes from reading the decimal as 412/999. Hence, option 1.Q12. Let S be the sum of 20 consecutive integers, the smallest of which is the integer k. Which of the following must be true?Correct Option 3 … Explanation: S = k + (k + 1) + … + (k + 19) = 20k + (0 + 1 + … + 19) = 20k + 190, which simplifies to 10(2k + 19). Since 2k + 19 is always odd, S is always an odd multiple of 10. Options 1, 2 and 4 all fail because 10 × (an odd number) contains only one factor of 2. Option 5 fails at k = 0, since 190 is not a multiple of 19 × anything that holds generally (S = 190 at k = 0, but S = 210 at k = 1, and 210 is not a multiple of 19). Hence, option 3.Q13. How many pairs of positive integers (m, n) satisfy m² − n² = 45?Correct Option 4 … Explanation: Factor the equation as (m − n)(m + n) = 45. Both factors must be positive, and m − n must be smaller than m + n. The possible factor pairs are (1, 45), (3, 15) and (5, 9). These give (m, n) = (23, 22), (9, 6) and (7, 2). The pair (45, 1) is excluded because m − n cannot exceed m + n, and a factor pair giving n = 0 would not be allowed since n must be positive. That leaves exactly 3 pairs. Option 3 (6) counts each factor pair twice. Hence, option 4.Q14. If a is a rational number and b is an irrational number, which of the following must be irrational?Correct Option 5 … Explanation: Suppose a + b = r, where r is rational. Then b = r − a, which would be a difference of two rationals and hence rational, contradicting the fact that b is irrational. So a + b is always irrational. The other options can each be rational. Option 1: b = ∛2 gives b³ = 2. Options 2 and 3: a = 0 gives a/b = 0 and ab = 0. Option 4: b = √2 gives b² = 2. Hence, option 5.Q15. How many six-digit numbers of the form 3A5B72, where A and B are single digits (0–9), are divisible by 33?Correct Option 1 … Explanation: A number is divisible by 33 exactly when it is divisible by both 3 and 11. For 11, the alternating digit sum must be a multiple of 11: (3 + 5 + 7) − (A + B + 2) = 13 − (A + B). So A + B = 2 or 13. For 3, the digit sum 17 + A + B must be divisible by 3. A + B = 2 gives 19, which fails. A + B = 13 gives 30, which works. The digit pairs with A + B = 13 are (4, 9), (5, 8), (6, 7), (7, 6), (8, 5) and (9, 4), which is 6 numbers. Option 2 (4) undercounts these pairs. Hence, option 1.Q16. How many integers between 100 and 500, inclusive, are divisible by 6 but not by 8?Correct Option 3 … Explanation: The multiples of 6 in the range run from 102 to 498, which gives (498 − 102)/6 + 1 = 67 numbers. A number divisible by both 6 and 8 is a multiple of their LCM, 24. The multiples of 24 run from 120 to 480, which gives (480 − 120)/24 + 1 = 16 numbers. So the answer is 67 − 16 = 51. Option 4 (67) forgets to remove the multiples of 8. Options 1 and 5 come from off-by-one errors in the counting. Hence, option 3.Q17. What is the smallest positive integer that leaves a remainder of 1 when divided by 4, by 5 and by 6, and is also divisible by 7?Correct Option 4 … Explanation: The LCM of 4, 5 and 6 is 60, so the number has the form 60k + 1. For divisibility by 7, note that 60 ≡ 4 (mod 7), so we need 4k + 1 ≡ 0 (mod 7). That means 4k ≡ 6 (mod 7), so k ≡ 5 (mod 7). The smallest such k is 5, giving 60 × 5 + 1 = 301 = 7 × 43. Options 1, 2, 3 and 5 all have the form 60k + 1, but none of them is divisible by 7. Hence, option 4.Q18. How many ordered pairs (a, b) of positive integers have greatest common divisor 6 and least common multiple 180?Correct Option 5 … Explanation: Write a = 6x and b = 6y, where x and y are coprime. Since ab = GCD × LCM = 6 × 180, we get 36xy = 1080, so xy = 30 = 2 × 3 × 5. Because x and y must be coprime, each of the three primes must go entirely to x or entirely to y. That gives 2³ = 8 ordered pairs. Option 3 (4) counts unordered pairs only. Hence, option 5.Q19. How many two-digit positive integers have the property that the sum of the number and the number formed by reversing its digits is a perfect square?Correct Option 2 … Explanation: For the number 10a + b, adding its reverse gives (10a + b) + (10b + a) = 11(a + b). For this to be a perfect square, a + b must be 11 times a perfect square. Since 1 ≤ a + b ≤ 18, the only possibility is a + b = 11. The numbers are 29, 38, 47, 56, 65, 74, 83 and 92, which is 8 in total. Option 1 (9) wrongly includes a pair such as (1, 10), where 10 is not a single digit. Hence, option 2.Q20. If ab < 0, bc > 0 and a²c < 0, which of the following must be true?Correct Option 4 … Explanation: Since a² > 0 and a²c < 0, we get c < 0. Since bc > 0, b must have the same sign as c, so b < 0. Since ab < 0, a must have the opposite sign to b, so a > 0. Then b + c < 0, and a(b + c) is a positive number times a negative number, which is negative. Option 1 depends on the actual values of b and c. Option 2 is false because a/c < 0. Option 3 is false because abc = (positive)(negative)(negative) > 0. Option 5 depends on the actual magnitudes. Hence, option 4.
