Question 1Number
TMUA Number — Practice Questions by Topic
These are SummitPapers original questions, not official past paper questions. Official TMUA questions sorted by topic are on TMUA past papers by topic.
- M2Number
- 63 questions30 on Paper 1 · 33 on Paper 2
- 8 free solutionsThe rest show the correct letter only
Covers: remainders and divisibility, factors, primes and factorials, counting digits or integers in a range, comparing the sizes of numbers.
What this topic tests
Number here means remainders and divisibility, factors, primes and factorials, counting digits or integers in a range, and comparing the sizes of numbers.
How it is assessed
This is Section 1, on both papers in this set. Each question has five options. A paper is 20 questions in 75 minutes, with no calculator.
This note is written for SummitPapers. The official list of what can be examined is the specification, together with the Notes on Mathematics and, for Paper 2, the Notes on Logic and Proof.
Key methods
Divisibility
a divides b when b = ka for some integer k. If a prime divides a product, it divides one of the factors. A remainder on division by m is one of 0, 1, …, m − 1.
Comparing sizes
For positive numbers, comparing a^b and c^d is often easier after a logarithm, or after writing both as powers of the same base. The order of positive numbers is preserved by squaring.
Common mistakes
A remainder equal to the divisor
The remainder when dividing by m is never m. It reduces to 0.
Counting both ends
The integers from a to b inclusive number b − a + 1. Missing the +1, or excluding one end, is an off-by-one.
Worked example
This question is also in the list below, with the solution folded. It is opened here so the method is on the page before the other questions.
What is the smallest positive integer n such that n! is divisible by 107?
Answer: C. 30
Worked solution. Since 107 = 27 × 57, n! must contain at least seven factors of 5 and at least seven factors of 2. Factors of 2 are always at least as common as factors of 5, so it is enough to count fives. The number of fives in n! is floor(n/5) + floor(n/25) (for n < 125). For n = 29 this is 5 + 1 = 6, which is too few, and for n = 30 it is 6 + 1 = 7. So the smallest such n is 30.
Why the other options look right. A counts both factors of 5 in 25 on top of counting 25 among the multiples of 5, so 25! seems to have 5 + 2 = 7 fives and it stops at 25; in fact 25! has only floor(25/5) + floor(25/25) = 5 + 1 = 6. B solves n/5 + n/25 = 7 without floor functions, getting n = 175/6 ≈ 29.2, and rounds to 29; but 29! has only 5 + 1 = 6 factors of 5. D takes one multiple of 5 for each factor of 5, 5 × 7 = 35, ignoring the extra five from 25. E assumes each factor 10 must come from a multiple of 10, so it needs 10, 20, …, 70 and gives 70.
Paper 1 questions
Paper 1 is Applications of Mathematical Knowledge. Calculators are not allowed.
Q1. What is the smallest positive integer n such that n! is divisible by 107?
Free · worked solution included
Answer and worked solution
Answer: C. 30
Worked solution. Since 107 = 27 × 57, n! must contain at least seven factors of 5 and at least seven factors of 2. Factors of 2 are always at least as common as factors of 5, so it is enough to count fives. The number of fives in n! is floor(n/5) + floor(n/25) (for n < 125). For n = 29 this is 5 + 1 = 6, which is too few, and for n = 30 it is 6 + 1 = 7. So the smallest such n is 30.
Why the other options look right. A counts both factors of 5 in 25 on top of counting 25 among the multiples of 5, so 25! seems to have 5 + 2 = 7 fives and it stops at 25; in fact 25! has only floor(25/5) + floor(25/25) = 5 + 1 = 6. B solves n/5 + n/25 = 7 without floor functions, getting n = 175/6 ≈ 29.2, and rounds to 29; but 29! has only 5 + 1 = 6 factors of 5. D takes one multiple of 5 for each factor of 5, 5 × 7 = 35, ignoring the extra five from 25. E assumes each factor 10 must come from a multiple of 10, so it needs 10, 20, …, 70 and gives 70.
Q2. What is the largest integer n ≤ 70 such that n3 + 6n is divisible by 24?
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Answer and worked solution
Answer: B. 60
Worked solution. Factor n3 + 6n = n(n2 + 6). Modulo 3 the expression is congruent to n, so 3 divides it exactly when 3 divides n. Modulo 8 it is divisible by 8 exactly when n is divisible by 4. Thus 24 divides it exactly when 12 divides n. The largest multiple of 12 that is at most 70 is 60. As a check, 60(3600 + 6) = 216360 and 216360/24 = 9015. The largest such integer is 60.
Why the other options look right. A requires n itself to be a multiple of 24, and the largest such integer at most 70 is 48. C takes the largest multiple of 6, which is 66; 66 is not a multiple of 4, so the expression is not divisible by 8. D takes the largest multiple of 4, which is 68, and the product is then not divisible by 3. E answers with the upper bound 70, which is not a multiple of 12.
Q3. Each cell of a 3 by 3 grid is filled with 0 or 1 so that every row and every column contains an even number of 1s (zero counts as even). How many such grids are there?
Free · worked solution included
Answer and worked solution
Answer: A. 16
Worked solution. Fill the top-left 2 by 2 block freely: 24 = 16 ways. The third entry of each of the first two rows is then forced, since it must make that row's number of 1s even, and the bottom entry of each of the first two columns is forced in the same way. The bottom-right entry must make the third row even and also the third column even. Both conditions ask for the same value: the first two entries of the third row and the first two entries of the third column each have the same parity as the number of 1s in the 2 by 2 block. So the corner is forced and consistent, and there are 16 grids.
Why the other options look right. B treats the bottom-right cell as a free choice as well, giving 25 = 32. C imposes only the row condition: each row is one of the four rows 000, 110, 101, 011, giving 43 = 64. D treats only the bottom-right cell as forced, giving 28 = 256. E ignores both parity conditions and counts all 29 = 512 grids.
Q4. What is the remainder when 65 is divided by 100?
Free · worked solution included
Answer and worked solution
Answer: D. 76
Worked solution. The last two digits can be tracked one power at a time. 62 = 36. Then 63 = 216, which leaves remainder 16. Then 64 leaves the same remainder as 6 × 16 = 96. Then 65 leaves the same remainder as 6 × 96 = 576, and 576 divided by 100 leaves 76. So the remainder is 76. Directly, 65 = 7776.
Why the other options look right. A notes that every power of 6 ends in the digit 6 and wrongly assumes that the remainder on division by 100 is therefore 6. B stops two multiplications early and reports the remainder of 63 = 216. C stops three multiplications early and reports the remainder of 62 = 36. E is the remainder of 64, which is 96, one multiplication short of 65.
Q5. How many pairs of integers (x, y) satisfy x2 − y2 = 105?
Free · worked solution included
Answer and worked solution
Answer: C. 16
Worked solution. Factor the left side as (x − y)(x + y) = 105. Every divisor d of 105 can be x − y, with x + y = 105/d. The positive divisors are 1, 3, 5, 7, 15, 21, 35 and 105, so there are 16 divisors once negative divisors are included. Each of them is odd, and so is 105/d, hence x = (d + 105/d)/2 and y = (105/d − d)/2 are integers. Distinct divisors give distinct pairs, so there are 16 solutions.
Why the other options look right. A counts only the four solutions with x > 0 and y > 0, coming from the factor pairs (1, 105), (3, 35), (5, 21) and (7, 15). B counts only the eight pairs of positive factors. D first lists the 8 solutions from the positive divisors d and then adds all 16 solutions from the signed divisors, counting the positive-divisor solutions twice: 8 + 16 = 24. E allows all four sign patterns on the eight positive factor pairs, but opposite signs multiply to −105 rather than 105.
Q6. What is the remainder when 28 + 38 is divided by 7?
Free · worked solution included
Answer and worked solution
Answer: E. 6
Worked solution. Modulo 7, the powers of 2 repeat every 3 steps because 23 = 8 ≡ 1. Thus 28 = 26+2 ≡ 22 ≡ 4. The powers of 3 satisfy 31 ≡ 3, 32 ≡ 2 and 33 ≡ −1, so 36 ≡ 1 and 38 ≡ 32 ≡ 2. Adding these remainders gives 4 + 2 = 6, which is already smaller than 7. The remainder is 6.
Why the other options look right. A multiplies the two powers instead of adding them: 28 × 38 = 68 ≡ (−1)8 = 1. B finds the remainder of 38 and forgets to add the remainder of 28. C finds the remainder of 28 and forgets to add the remainder of 38. D averages those two remainders, (4 + 2)/2 = 3, instead of adding them.
Q7. What is the remainder when 32024 is divided by 10?
Free · worked solution included
Answer and worked solution
Answer: A. 1
Worked solution. The last digits of the powers of 3 cycle every 4: 3, 9, 7, 1. Since 2024 = 4 × 506, the exponent is a multiple of the cycle length, so 32024 has the same last digit as 34, which is 1. The remainder on division by 10 is 1.
Why the other options look right. B is the last digit of 31, as if the remainder of the exponent on division by 4 were 1. C is the last digit of 33, as if that remainder were 3. D is the last digit of 32, as if that remainder were 2. E finds that 2024 leaves remainder 0 on division by 4 and reports that remainder as the last digit.
Q8. The numbers m and n are non-zero integers. For which condition is (6m − n × 152m) / (10m × 9m + n) always an integer?
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Answer
Answer: E. m > 0 and n < 0
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Q9. Which of the following numbers is the smallest?
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Answer
Answer: A. the square of (log base 4 of the square root of 64)
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Q10. When the integers from 1 to 300 are written out in decimal, how many times is the digit 2 written in total?
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Answer
Answer: E. 160
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Q11. For a positive integer n, let floor(log2 n) denote the greatest integer less than or equal to log2 n. What is floor(log2 1) + floor(log2 2) + ... + floor(log2 100)?
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Answer
Answer: C. 480
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Q12. Which of these numbers is the greatest?
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Answer
Answer: D. 5√2
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Q13. Write 1/(√11 − 3) so that the denominator is rational.
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Answer
Answer: B. (√11 + 3)/2
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Q14. Which of these numbers is the largest?
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Answer
Answer: A. 3√6
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Q15. How many integers k satisfy 4√3 < k√2 < 6√5?
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Answer
Answer: C. 5
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Q16. Which of these five numbers is greatest?
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Answer
Answer: D. 31/3
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Q17. Write 4/(√7 − √3) with a rational denominator. The simplified result is
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Answer
Answer: B. √7 + √3
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Q18. Which number below is the greatest?
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Answer
Answer: E. 55/6
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Q19. Write 1/(√19 − 3) with a rational denominator.
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Answer
Answer: D. (√19 + 3)/10
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Q20. Which one of these five numbers is the greatest?
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Answer
Answer: D. 2 √15
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Q21. In rationalised form, what is 1/(√23 − 3)?
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Answer
Answer: A. (√23 + 3)/14
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Q22. Which of these five numbers has the greatest value?
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Answer
Answer: B. 31/4
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Q23. Let m and n be integers. For which condition is 2m + n × 32m − n × 7n / (6m × 14n − m × 9m) an integer for every pair m, n satisfying the condition?
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Answer
Answer: C. m ≥ 0 and m + n ≤ 0
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Q24. Which of the following numbers is the least?
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Answer
Answer: E. (√2)5
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Q25. How many of the integers 1, 2, 3, …, 600 are divisible by exactly one of the numbers 4, 6 and 10?
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Answer
Answer: A. 140
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Q26. What is the sum of the reciprocals of all positive factors of 180?
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Answer
Answer: E. 91/30
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Q27. The digital root of a positive integer is obtained by summing its digits repeatedly until one digit remains. What is the digital root of 420?
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Answer
Answer: D. 7
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Q28. In how many ways can three different numbers be chosen from 1, 2, 3, ..., 20 so that their sum is a multiple of 3? The order of the choice does not matter.
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Answer
Answer: E. 384
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Q29. Which of these numbers is closest to √160001 − √160000?
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Answer
Answer: B. 1/800
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Q30. How many pairs of positive integers (a, b) satisfy a + 2b ≤ 18?
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Answer
Answer: E. 72
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Paper 2 questions
Paper 2 is Mathematical Reasoning. It can test this same topic. Argument, proof, and identifying errors are the topics that appear on Paper 2 only.
Q1. The positive real numbers a × 101, b × 102 and c × 102 are each in standard form, and (a × 101) + (b × 102) = c × 102. Which of the following must be true? (I) a > 5; (II) c > b; (III) a < c; (IV) b > 9.
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Answer and worked solution
Answer: B. II only
Worked solution. Standard form means 1 ≤ a, b, c < 10. The equation is a + 10b = 10c, so a = 10(c − b). Then 1 ≤ 10(c − b) < 10, so 0.1 ≤ c − b < 1. In particular c > b, which is II. The values a = 2, b = 1 and c = 1.2 satisfy the equation and standard form, and they make I, III and IV false. Only II must be true, which is option B.
Why the other options look right. A keeps only claim I, but a = 2, b = 1 and c = 1.2 is a counterexample to I. C includes I as well as II, and that same triple makes I false. D includes IV, but b can be 1, so IV is not forced. E requires all four claims, while III fails for a = 2 and c = 1.2 and IV fails for b = 1. Only II must be true.
Q2. A positive integer is a squaresum if and only if it is the sum of the squares of two integers. A prime is awkward if and only if it leaves remainder 3 when divided by 4. A theorem of Fermat says that a positive integer is a squaresum if and only if every awkward prime factor occurs to an even power. Which one of the following numbers is a squaresum?
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Answer
Answer: C. 1274
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Q3. For which one of the following statements can the fact that 211 − 1 = 2047 = 23 × 89 be used to produce a counterexample?
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Answer
Answer: B. If n is prime, then 2n − 1 is prime.
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Q4. In base 3, the recurring expansion 0.020202… means 0 × 3⁻¹ + 2 × 3⁻² + 0 × 3⁻³ + 2 × 3⁻⁴ + … . What is its value?
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Answer
Answer: D. 1/4
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Q5. The area of a rectangle is measured as 2400 cm2, correct to 2 significant figures. The width is measured as 50 cm, correct to the nearest centimetre. What is the upper bound for the possible length of the rectangle?
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Answer
Answer: E. 2450/49.5 cm
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Q6. For which positive integers n is 4ⁿ + 1 a multiple of 17?
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Answer
Answer: C. Exactly the values of n that leave remainder 2 when divided by 4
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Q7. The positive integers a, b and c satisfy a ≤ b ≤ c and 1/a + 1/b + 1/c = 3/4. Which of the following must be true? I. a ≤ 3. II. c ≥ 6. III. b ≥ 4.
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Answer
Answer: E. none of them
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Q8. How many pairs of integers (x, y) satisfy (x² − 1)(y − 3) ≤ 2(y − 3), with −2 ≤ x ≤ 3 and 1 ≤ y ≤ 5?
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Answer
Answer: D. 18
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Q9. Let n be an integer greater than 1, and let P(n) be the greatest prime factor of n. For example, P(33) = 11. Which statement is true?
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Answer
Answer: E. If P(n) = 3, then n² ends in 1, 4, 6 or 9.
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Q10. For a positive integer n, let G(n) be the product of the digits of n. For example, G(305) = 3 × 0 × 5 = 0 and G(47) = 4 × 7 = 28. Find G(1) + G(2) + ... + G(99).
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Answer
Answer: E. 2070
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Q11. The positive integers a, b and c satisfy a ≤ b ≤ c and abc = 36. Which of the following must be true? I: a + b + c ≤ 16. II: b ≤ 4. III: c ≥ 4.
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Answer
Answer: C. III only
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Q12. A positive integer n leaves remainder 5 when divided by 13. What remainder does 4n leave when divided by 13?
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Answer
Answer: D. 7
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Q13. The expression n² + n + 17 is evaluated at each positive integer below. For which value of n is the result composite?
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Answer
Answer: E. 16
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Q14. Positive integers a, b and c satisfy a ≤ b ≤ c and abc = 60. Which of the following must be true?
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Answer
Answer: E. b ≤ 6
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Q15. The positive integer n leaves remainder 5 when it is divided by 11. What is the remainder when 4n is divided by 11?
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Answer
Answer: D. 9
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Q16. Positive integers a, b and c satisfy a ≤ b ≤ c and abc = 36. Which of these must be true? I. a ≤ 3. II. b ≤ 4. III. c ≥ 6.
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Answer
Answer: A. I only
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Q17. A positive integer leaves remainder 5 when it is divided by 8. What remainder does five times that integer leave when it is divided by 8?
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Answer
Answer: A. 1
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Q18. Positive integers x, y and z satisfy x ≤ y ≤ z and x × y × z = 60. Which of the following statements must be true? I. x ≤ 3 II. y ≤ 5 III. z ≥ 5
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Answer
Answer: D. I and III only
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Q19. Which one of the following claims about integers m and n is not always true?
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Answer
Answer: C. If 6 divides mn, then 6 divides m or 6 divides n.
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Q20. When the positive integer n is divided by 11, the remainder is 5. What remainder is left when the product 7n is divided by 11?
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Answer
Answer: E. 2
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Q21. A positive integer n leaves remainder 4 when divided by 9. What remainder does 5n leave when divided by 9?
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Answer
Answer: B. 2
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Q22. How many positive integers n satisfy 2n < n³?
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Answer
Answer: B. 8
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Q23. A positive integer n leaves remainder 5 on division by 8. What remainder does 7n leave on division by 8?
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Answer
Answer: C. 3
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Q24. The positive integer n leaves remainder 4 when divided by 9. What remainder does 4n leave when divided by 9?
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Answer
Answer: C. 7
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Q25. For each positive integer n, let f(n) = 0 when n is a multiple of 4 and f(n) = 1 otherwise, and let g(n) = 0 when n is a multiple of 6 and g(n) = 1 otherwise. Define h(n) = (1 − f(n))(1 − g(n)). What is h(3) + h(6) + h(9) + … + h(180)?
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Answer
Answer: C. 15
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Q26. What is the value of √(28 + 10√3) − 2√(4 − 2√3)?
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Answer
Answer: D. 7 − √3
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Q27. Let n be an integer greater than 1. Let P(n) be the greatest prime factor of n, and let d(n) be the final digit of n. Which statement is true?
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Answer
Answer: E. If P(n) = 5, then d(n) is 0 or 5.
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Q28. Let s(n) be the number of ones in the binary expansion of the positive integer n. For example, s(13) = 3 because 13 = 1101 in binary. What is s(1) + s(2) + … + s(63)?
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Answer
Answer: D. 192
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Q29. For a positive integer n, let F(n) be the sum of the squares of the decimal digits of n. For example, F(418) = 16 + 1 + 64 = 81. For how many integers n with 1 ≤ n ≤ 99 is F(n) a perfect square?
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Answer
Answer: E. 22
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Q30. Which statement about 65 is true?
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Answer
Answer: B. 65 = 8² + 1² and also 65 = 7² + 4².
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Q31. A repunit is a positive integer whose digits are all 1. A repunit with n digits is divisible by 9 if and only if which condition holds?
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Answer
Answer: B. n is a multiple of 9
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Q32. A word is a string of length 4 using only the letters P, Q, R and S. How many such words contain exactly two P's and at least one Q?
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Answer
Answer: A. 30
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Q33. All 120 permutations of the letters of MATHS are listed in alphabetical order. In which position is MATHS?
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Answer
Answer: C. 53rd
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