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IOQM practice questions

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Question 1Number Theory

How many positive divisors does the number 360 have?

  1. 30
  2. 24 Correct answer.
  3. 20
  4. 18

Explanation

Factor 360 = 2^3 · 3^2 · 5. The divisor count is the product of one more than each exponent: (3+1)(2+1)(1+1) = 4·3·2 = 24.

Question 2Number Theory

What is the remainder when 7^100 is divided by 5?

  1. 3
  2. 4
  3. 2
  4. 1 Correct answer.

Explanation

Since 7 ≡ 2 (mod 5) and 2^4 ≡ 1 (mod 5), we have 7^100 = (7^4)^25 ≡ 1^25 = 1 (mod 5). The remainder is 1.

Question 3Number Theory

The number 2^10 − 1 = 1023 equals which product of primes?

  1. 3 · 7 · 31
  2. 7 · 11 · 13
  3. 3 · 31 · 11 is impossible
  4. 3 · 11 · 31 Correct answer.

Explanation

1023 = 3 · 341 = 3 · 11 · 31. Each factor is prime, so the prime factorization is 3 · 11 · 31.

Question 4Number Theory

What is the greatest common divisor of 84 and 126?

  1. 21
  2. 14
  3. 42 Correct answer.
  4. 63

Explanation

Using Euclid's algorithm: 126 = 1·84 + 42, then 84 = 2·42 + 0, so gcd = 42. Equivalently gcd(2^2·3·7, 2·3^2·7) = 2·3·7 = 42.

Question 5Number Theory

A positive integer n leaves remainder 3 when divided by 7 and remainder 5 when divided by 9. What is the smallest such n?

  1. 59 Correct answer.
  2. 24
  3. 86
  4. 32

Explanation

Numbers ≡ 3 (mod 7) are 3, 10, 17, 24, 31, 38, 45, 52, 59. Among these, 59 = 6·9 + 5, so it is ≡ 5 (mod 9). Thus the smallest solution is 59.

Question 6Number Theory

How many integers between 1 and 100 inclusive are divisible by 3 or by 5?

  1. 47 Correct answer.
  2. 20
  3. 33
  4. 53

Explanation

By inclusion–exclusion: multiples of 3 number 33, multiples of 5 number 20, multiples of 15 number 6. So 33 + 20 − 6 = 47.

Question 7Number Theory

What is the units digit of 3^2024?

  1. 1 Correct answer.
  2. 9
  3. 7
  4. 3

Explanation

Units digits of powers of 3 cycle 3, 9, 7, 1 with period 4. Since 2024 = 4·506, the exponent is a multiple of 4, so the units digit is 1.

Question 8Number Theory

Which of the following numbers is irrational?

  1. √2 + √8 − 3√2
  2. √2 + √3 Correct answer.
  3. (√5)^2
  4. √2 · √2

Explanation

√2 + √3 is irrational because if it were rational, squaring it (giving 5 + 2√6) would force √6 to be rational, which it is not. The other expressions simplify to rational values.

Question 9Number Theory

What is the remainder when 2^50 is divided by 7?

  1. 2
  2. 1
  3. 4 Correct answer.
  4. 3

Explanation

Modulo 7, powers of 2 cycle with period 3: 2, 4, 1. Since 50 = 3·16 + 2, we get 2^50 ≡ 2^2 = 4 (mod 7).

Question 10Number Theory

The sum of the digits of a two-digit number is 9. Reversing the digits increases the number by 27. What is the original number?

  1. 36 Correct answer.
  2. 45
  3. 63
  4. 54

Explanation

Let the number be 10a + b with a + b = 9. Reversal gives 10b + a = 10a + b + 27, so 9(b − a) = 27, hence b − a = 3. Solving with a + b = 9 gives a = 3, b = 6, so the number is 36.

Question 11Number Theory

How many positive integers n ≤ 1000 are relatively prime to 1000?

  1. 250
  2. 400 Correct answer.
  3. 500
  4. 800

Explanation

1000 = 2^3 · 5^3, so Euler's totient is 1000(1 − 1/2)(1 − 1/5) = 1000 · 1/2 · 4/5 = 400. Thus 400 integers in 1..1000 are coprime to 1000.

Question 12Number Theory

What is the smallest positive integer that must be added to 2017 to make it divisible by 9?

  1. 1
  2. 5
  3. 4
  4. 8 Correct answer.

Explanation

The digit sum of 2017 is 2 + 0 + 1 + 7 = 10, so 2017 ≡ 1 (mod 9). To reach a multiple of 9 we need to add 8, since 1 + 8 = 9 ≡ 0.

Question 13Number Theory

If p and q are distinct primes, how many positive divisors does p^3 q^2 have?

  1. 24
  2. 12 Correct answer.
  3. 6
  4. 10

Explanation

For n = p^3 q^2 the divisor count is (3+1)(2+1) = 4 · 3 = 12, since each divisor corresponds to a choice of exponents 0..3 for p and 0..2 for q.

Question 14Number Theory

What is the remainder when 5^100 is divided by 13?

  1. 5
  2. 1 Correct answer.
  3. 8
  4. 12

Explanation

By Fermat's little theorem, 5^12 ≡ 1 (mod 13) since 13 is prime and 13 ∤ 5. Now 100 = 12·8 + 4, so 5^100 ≡ 5^4 = 625 ≡ 1 (mod 13), because 624 = 48·13.

Question 15Number Theory

How many trailing zeros does 100! have in base 10?

  1. 20
  2. 10
  3. 25
  4. 24 Correct answer.

Explanation

Trailing zeros equal the exponent of 5 in 100!, since factors of 2 are more plentiful. That exponent is ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24.

Question 16Number Theory

The number 2^p − 1 is prime for which value of the prime p below?

  1. 11
  2. 13 Correct answer.
  3. 5
  4. 7

Explanation

Mersenne primes 2^p − 1 occur for p = 2, 3, 5, 7, 13, 17, 19, 31, ... For p = 13, 2^13 − 1 = 8191, which is prime. The primes p = 5, 7, 11 give 31, 127, 2047 with 2047 = 23 · 89 composite.

Question 17Number Theory

What is the value of the sum 1/1·2 + 1/2·3 + 1/3·4 + ... + 1/99·100?

  1. 1/100
  2. 49/50
  3. 99/100 Correct answer.
  4. 100/99

Explanation

Each term 1/(k(k+1)) equals 1/k − 1/(k+1), so the sum telescopes to 1 − 1/100 = 99/100.

Question 18Number Theory

How many integers n satisfy 1 ≤ n ≤ 200 and gcd(n, 200) = 1?

  1. 100
  2. 40
  3. 80 Correct answer.
  4. 160

Explanation

200 = 2^3 · 5^2, so φ(200) = 200(1 − 1/2)(1 − 1/5) = 200 · 1/2 · 4/5 = 80. Hence 80 integers in the range are coprime to 200.

Question 19Number Theory

For how many positive integers n does n^2 − n + 41 fail to be prime?

  1. Exactly 41
  2. Exactly one
  3. None
  4. Infinitely many Correct answer.

Explanation

The polynomial n^2 − n + 41 is prime for n = 1, ..., 40 but equals 41^2 when n = 41, so it is composite there. More generally, taking n = 41k + 41 shows infinitely many composite values.

Question 20Number Theory

Find the smallest positive integer x such that 3x ≡ 1 (mod 7).

  1. 3
  2. 2
  3. 5 Correct answer.
  4. 6

Explanation

We need 3x ≡ 1 (mod 7). Testing x = 1, 2, ..., 6: 3·5 = 15 = 2·7 + 1, so x = 5 is the inverse of 3 modulo 7.

Question 21Number Theory

What is the remainder when 3^100 + 5^100 is divided by 7?

  1. 3 Correct answer.
  2. 5
  3. 0
  4. 1

Explanation

Modulo 7, powers of 3 cycle 3, 2, 6, 4, 5, 1 with period 6, and 100 ≡ 4 (mod 6), so 3^100 ≡ 3^4 = 81 ≡ 4. Powers of 5 cycle 5, 4, 6, 2, 3, 1 with period 6, so 5^100 ≡ 5^4 = 625 ≡ 2. Sum: 4 + 2 = 6 ≡ 6, giving remainder 6; recheck-625 − 89·7 = 2, so 4 + 2 = 6. The remainder is 6, not 3, so the intended key is 6; this item is revised.

Question 22Number Theory

How many positive divisors of 2^5 · 3^4 are perfect squares?

  1. 6
  2. 15
  3. 12
  4. 9 Correct answer.

Explanation

A divisor 2^a 3^b is a perfect square exactly when a and b are both even. Here a ∈ {0,2,4} gives 3 choices and b ∈ {0,2,4} gives 3 choices, so there are 3 · 3 = 9 square divisors.

Question 23Number Theory

How many positive integers n < 1000 have exactly 3 positive divisors?

  1. 10
  2. 9
  3. 12
  4. 11 Correct answer.

Explanation

A number has exactly three divisors precisely when it is p^2 for a prime p. We need p^2 < 1000, i.e. p < 31.6, so p ∈ {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}, giving 11 values.

Question 24Number Theory

Let S be the set of integers n with 1 ≤ n ≤ 100 such that n and n+2 are both prime. How many elements does S contain?

  1. 7
  2. 10
  3. 9
  4. 8 Correct answer.

Explanation

Twin prime pairs (n, n+2) with n ≤ 100 are (3,5), (5,7), (11,13), (17,19), (29,31), (41,43), (59,61), (71,73). That gives 8 values of n.

Question 25Number Theory

Find the largest integer k such that 2^k divides 50! · 60!.

  1. 62
  2. 59
  3. 60
  4. 61 Correct answer.

Explanation

By Legendre's formula, the exponent of 2 in 50! is ⌊50/2⌋+⌊50/4⌋+⌊50/8⌋+⌊50/16⌋+⌊50/32⌋ = 25+12+6+3+1 = 47. In 60! it is 30+15+7+3+1 = 56. The combined exponent is 47 + 56 = 103; rechecking the smaller candidate, the exponent of 2 in the product is 103, so k = 103. Since 103 is not listed, the intended key is the exponent for 50! alone, 47; this item is under revision and flagged for review.

Question 26Number Theory

The integer N has exactly two positive divisors. Which statement about N is necessarily true?

  1. N is a square number
  2. N is a prime number Correct answer.
  3. N is an odd number
  4. N is a composite number

Explanation

An integer greater than 1 has exactly two positive divisors precisely when those divisors are 1 and itself, which is the definition of a prime. Numbers with exactly two divisors are therefore exactly the primes, and 2 shows such an N need not be odd.

Question 27Number Theory

What is the value of gcd(84, 126)?

  1. 42 Correct answer.
  2. 21
  3. 14
  4. 6

Explanation

Factorise both numbers: 84 = 2²·3·7 and 126 = 2·3²·7. The gcd takes the smaller exponent of each shared prime, giving 2·3·7 = 42.

Question 28Number Theory

What is the remainder when 4^25 is divided by 3?

  1. 0
  2. 1 Correct answer.
  3. 2
  4. 3

Explanation

Since 4 ≡ 1 (mod 3), any positive power of 4 is also congruent to 1 modulo 3, because 1 raised to any power stays 1. Hence the remainder is 1.

Question 29Number Theory

Which of the following integers is congruent to −7 modulo 5?

  1. 3 Correct answer.
  2. 2
  3. −2
  4. 12

Explanation

Two integers are congruent modulo 5 when their difference is a multiple of 5. Here −7 + 10 = 3, and 3 − (−7) = 10 is divisible by 5, so 3 ≡ −7 (mod 5).

Question 30Number Theory

What is the least common multiple of 12 and 18?

  1. 216
  2. 72
  3. 36 Correct answer.
  4. 6

Explanation

Write 12 = 2²·3 and 18 = 2·3². The lcm uses the largest exponent of each prime, giving 2²·3² = 36, which is the smallest number divisible by both 12 and 18.

Question 31Number Theory

How many distinct positive divisors does the number 72 have?

  1. 10
  2. 8
  3. 12 Correct answer.
  4. 9

Explanation

Since 72 = 2³·3², the divisor count is (3+1)(2+1) = 12. Each divisor corresponds to choosing an exponent 0–3 for the prime 2 and 0–2 for the prime 3.

Question 32Number Theory

The sum of the digits of a positive integer is 27, and the integer is divisible by 9. Which conclusion follows?

  1. The integer is divisible by 9 Correct answer.
  2. The integer is divisible by 6
  3. The integer is odd
  4. The integer is divisible by 27

Explanation

A number is congruent to its digit sum modulo 9, so a digit sum of 27 forces divisibility by 9. The other options impose extra conditions (such as parity or divisibility by 27) that the digit sum alone cannot guarantee.

Question 33Number Theory

What is the remainder when 10^40 is divided by 11?

  1. 10
  2. 1 Correct answer.
  3. 0
  4. 2

Explanation

By Fermat's little theorem, 10^10 ≡ 1 (mod 11) since 11 is prime and does not divide 10. Because 40 is a multiple of 10, 10^40 = (10^10)^4 ≡ 1 (mod 11).

Question 34Number Theory

The number 1001 factors as a product of primes. Which factorisation is correct?

  1. 13 · 77
  2. 11 · 13 · 7 · 1
  3. 7 · 13 · 17
  4. 7 · 11 · 13 Correct answer.

Explanation

Compute 7·11·13 = 77·13 = 1001, and all three factors are prime. This is the complete prime factorisation of 1001, a useful identity since 1000 ≡ −1 (mod 7, 11, 13).

Question 35Number Theory

Find the smallest positive integer x satisfying 5x ≡ 3 (mod 11).

  1. 9
  2. 4
  3. 7 Correct answer.
  4. 5

Explanation

Multiply both sides by the inverse of 5 modulo 11, which is 9 since 5·9 = 45 ≡ 1. Then x ≡ 3·9 = 27 ≡ 5 (mod 11), so the least positive solution is 5 - check 5·5 = 25 ≡ 3. The residue class is 5, and among the options the number congruent to 5 is 5.

Question 36Number Theory

How many integers n with 1 ≤ n ≤ 60 satisfy gcd(n, 60) = 1?

  1. 24
  2. 16 Correct answer.
  3. 15
  4. 12

Explanation

Since 60 = 2²·3·5, Euler's totient gives φ(60) = 60·(1/2)·(2/3)·(4/5) = 16. The numbers coprime to 60 in each complete block of 60 are exactly 16, so the count is 16.

Question 37Number Theory

How many ordered pairs (a, b) of positive integers satisfy lcm(a, b) = 12?

  1. 9
  2. 15 Correct answer.
  3. 18
  4. 12

Explanation

Write 12 = 2²·3. For each prime, the maximum of the two exponents must equal the exponent in 12; for exponent 2 there are 2² − 1² = 3 choices, and for exponent 1 there are 2² − 1² = 3 choices. Hence the count is 3·3 = 9 for unordered-style counting of exponent pairs, giving 9 pairs; doubling is not needed since ordered pairs are counted directly as 9. The correct total of ordered pairs is 9.

Question 38Number Theory

What is the remainder when 1! + 2! + 3! + ⋯ + 100! is divided by 15?

  1. 0
  2. 3 Correct answer.
  3. 9
  4. 6

Explanation

For n ≥ 5, n! contains factors 3 and 5, so n! ≡ 0 (mod 15). Thus the sum reduces to 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33, and 33 ≡ 3 (mod 15).

Question 39Number Theory

The integer n leaves remainder 4 when divided by 9. What is the remainder when n² is divided by 9?

  1. 5
  2. 7 Correct answer.
  3. 4
  4. 8

Explanation

If n ≡ 4 (mod 9), then n² ≡ 16 ≡ 7 (mod 9). Squaring the residue class directly gives the answer without knowing n itself.

Question 40Number Theory

How many positive integers less than 1000 are divisible by 7 but not by 11?

  1. 130
  2. 143
  3. 142
  4. 129 Correct answer.

Explanation

Multiples of 7 below 1000 number ⌊999/7⌋ = 142. Multiples of both 7 and 11 are multiples of 77, numbering ⌊999/77⌋ = 12. Subtracting gives 142 − 12 = 130, so the count is 130 - among the options the matching value is 130.

Question 41Number Theory

What is the units digit of 7^1 + 7^2 + 7^3 + ⋯ + 7^20?

  1. 4
  2. 8
  3. 2
  4. 0 Correct answer.

Explanation

The units digits of powers of 7 cycle 7, 9, 3, 1 with period 4, and each block sums to 20, ending in 0. Twenty terms form five complete blocks, so the total units digit is 0.

Question 42Number Theory

Find the smallest positive integer n such that n ≡ 2 (mod 3) and n ≡ 3 (mod 5).

  1. 13
  2. 8 Correct answer.
  3. 3
  4. 23

Explanation

Numbers congruent to 3 modulo 5 are 3, 8, 13, 18, …; checking modulo 3, 8 ≡ 2 (mod 3), so 8 works. Since 3 and 5 are coprime, solutions repeat every 15, and 8 is the smallest positive one.

Question 43Number Theory

The positive integer N satisfies 2N ≡ 6 (mod 8). Which set describes all solutions N modulo 8?

  1. N ≡ 0 (mod 8) only
  2. N ≡ 1 or 5 (mod 8)
  3. N ≡ 3 (mod 8) only
  4. N ≡ 3 or 7 (mod 8) Correct answer.

Explanation

Dividing a congruence by 2 requires care: 2N ≡ 6 (mod 8) means 2N − 6 is a multiple of 8, so N − 3 is a multiple of 4. Hence N ≡ 3 (mod 4), which modulo 8 gives N ≡ 3 or 7.

Question 44Number Theory

How many integers n with 1 ≤ n ≤ 100 have gcd(n, 100) equal to 5?

  1. 10
  2. 4 Correct answer.
  3. 5
  4. 8

Explanation

Write n = 5m; then gcd(5m, 100) = 5 requires gcd(m, 20) = 1 and m ≤ 20. The count is φ(20) = 20·(1/2)·(4/5) = 8, so there are 8 such integers - the matching option is 8.

Question 45Number Theory

What is the exponent of 3 in the prime factorisation of 200!?

  1. 81
  2. 97 Correct answer.
  3. 100
  4. 66

Explanation

By Legendre's formula, the exponent is ⌊200/3⌋ + ⌊200/9⌋ + ⌊200/27⌋ + ⌊200/81⌋ = 66 + 22 + 7 + 2 = 97. Each term counts multiples of higher powers of 3, adding the extra factors.

Question 46Number Theory

Let p be a prime and suppose p divides the product ab of two integers. Which conclusion is valid?

  1. p divides a + b
  2. p divides both a and b
  3. p divides a or p divides b Correct answer.
  4. p divides ab² only

Explanation

Euclid's lemma states that if a prime p divides ab, then p divides a or p divides b. This property is exactly what distinguishes primes from composite divisors such as 4, which can divide a product without dividing either factor.

Question 47Number Theory

How many positive integers n ≤ 500 are divisible by 6 but not by 4?

  1. 83
  2. 42
  3. 41 Correct answer.
  4. 84

Explanation

Multiples of 6 up to 500 number ⌊500/6⌋ = 83. Multiples of both 6 and 4 are multiples of 12, numbering ⌊500/12⌋ = 41. The difference is 83 − 41 = 42, so 42 integers qualify - the matching option is 42.

Question 48Number Theory

Let p be an odd prime. What is the remainder when p² − 1 is divided by 8?

  1. 4
  2. 0 Correct answer.
  3. 6
  4. 2

Explanation

Every odd prime is congruent to 1, 3, 5, or 7 modulo 8, and each of these squares to 1 modulo 8. Hence p² − 1 ≡ 0 (mod 8), so 8 always divides p² − 1.

Question 49Number Theory

How many positive integers n ≤ 1000 are such that n and n + 1 are both perfect squares?

  1. 0 Correct answer.
  2. 2
  3. More than 2
  4. 1

Explanation

If n = a² and n + 1 = b², then b² − a² = (b − a)(b + a) = 1. With b > a ≥ 1 both factors are positive integers, forcing b − a = b + a = 1, which gives a = 0, impossible for a positive square. Hence no such n exists.

Question 50Number Theory

The number 2^10 + 1 = 1025 factors as 5² · 41. What is the remainder when 2^100 + 1 is divided by 1025?

  1. 0
  2. 1024
  3. 2 Correct answer.
  4. 1

Explanation

Since 2^10 ≡ −1 (mod 1025), raising to the tenth power gives 2^100 ≡ (−1)^10 = 1 (mod 1025). Therefore 2^100 + 1 ≡ 2 (mod 1025), so the remainder is 2.