Divisors and Factors
How many positive integer divisors does $120$ have?
The prime factorization of 120 is 2^3 × 3^1 × 5^1. Using the divisor count formula, add 1 to each exponent and multiply: (3 + 1)(1 + 1)(1 + 1) = 4 × 2 × 2 = 16. The answer is 16.
What is the largest prime factor of 9999?
Factor 9999 as 9 × 1111. Then 1111 factors into 11 × 101. The prime 101 has no factors up to sqrt(101) ≈ 10, so it is prime. The full factorization is 3^2 × 11 × 101. The largest prime factor is 101.
What is the remainder when 2^20 is divided by 3?
Note that 2 ≡ -1 (mod 3). Raising both sides to the 20th power gives 2^20 ≡ (-1)^20 ≡ 1 (mod 3). The remainder is 1.
How many pairs of positive integers (x, y) satisfy x + y = 10?
The positive integer choices for x are 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each choice determines a unique positive integer y = 10 - x. There are 9 pairs.
The number 101 in base 2 is equal to what number in base 10?
Expand using powers of 2: 1 × 2^2 + 0 × 2^1 + 1 × 2^0 = 4 + 0 + 1 = 5. The answer is 5.
What is the smallest positive integer that has exactly 6 positive integer divisors?
A number with 6 divisors must have a prime factorization of the form p^5 or p^2 × q^1, where p and q are distinct primes. The smallest number of the form p^5 is 2^5 = 32. The smallest number of the form p^2 × q^1 is 2^2 × 3^1 = 12. Comparing the two, 12 is smaller. The answer is 12.
How many terminal zeros does 25! end with when written in base 10?
Terminal zeros are determined by pairs of factors 2 and 5 in 25!. Since 2s are abundant, count factors of 5 using Legendre's Formula: floor(25/5) + floor(25/25) = 5 + 1 = 6. The answer is 6.
What is the units digit of 7^2023?
The units digits of powers of 7 repeat in a 4-step cycle: 7, 9, 3, 1. Divide the exponent by 4: 2023 ÷ 4 = 505 with a remainder of 3. The remainder 3 points to the 3rd term in the cycle, which is 3. The answer is 3.
Find the only pair of positive integers (x, y) that satisfies xy = x + y.
Rearrange as xy - x - y = 0. Add 1 to both sides to factor: (x - 1)(y - 1) = 1. Since x and y are positive integers, x - 1 and y - 1 must be positive integer factors of 1. Thus x - 1 = 1 and y - 1 = 1, giving x = 2 and y = 2. The answer is (2, 2).
Find the sum of the digits of the largest 3-digit number (in base 10) that is a multiple of 9.
The largest 3-digit number is 999. Since 999 = 9 × 111, it is a multiple of 9. The sum of its digits is 9 + 9 + 9 = 27.
How many odd positive integer divisors does 2000 have?
The prime factorization of 2000 is 2^4 × 5^3. For a divisor to be odd, it cannot contain any factors of 2. Therefore, the odd divisors come entirely from the 5^3 term. The number of such divisors is (3 + 1) = 4. The answer is 4.
What is the greatest power of 2 that divides 10! + 11!?
Factor out common terms: 10! + 11! = 10!(1 + 11) = 10! × 12. Count factors of 2 in 10! using Legendre's formula: floor(10/2) + floor(10/4) + floor(10/8) = 5 + 2 + 1 = 8. The term 12 contributes 2^2 × 3, adding 2 more factors of 2. The total exponent of 2 is 8 + 2 = 10. The answer is 10.
What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 12?
For any n ≥ 4, n! contains both 3 and 4 as factors, so n! is divisible by 12 (i.e., n! ≡ 0 mod 12). Thus, we only sum the terms before 4!: 1! + 2! + 3! = 1 + 2 + 6 = 9. Since 9 < 12, the remainder is 9.
How many pairs of positive integers (a, b) satisfy 2a + 3b = 30?
Rearrange to 2a = 30 - 3b. For a to be a positive integer, 30 - 3b must be positive and divisible by 2. Thus 3b must be even and less than 30, meaning b must be an even positive integer less than 10. The choices for b are 2, 4, 6, and 8. Each yields a valid positive integer a. There are 4 pairs.
What is the largest two-digit integer that is equal to exactly 7 times the sum of its digits?
Let the number be 10a + b. The condition states 10a + b = 7(a + b). Expanding gives 10a + b = 7a + 7b => 3a = 6b => a = 2b. The digit pairs (a, b) satisfying this are (2, 1), (4, 2), (6, 3), and (8, 4), yielding the numbers 21, 42, 63, and 84. The largest is 84.
The number N = 2^3 × 3^a × 5^b has exactly 60 positive integer divisors. What is the minimum possible value of N?
Using the divisor formula: (3 + 1)(a + 1)(b + 1) = 60, which simplifies to 4(a + 1)(b + 1) = 60, so (a + 1)(b + 1) = 15. The factor pairs of 15 are (1, 15) and (3, 5). The possible pairs for (a, b) are (0, 14), (14, 0), (2, 4), or (4, 2). To minimize N, place larger exponents on smaller prime bases. Comparing (a,b) = (4,2) and (2,4): 2^3 × 3^4 × 5^2 = 8 × 81 × 25 = 16200, whereas 2^3 × 3^2 × 5^4 = 8 × 9 × 625 = 45000. The minimum value is 16200.
Find the smallest positive integer n such that n! is a multiple of 3^7.
We need the exponent of 3 in the prime factorization of n! to be at least 7. Checking multiples of 3: For n = 9: floor(9/3) + floor(9/9) = 3 + 1 = 4. For n = 15: floor(15/3) + floor(15/9) = 5 + 1 = 6. For n = 18: floor(18/3) + floor(18/9) = 6 + 2 = 8. The value n = 15 yields an exponent of 6, while n = 18 reaches an exponent of 8 (which is ≥ 7). The answer is 18.
What is the smallest positive integer x such that x leaves a remainder of 2 when divided by 5, and a remainder of 3 when divided by 7?
Set up the system of congruences: x ≡ 2 (mod 5) and x ≡ 3 (mod 7). From x ≡ 2 (mod 5), write x = 5k + 2. Substitute into the second congruence: 5k + 2 ≡ 3 (mod 7) => 5k ≡ 1 (mod 7). Multiplying by 3 gives 15k ≡ 3 (mod 7) => k ≡ 3 (mod 7). The smallest positive k is 3. Plugging back in yields x = 5(3) + 2 = 17. The answer is 17.
Find the sum of all positive integers n such that (n + 10) / (n - 2) is an integer.
Rewrite (n + 10) / (n - 2) as 1 + 12 / (n - 2). For this to be an integer, n - 2 must be a divisor of 12. The divisors of 12 are ±1, ±2, ±3, ±4, ±6, ±12. Solving for n = d + 2 and keeping only positive n: d = -1 => n = 1 d = 1 => n = 3 d = 2 => n = 4 d = 3 => n = 5 d = 4 => n = 6 d = 6 => n = 8 d = 12 => n = 14 The positive values of n are 1, 3, 4, 5, 6, 8, and 14. Their sum is 1 + 3 + 4 + 5 + 6 + 8 + 14 = 41.
In what base b does the equation 3 × 4 = 15 (in base b) hold true?
In base 10, 3 × 4 = 12. In base b, 15 is expressed as 1 × b^1 + 5 × b^0 = b + 5. Equating the values gives b + 5 = 12 => b = 7. The answer is 7.
Let d(n) be the number of positive integer divisors of n. How many positive integers n ≤ 2026 have an odd value for d(n)?
A number has an odd number of divisors if and only if it is a perfect square, because factors come in pairs unless a factor pairs with itself. The perfect squares less than or equal to 100 are 1^2, 2^2, 3^2, ..., 45^2. There are 45 such numbers. The answer is 45.
Three distinct prime numbers p, q, and r satisfy p × q × r = 7(p + q + r). What is the sum p + q + r?
Since p, q, and r are prime, one of them must be 7. Let r = 7. The equation becomes 7 × p × q = 7(p + q + 7), which simplifies to p × q = p + q + 7. Rearranging gives p × q - p - q = 7. Adding 1 to both sides factors it as (p - 1)(q - 1) = 8. The factor pairs of 8 are (1, 8) and (2, 4). Case 1: p - 1 = 1 => p = 2; q - 1 = 8 => q = 9 (not prime). Case 2: p - 1 = 2 => p = 3; q - 1 = 4 => q = 5 (both prime). The three primes are 3, 5, and 7. Their sum is 3 + 5 + 7 = 15. The answer is 15.
What is the remainder when 5^202 is divided by 7?
By Fermat's Little Theorem, 5^6 ≡ 1 (mod 7). Divide the exponent by 6: 202 = 6 × 33 + 4. Therefore, 5^202 ≡ 5^4 (mod 7). Calculate 5^4 mod 7: 5^2 = 25 ≡ 4 (mod 7), so 5^4 ≡ 4^2 = 16 ≡ 2 (mod 7). The remainder is 2.
How many ordered pairs of positive integers (x, y) satisfy 1/x + 1/y = 1/6?
Clear fractions by multiplying by 6xy: 6y + 6x = xy. Rearrange to xy - 6x - 6y = 0. Add 36 to both sides to factor: (x - 6)(y - 6) = 36. Since x and y are positive integers, (x - 6) and (y - 6) form positive integer factor pairs of 36. The number of such pairs equals the number of divisors of 36. Since 36 = 2^2 × 3^2, it has (2 + 1)(2 + 1) = 9 divisors. The answer is 9.
A 3-digit number ABC in base 10 is equal to the 3-digit number CBA written in base 9. What is the base 10 value of ABC?
Expand both representations: 100A + 10B + C = C × 9^2 + B × 9^1 + A × 9^0 100A + 10B + C = 81C + 9B + A Rearranging yields 99A + B = 80C. Since CBA is a base 9 number, C must be between 1 and 8, and A, B must be between 0 and 8 (with A ≠ 0). For 99A + B = 80C with 0 ≤ B ≤ 8, 80C must be slightly greater than 99A. Testing values gives A = 4 and C = 5: 99(4) + B = 80(5) => 396 + B = 400 => B = 4. Since B = 4 is a valid digit (< 9), the digits are A = 4, B = 4, C = 5. The base 10 number ABC is 445.