Divisibility, remainders, unit digits, factors and types of numbers.
47 solved questions · 18 easy · 20 medium · 9 hard
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What is the unit digit of 7^75?
Answer: A. 3
Solution
Unit digits of powers of 7: 7, 9, 3, 1 (cycle of 4). 75 mod 4 = 3, so we take position 3 of the cycle. Unit digit = 3. Answer: 3
Shortcut: Divide the power by 4 and use the remainder (4 if it is 0) as the new power.
What is the unit digit of 3^50?
Answer: C. 9
Solution
Unit digits of powers of 3: 3, 9, 7, 1 (cycle of 4). 50 mod 4 = 2, so we take position 2 of the cycle. Unit digit = 9. Answer: 9
Shortcut: Divide the power by 4 and use the remainder (4 if it is 0) as the new power.
What is the unit digit of 2^43?
Answer: B. 8
Solution
Unit digits of powers of 2: 2, 4, 8, 6 (cycle of 4). 43 mod 4 = 3, so we take position 3 of the cycle. Unit digit = 8. Answer: 8
Shortcut: Divide the power by 4 and use the remainder (4 if it is 0) as the new power.
What is the unit digit of 8^102?
Answer: D. 4
Solution
Unit digits of powers of 8: 8, 4, 2, 6 (cycle of 4). 102 mod 4 = 2, so we take position 2 of the cycle. Unit digit = 4. Answer: 4
Shortcut: Divide the power by 4 and use the remainder (4 if it is 0) as the new power.
What is the unit digit of 9^25?
Answer: D. 9
Solution
Unit digits of powers of 9: 9, 1, 9, 1 (cycle of 4). 25 mod 4 = 1, so we take position 1 of the cycle. Unit digit = 9. Answer: 9
Shortcut: Divide the power by 4 and use the remainder (4 if it is 0) as the new power.
Which of the following numbers is divisible by 11?
Answer: B. 5555
Solution
Rule: (sum of digits at odd places) - (sum at even places) must be 0 or a multiple of 11. For 5555: difference = 0. 5555 / 11 = 505. Answer: 5555
How many factors does 72 have?
Answer: A. 12
Solution
72 = 2^3 x 3^2 Factors = (3+1)(2+1) = 12 Answer: 12
How many factors does 180 have?
Answer: C. 18
Solution
180 = 2^2 x 3^2 x 5 Factors = (2+1)(2+1)(1+1) = 18 Answer: 18
What is the sum of the first 25 odd natural numbers?
Answer: C. 625
Solution
Sum of first n odd numbers = n^2 = 25^2 = 625. Answer: 625
Shortcut: Sum of the first n odd numbers is n squared.
What is the remainder when 748653 is divided by 9?
Answer: A. 6
Solution
A number leaves the same remainder mod 9 as its digit sum. Digit sum = 7+4+8+6+5+3 = 33 33 mod 9 = 6 Answer: 6
What is the smallest prime number greater than 90?
Answer: D. 97
Solution
91 = 7 x 13, 92 is even, 93 = 3 x 31, 94 is even, 95 = 5 x 19, 96 is even, so all are composite. 97 is not divisible by 2, 3, 5 or 7, so it is prime. Answer: 97
What is the largest 3-digit number divisible by 13?
Answer: B. 988
Solution
999 / 13 = 76 remainder 11. Largest multiple = 999 - 11 = 988. Answer: 988
Shortcut: Divide 999 by the number and subtract the remainder.
How many prime numbers are there from 1 to 50?
Answer: A. 15
Solution
Primes up to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Count = 15. Answer: 15
If the 5-digit number 4x521 is divisible by 3, which of the following can be the value of x?
Answer: C. 6
Solution
Digit sum = 4 + x + 5 + 2 + 1 = 12 + x. 12 + x must be a multiple of 3, so x is 0, 3, 6 or 9. From the options only 6 works (12 + 6 = 18). Answer: 6
How many multiples of 7 are there from 1 to 200?
Answer: B. 28
Solution
Count = floor(200/7) = 28. Answer: 28
Which of the following numbers is divisible by 8?
Answer: D. 7416
Solution
A number is divisible by 8 if its last three digits are divisible by 8. 416 / 8 = 52, so 7416 is divisible by 8. The others end in 412, 418, 414: 412/8 = 51.5, 418/8 = 52.25, 414/8 = 51.75, none is a whole number. Answer: 7416
What is the sum of the first 20 even natural numbers?
Answer: D. 420
Solution
Sum of first n even numbers = n(n+1) = 20 x 21 = 420. Answer: 420
Shortcut: n(n+1) for the first n even numbers.
Which of the following is a prime number?
Answer: B. 83
Solution
91 = 7 x 13, 87 = 3 x 29, 93 = 3 x 31, 51 = 3 x 17. 83 is not divisible by 2, 3, 5 or 7, so it is prime. Answer: 83
What is the unit digit of 123 x 457 x 789?
Answer: A. 9
Solution
Only unit digits matter: 3 x 7 x 9. 3 x 7 = 21, unit digit 1 1 x 9 = 9 Answer: 9
What is the remainder when 7^100 is divided by 5?
Answer: C. 1
Solution
Remainders of 7^k mod 5 for k = 1, 2, ...: 2, 4, 3, 1 and then repeat (cycle length 4). 100 mod 4 = 0, so remainder = 7^0 mod 5 = 1. Answer: 1
What is the remainder when 3^81 is divided by 7?
Answer: C. 6
Solution
Remainders of 3^k mod 7 for k = 1, 2, ...: 3, 2, 6, 4, 5, 1 and then repeat (cycle length 6). 81 mod 6 = 3, so remainder = 3^3 mod 7 = 6. Answer: 6
What is the remainder when 2^50 is divided by 9?
Answer: A. 4
Solution
Remainders of 2^k mod 9 for k = 1, 2, ...: 2, 4, 8, 7, 5, 1 and then repeat (cycle length 6). 50 mod 6 = 2, so remainder = 2^2 mod 9 = 4. Answer: 4
Find the smallest positive number which leaves remainder 3 when divided by 5 and remainder 4 when divided by 7.
Answer: D. 18
Solution
Numbers leaving remainder 3 on division by 5: 3, 8, 13, 18, 23, ... Check which leaves 4 on division by 7: 3 (3), 8 (1), 13 (6), 18 (4). Yes. Answer: 18
How many zeros are there at the end of 100!?
Answer: B. 24
Solution
Zeros = number of 5s in the prime factorisation = floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24. Answer: 24
Shortcut: Keep dividing by 5 and add the quotients.
How many zeros are there at the end of 60!?
Answer: A. 14
Solution
Zeros = number of 5s in the prime factorisation = floor(60/5) + floor(60/25) + floor(60/125) = 12 + 2 + 0 = 14. Answer: 14
Shortcut: Keep dividing by 5 and add the quotients.
How many factors of 2^4 x 3^3 x 5^2 are perfect squares?
Answer: C. 12
Solution
A perfect-square factor has even powers. 2 can have power 0, 2, 4 (3 choices); 3 can have 0, 2 (2 choices); 5 can have 0, 2 (2 choices). Count = 3 x 2 x 2 = 12. Answer: 12
How many even factors does 360 have?
Answer: B. 18
Solution
360 = 2^3 x 3^2 x 5. Total factors = 4 x 3 x 2 = 24. Odd factors = factors of 3^2 x 5 = 3 x 2 = 6. Even factors = 24 - 6 = 18. Answer: 18
Shortcut: Even factors = 3 (powers 1 to 3 of 2) x 3 x 2 = 18.
What is the sum of all factors of 180?
Answer: D. 546
Solution
180 = 2^2 x 3^2 x 5. Sum = (1+2+4)(1+3+9)(1+5) = 7 x 13 x 6 = 546. Answer: 546
A number when divided by 357 leaves remainder 39. What is the remainder when the same number is divided by 17?
Answer: D. 5
Solution
Number = 357q + 39. Since 357 = 17 x 21, the first term is a multiple of 17. Remainder = 39 mod 17 = 5. Answer: 5
How many numbers from 1 to 500 are divisible by 3 or 5?
Answer: B. 233
Solution
Divisible by 3: floor(500/3) = 166 Divisible by 5: 100 Divisible by both (15): floor(500/15) = 33 Inclusion-exclusion: 166 + 100 - 33 = 233. Answer: 233
If the number 37x2 is divisible by 9, what is the value of x?
Answer: A. 6
Solution
Digit sum = 3 + 7 + x + 2 = 12 + x. For divisibility by 9, 12 + x must be 18 (the only multiple of 9 reachable by a single digit), so x = 6. Answer: 6
What is the least number that must be added to 7,586 so that the sum is exactly divisible by 13?
Answer: C. 6
Solution
7,586 / 13 leaves remainder 7. Number to add = 13 - 7 = 6. Check: 7586 + 6 = 7592 = 13 x 584. Answer: 6
What is the least number that must be added to 9,500 so that the sum is exactly divisible by 17?
Answer: C. 3
Solution
9,500 / 17 leaves remainder 14. Number to add = 17 - 14 = 3. Check: 9500 + 3 = 9503 = 17 x 559. Answer: 3
What is the least number that must be subtracted from 5000 so that the result is divisible by 24?
Answer: A. 8
Solution
5000 / 24 = 208 remainder 8. Subtract the remainder: 8. Answer: 8
How many 2-digit numbers are divisible by 4?
Answer: D. 22
Solution
2-digit multiples of 4: 12, 16, ..., 96. Count = (96 - 12)/4 + 1 = 22. Answer: 22
What are the last two digits of 7^100?
Answer: B. 01
Solution
7^2 = 49, 7^4 = 49 x 49 = 2401, whose last two digits are 01. 7^100 = (7^4)^25, and 01 raised to any power ends in 01. Answer: 01
What is the remainder when 1! + 2! + 3! + ... + 30! is divided by 5?
Answer: A. 3
Solution
From 5! onwards every factorial is divisible by 5, so only 1! + 2! + 3! + 4! matters. 1 + 2 + 6 + 24 = 33. 33 mod 5 = 3. Answer: 3
What is the remainder when 2^10 is divided by 11?
Answer: C. 1
Solution
2^5 = 32 = 2 x 11 + 10, so 2^5 is -1 mod 11. 2^10 = (2^5)^2 = (-1)^2 = 1 mod 11. Answer: 1
Shortcut: Fermat: 2^10 mod 11 = 1 since 11 is prime.
What are the last two digits of 13^50?
Answer: B. 49
Solution
13^2 = 169 -> 69 13^4 = 69^2 = 4761 -> 61 13^5 = 61 x 13 = 793 -> 93 13^10 = 93^2 = 8649 -> 49 13^20 = 49^2 = 2401 -> 01 13^50 = 13^20 x 13^20 x 13^10 -> 01 x 01 x 49 = 49. Answer: 49
How many numbers from 1 to 1000 are divisible by none of 2, 3 and 5?
Answer: D. 266
Solution
Inclusion-exclusion on numbers divisible by 2, 3 or 5. By 2: 500, by 3: 333, by 5: 200. By 6: 166, by 10: 100, by 15: 66. By 30: 33. Union = 500 + 333 + 200 - 166 - 100 - 66 + 33 = 734. None = 1000 - 734 = 266. Answer: 266
What is the sum of all 3-digit numbers divisible by 7?
Answer: D. 70,336
Solution
Smallest = 105 (7 x 15), largest = 994 (7 x 142). Number of terms = 142 - 15 + 1 = 128. Sum = 128 x (105 + 994)/2 = 64 x 1099 = 70,336. Answer: 70,336
What is the smallest natural number having exactly 15 factors?
Answer: B. 144
Solution
15 = 15 x 1 or 5 x 3, so N = p^14 or p^4 x q^2. Smallest: 2^4 x 3^2 = 16 x 9 = 144 (2^14 = 16384 is much larger). Answer: 144
What is the remainder when 1! + 2! + 3! + ... + 50! is divided by 7?
Answer: A. 5
Solution
From 7! onwards every term is divisible by 7. 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873. 873 = 7 x 124 + 5. Answer: 5
What is the remainder when 3^100 is divided by 13?
Answer: C. 3
Solution
3^3 = 27 = 2 x 13 + 1, so 3^3 leaves remainder 1. 3^100 = (3^3)^33 x 3 leaves 1^33 x 3 = 3. Answer: 3
In how many ways can 360 be written as a product of two coprime factors (order does not matter; 1 x 360 counts)?
Answer: C. 4
Solution
360 = 2^3 x 3^2 x 5 has 3 distinct prime factors. Unordered coprime pairs = 2^(3-1) = 4. They are 1 x 360, 8 x 45, 9 x 40, 5 x 72. Answer: 4
For how many positive integers n is n^2 + 100 a perfect square?
Answer: A. 1
Solution
Let n^2 + 100 = m^2, so (m - n)(m + n) = 100. Both factors must have the same parity, hence both even: 2 x 50 or 10 x 10. 2 x 50 gives m = 26, n = 24. 10 x 10 gives n = 0, which is not positive. So exactly 1 value (n = 24). Answer: 1
How many numbers from 1 to 1000 leave remainder 3 when divided by 4, remainder 5 when divided by 6 and remainder 8 when divided by 9?
Answer: D. 27
Solution
Each remainder is 1 less than the divisor, so N + 1 is a multiple of LCM(4, 6, 9) = 36. N = 36k - 1, and N <= 1000 means 36k <= 1001, so k <= 27.8. k = 1 to 27 gives 27 numbers (35, 71, ..., 971). Answer: 27
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