Number system is the base of the whole quantitative paper, so the time you put here pays you back in every other chapter. Do not try to learn it as a list of tricks. Learn the divisibility rules first, because they turn a two minute division into a five second check. Then learn prime factorisation properly, because factors, HCF, LCM and remainders all come out of that one skill. The chapter that scares students is remainders, and it should not, because most of those questions are just cyclicity in disguise. Find the pattern of the last digit, count the cycle, and the answer is there. Work through these questions slowly the first time and quickly the second, and you will notice your speed in the whole paper going up.
Number System Formulas
Divisibility rules
By 2: last digit is even
By 3: sum of digits divisible by 3
By 4: last two digits divisible by 4
By 5: last digit is 0 or 5
By 6: divisible by both 2 and 3
By 8: last three digits divisible by 8
By 9: sum of digits divisible by 9
By 11: difference of the sum of odd place digits and the sum of even place digits is 0 or divisible by 11
The division formula
Dividend = Divisor × Quotient + Remainder
Remainder is always less than the divisor.
The standard sums
Sum of first n natural numbers = n (n + 1) ÷ 2
Sum of first n even numbers = n (n + 1)
Sum of first n odd numbers = n²
Sum of squares of first n natural numbers = n (n + 1) (2n + 1) ÷ 6
Sum of cubes of first n natural numbers = [n (n + 1) ÷ 2]²
Factors, from prime factorisation
Write N = aᵖ × bᑫ × cʳ where a, b and c are prime.
Number of factors = (p + 1) (q + 1) (r + 1)
HCF and LCM
HCF × LCM = Product of the two numbers. This holds for two numbers only.
HCF of fractions = HCF of numerators ÷ LCM of denominators
LCM of fractions = LCM of numerators ÷ HCF of denominators
Unit digit and cyclicity
Divide the power by 4 and use the remainder.
1, 5, 6 and 0 always end in themselves. 4 and 9 repeat every 2.
Cycle of 2: 2, 4, 8, 6. Cycle of 3: 3, 9, 7, 1. Cycle of 7: 7, 9, 3, 1. Cycle of 8: 8, 4, 2, 6.
If the remainder is 0, take the fourth value in the cycle.
Zeros at the end of a factorial
Zeros in n! = [n ÷ 5] + [n ÷ 25] + [n ÷ 125] and so on, taking only the whole number part each time.
The identities
a² − b² = (a + b) (a − b)
a³ + b³ = (a + b) (a² − ab + b²)
a³ − b³ = (a − b) (a² + ab + b²)
(aⁿ − bⁿ) is always divisible by (a − b)
(aⁿ + bⁿ) is divisible by (a + b) only when n is odd
60 Number System Aptitude Questions and Answers (Solved MCQs)
Question 1. Which of the following correctly describes √2 + √8?
a) A natural number
b) A rational number
c) An irrational number
d) A whole number
Answer:
c) An irrational number — Since √8 = 2√2, we have √2 + √8 = 3√2. As √2 is irrational, 3√2 is also irrational.
Question 2. What is the smallest five-digit number exactly divisible by 72?
a) 10,008
b) 10,016
c) 10,044
d) 10,080
Answer:
a) 10,008 — The smallest five-digit number is 10,000. Dividing it by 72 gives a quotient of 138 and a remainder of 64. The next multiple is 72 × 139 = 10,008.
Question 3. What is the greatest six-digit number exactly divisible by 125?
a) 9,99,625
b) 9,99,750
c) 9,99,875
d) 9,99,975
Answer:
c) 9,99,875 — The greatest six-digit number is 9,99,999. The greatest multiple of 125 not exceeding it is 125 × 7,999 = 9,99,875.
Question 4. The sum of three consecutive positive integers is 156. What is the largest integer?
a) 51
b) 52
c) 53
d) 54
Answer:
c) 53 — Let the consecutive integers be x – 1, x and x + 1. Their sum is 3x = 156, so x = 52. Therefore, the integers are 51, 52 and 53.
Question 5. The sum of the digits of a two-digit number is 11. When its digits are reversed, the number increases by 27. What is the original number?
a) 38
b) 47
c) 56
d) 74
Answer:
b) 47 — Let the tens digit be x and the units digit be y. Then x + y = 11. Reversing the digits increases the number by 27, so 9(y – x) = 27 and y – x = 3. Solving the equations gives x = 4 and y = 7. Hence, the number is 47.
Question 6. The three-digit number 4ab is exactly divisible by both 9 and 11. If a and b are non-zero digits, what is the value of a + b?
a) 10
b) 12
c) 14
d) 16
Answer:
c) 14 — Divisibility by 11 requires 4 – a + b = 0 or a – b = 4. Divisibility by 9 requires 4 + a + b to be a multiple of 9. The digits a = 9 and b = 5 satisfy both conditions, giving the number 495. Therefore, a + b = 14.
Question 7. How many prime numbers are there between 50 and 100?
a) 8
b) 9
c) 10
d) 11
Answer:
c) 10 — The prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Therefore, there are 10 prime numbers.
Question 8. How many positive factors does 360 have?
a) 18
b) 20
c) 24
d) 30
Answer:
c) 24 — Prime factorisation gives 360 = 2³ × 3² × 5. Therefore, the number of positive factors is (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24.
Question 9. What is the least positive number that is exactly divisible by 12, 15, 18 and 20?
a) 120
b) 180
c) 240
d) 360
Answer:
b) 180 — We have 12 = 2² × 3, 15 = 3 × 5, 18 = 2 × 3² and 20 = 2² × 5. Their LCM is 2² × 3² × 5 = 180.
Question 10. What is the greatest number that divides 187, 233 and 325 leaving the same remainder in each case?
a) 23
b) 32
c) 46
d) 69
Answer:
c) 46 — The required divisor must exactly divide the differences 233 – 187 = 46, 325 – 233 = 92 and 325 – 187 = 138. Their HCF is 46. Each original number leaves a remainder of 3 when divided by 46.
Question 11. What is the sum of the first 50 positive odd numbers?
a) 2,400
b) 2,450
c) 2,500
d) 2,550
Answer:
c) 2,500 — The sum of the first n positive odd numbers is n². Therefore, the sum of the first 50 odd numbers is 50² = 2,500.
Question 12. A number gives a quotient of 12 and a remainder of 5 when divided by 7. What remainder will it leave when divided by 9?
a) 5
b) 6
c) 7
d) 8
Answer:
d) 8 — The number is 7 × 12 + 5 = 89. Dividing 89 by 9 gives a quotient of 9 and a remainder of 8.
Question 13. A perfect square can never have which of the following as its units digit?
a) 1
b) 4
c) 6
d) 8
Answer:
d) 8 — The possible units digits of a perfect square are 0, 1, 4, 5, 6 and 9. Therefore, a perfect square can never end in 8.
Question 14. How many three-digit positive integers are exactly divisible by 7?
a) 126
b) 127
c) 128
d) 129
Answer:
c) 128 — The smallest three-digit multiple of 7 is 105 = 7 × 15, and the greatest is 994 = 7 × 142. Therefore, the number of multiples is 142 – 15 + 1 = 128.
Question 15. What is the smallest prime factor of 2¹⁶ – 1?
a) 3
b) 5
c) 7
d) 11
Answer:
a) 3 — Since 2 ≡ -1 (mod 3), we have 2¹⁶ ≡ (-1)¹⁶ ≡ 1 (mod 3). Therefore, 2¹⁶ – 1 is divisible by 3. As 3 is prime, it is the smallest prime factor.
Question 16. What is the remainder when 7¹⁰³ is divided by 10?
a) 1
b) 3
c) 7
d) 9
Answer:
b) 3 — The units digits of powers of 7 follow the cycle 7, 9, 3, 1. Since 103 leaves a remainder of 3 when divided by 4, the required units digit is the third number in the cycle, which is 3.
Question 17. What is the remainder when 3¹⁰⁰ is divided by 7?
a) 1
b) 2
c) 4
d) 6
Answer:
c) 4 — Powers of 3 modulo 7 repeat after six terms. Since 100 leaves a remainder of 4 when divided by 6, 3¹⁰⁰ has the same remainder as 3⁴. Now, 3⁴ = 81, which leaves a remainder of 4 when divided by 7.
Question 18. What is the remainder when 2⁵⁰ + 3⁵⁰ is divided by 5?
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3 — Powers of both 2 and 3 repeat modulo 5 in cycles of 4. Since 50 leaves a remainder of 2 when divided by 4, 2⁵⁰ ≡ 2² ≡ 4 and 3⁵⁰ ≡ 3² ≡ 4 modulo 5. Their sum is 8, which leaves a remainder of 3.
Question 19. How many positive factors does 7,560 have?
a) 48
b) 56
c) 64
d) 72
Answer:
c) 64 — Prime factorisation gives 7,560 = 2³ × 3³ × 5 × 7. Therefore, the number of positive factors is (3 + 1)(3 + 1)(1 + 1)(1 + 1) = 4 × 4 × 2 × 2 = 64.
Question 20. What is the sum of all the positive factors of 360?
a) 1,080
b) 1,170
c) 1,260
d) 1,350
Answer:
b) 1,170 — Since 360 = 2³ × 3² × 5, the sum of its factors is (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5) = 15 × 13 × 6 = 1,170.
Question 21. What is the highest power of 5 that exactly divides 100!?
a) 20
b) 22
c) 24
d) 25
Answer:
c) 24 — The exponent of 5 in 100! is ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24. Therefore, 5²⁴ exactly divides 100!.
Question 22. How many trailing zeroes are there in 250!?
a) 60
b) 61
c) 62
d) 63
Answer:
c) 62 — The number of trailing zeroes is determined by the exponent of 5. Thus, ⌊250/5⌋ + ⌊250/25⌋ + ⌊250/125⌋ = 50 + 10 + 2 = 62.
Question 23. What is the highest power of 2 that exactly divides 100!?
a) 94
b) 96
c) 97
d) 99
Answer:
c) 97 — The exponent of 2 in 100! is ⌊100/2⌋ + ⌊100/4⌋ + ⌊100/8⌋ + ⌊100/16⌋ + ⌊100/32⌋ + ⌊100/64⌋ = 50 + 25 + 12 + 6 + 3 + 1 = 97.
Question 24. If the five-digit number 53×28 is exactly divisible by 72, what is the value of x?
a) 3
b) 5
c) 7
d) 9
Answer:
d) 9 — A number divisible by 72 must be divisible by both 8 and 9. The last three digits x28 are divisible by 8 when x is odd. For divisibility by 9, 5 + 3 + x + 2 + 8 = 18 + x must be divisible by 9. Among the given odd digits, only x = 9 satisfies this condition.
Question 25. What is the greatest number that divides 1,223 and 2,351, leaving a remainder of 5 in each case?
a) 6
b) 9
c) 12
d) 18
Answer:
a) 6 — Subtract the common remainder from both numbers: 1,223 – 5 = 1,218 and 2,351 – 5 = 2,346. Their HCF is 6. Therefore, the greatest required divisor is 6.
Question 26. What is the smallest positive number greater than 5 that leaves a remainder of 5 when divided by 12, 15 and 18?
a) 175
b) 180
c) 185
d) 195
Answer:
c) 185 — The required number minus 5 must be divisible by 12, 15 and 18. Their LCM is 180. Therefore, the smallest required number greater than 5 is 180 + 5 = 185.
Question 27. What is the remainder when 1! + 2! + 3! + ⋯ + 10! is divided by 10?
a) 1
b) 3
c) 5
d) 7
Answer:
b) 3 — Every factorial from 5! onward is divisible by 10. Therefore, only the first four terms affect the remainder. Their sum is 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33, which leaves a remainder of 3.
Question 28. How many odd positive factors does 360 have?
a) 4
b) 6
c) 8
d) 12
Answer:
b) 6 — Since 360 = 2³ × 3² × 5, an odd factor cannot contain 2. Therefore, the number of odd factors is (2 + 1)(1 + 1) = 3 × 2 = 6.
Question 29. How many composite positive factors does 144 have?
a) 10
b) 11
c) 12
d) 13
Answer:
c) 12 — Prime factorisation gives 144 = 2⁴ × 3², so it has (4 + 1)(2 + 1) = 15 positive factors. Of these, 1 is neither prime nor composite, while 2 and 3 are prime. Therefore, the number of composite factors is 15 – 1 – 2 = 12.
Question 30. What is the least positive number that leaves remainders 2, 3 and 4 when divided by 3, 4 and 5, respectively?
a) 49
b) 54
c) 59
d) 64
Answer:
c) 59 — Each remainder is one less than the corresponding divisor. Therefore, the required number plus 1 must be divisible by 3, 4 and 5. Their LCM is 60, so the least required number is 60 – 1 = 59.
Question 31. What is the units digit of 17¹⁷ + 23²³?
a) 2
b) 4
c) 6
d) 8
Answer:
b) 4 — The units digits of powers of 7 repeat as 7, 9, 3, 1. Since 17 leaves a remainder of 1 when divided by 4, 17¹⁷ ends in 7. The units digits of powers of 3 repeat as 3, 9, 7, 1. Since 23 leaves a remainder of 3 when divided by 4, 23²³ ends in 7. Therefore, the sum ends in 7 + 7 = 14, giving a units digit of 4.
Question 32. What are the last two digits of 7²⁰?
a) 01
b) 07
c) 21
d) 49
Answer:
a) 01 — Since 7⁴ = 2,401, its last two digits are 01. Therefore, 7²⁰ = (7⁴)⁵ also ends in 01.
Question 33. How many trailing zeroes are there in the product 75! × 50!?
a) 28
b) 29
c) 30
d) 31
Answer:
c) 30 — The exponent of 5 in 75! is ⌊75/5⌋ + ⌊75/25⌋ = 15 + 3 = 18. The exponent of 5 in 50! is ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12. Therefore, the product has 18 + 12 = 30 trailing zeroes.
Question 34. What is the digital root of 7²⁰²⁶?
a) 1
b) 4
c) 7
d) 9
Answer:
c) 7 — Powers of 7 modulo 9 follow the cycle 7, 4, 1. Since 2,026 leaves a remainder of 1 when divided by 3, 7²⁰²⁶ leaves a remainder of 7 when divided by 9. Hence, its digital root is 7.
Question 35. What is the decimal value of the binary number (101101)₂?
a) 41
b) 43
c) 45
d) 47
Answer:
c) 45 — The decimal value is 1 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2 + 1 = 32 + 8 + 4 + 1 = 45.
Question 36. What is the decimal value of (243)₅?
a) 68
b) 71
c) 73
d) 78
Answer:
c) 73 — Converting from base 5 gives 2 × 5² + 4 × 5¹ + 3 × 5⁰ = 50 + 20 + 3 = 73.
Question 37. How is the decimal number 100 represented in base 3?
a) (10101)₃
b) (10201)₃
c) (11001)₃
d) (11201)₃
Answer:
b) (10201)₃ — We have 100 = 1 × 3⁴ + 0 × 3³ + 2 × 3² + 0 × 3 + 1 = 81 + 18 + 1. Therefore, 100 is written as (10201)₃.
Question 38. A three-digit number exceeds the number obtained by reversing its digits by 396. The sum of its digits is 16, and its tens digit is twice its units digit. What is the original number?
a) 673
b) 736
c) 763
d) 793
Answer:
c) 763 — Let the digits be a, b and c. The difference between the number and its reversal is 99(a – c) = 396, giving a – c = 4. Also, b = 2c and a + b + c = 16. Substituting a = c + 4 and b = 2c gives 4c + 4 = 16, so c = 3, b = 6 and a = 7. Hence, the number is 763.
Question 39. How many four-digit numbers divisible by 4 can be formed using the digits 1, 2, 3 and 4 exactly once?
a) 4
b) 6
c) 8
d) 12
Answer:
b) 6 — A number is divisible by 4 when its last two digits form a number divisible by 4. The possible endings are 12, 24 and 32. For each ending, the remaining two digits can be arranged in 2! = 2 ways. Therefore, the total number is 3 × 2 = 6.
Question 40. What is the smallest four-digit perfect square that is exactly divisible by 12?
a) 1,024
b) 1,156
c) 1,296
d) 1,444
Answer:
c) 1,296 — For a square to be divisible by 12 = 2² × 3, its square root must be divisible by both 2 and 3, and hence by 6. The smallest multiple of 6 whose square is four-digit is 36. Therefore, the required number is 36² = 1,296.
Question 41. What is the greatest four-digit perfect cube?
a) 8,000
b) 9,000
c) 9,261
d) 9,604
Answer:
c) 9,261 — Since 21³ = 9,261 and 22³ = 10,648, which is a five-digit number, the greatest four-digit perfect cube is 9,261.
Question 42. What is the highest power of 12 that exactly divides 100!?
a) 46
b) 47
c) 48
d) 49
Answer:
c) 48 — Since 12 = 2² × 3, calculate the exponents of 2 and 3 in 100!. We have v₂(100!) = 97 and v₃(100!) = 33 + 11 + 3 + 1 = 48. Therefore, the exponent of 12 is min(⌊97/2⌋, 48) = min(48, 48) = 48.
Question 43. How many trailing zeroes are there in 2⁵⁰ × 5³⁵ × 3²⁰?
a) 30
b) 35
c) 40
d) 50
Answer:
b) 35 — Each trailing zero requires one pair of 2 and 5. The expression contains fifty factors of 2 and thirty-five factors of 5. Therefore, it contains 35 pairs of 2 and 5 and has 35 trailing zeroes.
Question 44. A number consists of 100 digits, all of which are 1. What remainder does it leave when divided by 9?
a) 0
b) 1
c) 2
d) 8
Answer:
b) 1 — A number and the sum of its digits leave the same remainder when divided by 9. The sum of the 100 digits is 100, which leaves a remainder of 1 when divided by 9.
Question 45. The product of two positive integers is 13,824, and their HCF is 12. What is their LCM?
a) 864
b) 1,024
c) 1,152
d) 1,296
Answer:
c) 1,152 — For two positive integers, their product equals the product of their HCF and LCM. Therefore, LCM = 13,824 ÷ 12 = 1,152.
Question 46. What is the remainder when 2¹⁰⁰⁰ is divided by 17?
a) 0
b) 1
c) 2
d) 16
Answer:
b) 1 — Since 2⁸ = 256 and 256 leaves a remainder of 1 when divided by 17, we have 2⁸ ≡ 1 (mod 17). As 1,000 = 8 × 125, 2¹⁰⁰⁰ = (2⁸)¹²⁵ ≡ 1.
Question 47. What is the remainder when 3¹⁰⁰ + 4¹⁰⁰ is divided by 7?
a) 0
b) 1
c) 2
d) 6
Answer:
b) 1 — Powers of 3 modulo 7 repeat after six terms. Since 100 leaves a remainder of 4 when divided by 6, 3¹⁰⁰ ≡ 3⁴ ≡ 4 (mod 7). Powers of 4 repeat after three terms, and 100 leaves a remainder of 1 when divided by 3, so 4¹⁰⁰ ≡ 4. Their sum is 8, which leaves a remainder of 1.
Question 48. What is the highest power of 72 that exactly divides 200!?
a) 46
b) 47
c) 48
d) 49
Answer:
c) 48 — Since 72 = 2³ × 3², calculate the exponents of 2 and 3 in 200!. We get v₂(200!) = 197 and v₃(200!) = 97. Therefore, the exponent of 72 is min(⌊197/3⌋, ⌊97/2⌋) = min(65, 48) = 48.
Question 49. How many trailing zeroes does 1,000! have when written in base 12?
a) 496
b) 497
c) 498
d) 499
Answer:
b) 497 — Since 12 = 2² × 3, the number of trailing zeroes in base 12 is min(⌊v₂(1000!)/2⌋, v₃(1000!)). Here, v₂(1000!) = 994 and v₃(1000!) = 498. Thus, the answer is min(497, 498) = 497.
Question 50. What is the least positive number that leaves remainders 3, 5 and 7 when divided by 5, 7 and 9, respectively?
a) 303
b) 308
c) 313
d) 318
Answer:
c) 313 — Each remainder is two less than the corresponding divisor. Therefore, the required number plus 2 must be divisible by 5, 7 and 9. Their LCM is 315, so the least required number is 315 – 2 = 313.
Question 51. What is the remainder when 10¹⁰⁰ + 10⁵⁰ + 1 is divided by 37?
a) 0
b) 1
c) 10
d) 36
Answer:
a) 0 — Since 10³ = 1,000 ≡ 1 (mod 37), powers of 10 repeat modulo 37 after three terms. Thus, 10¹⁰⁰ ≡ 10 and 10⁵⁰ ≡ 10² ≡ 26. Therefore, 10¹⁰⁰ + 10⁵⁰ + 1 ≡ 10 + 26 + 1 = 37 ≡ 0.
Question 52. How many positive factors does 10! have?
a) 240
b) 252
c) 270
d) 288
Answer:
c) 270 — Prime factorisation gives 10! = 2⁸ × 3⁴ × 5² × 7. Therefore, the number of positive factors is (8 + 1)(4 + 1)(2 + 1)(1 + 1) = 9 × 5 × 3 × 2 = 270.
Question 53. What is the sum of the digits of (10⁵⁰ – 1)²?
a) 441
b) 450
c) 459
d) 500
Answer:
b) 450 — For a positive integer n, (10ⁿ – 1)² consists of n – 1 digits equal to 9, followed by 8, then n – 1 zeroes and finally 1. Its digit sum is 9(n – 1) + 8 + 1 = 9n. For n = 50, the sum is 9 × 50 = 450.
Question 54. What is the remainder when 2²⁰²⁶ is divided by 31?
a) 1
b) 2
c) 4
d) 16
Answer:
b) 2 — Since 2⁵ = 32 ≡ 1 (mod 31), powers repeat after five terms. As 2,026 leaves a remainder of 1 when divided by 5, 2²⁰²⁶ ≡ 2¹ ≡ 2.
Question 55. What is the units digit of 1! + 2! + 3! + ⋯ + 2,026!?
a) 1
b) 3
c) 5
d) 7
Answer:
b) 3 — Every factorial from 5! onward ends in zero. Therefore, only 1! + 2! + 3! + 4! affects the units digit. Their sum is 1 + 2 + 6 + 24 = 33, whose units digit is 3.
Question 56. How many integers x satisfying 0 ≤ x < 35 fulfil the congruence x² ≡ 1 (mod 35)?
a) 2
b) 3
c) 4
d) 6
Answer:
c) 4 — Since 35 = 5 × 7, x² ≡ 1 modulo both 5 and 7. Thus, x can be congruent to either 1 or -1 modulo each prime. These independent choices produce 2 × 2 = 4 solutions modulo 35.
Question 57. What is the smallest positive integer having exactly 24 positive factors?
a) 240
b) 300
c) 360
d) 420
Answer:
c) 360 — To minimise the number, assign larger exponents to smaller primes. Since 24 = 4 × 3 × 2, the smallest suitable number has the form 2³ × 3² × 5¹ = 8 × 9 × 5 = 360. It has (3 + 1)(2 + 1)(1 + 1) = 24 factors.
Question 58. What is the remainder when 1! + 2! + 3! + ⋯ + 50! is divided by 12?
a) 3
b) 6
c) 9
d) 11
Answer:
c) 9 — Every factorial from 4! onward is divisible by 12. Therefore, the required remainder comes from 1! + 2! + 3! = 1 + 2 + 6 = 9.
Question 59. What is the highest power of 30 that exactly divides 100!?
a) 22
b) 23
c) 24
d) 25
Answer:
c) 24 — Since 30 = 2 × 3 × 5, its exponent in 100! is the minimum of the exponents of 2, 3 and 5. These are 97, 48 and 24, respectively. Therefore, the highest power is 30²⁴.
Question 60. What is the least perfect square that is exactly divisible by 8, 12 and 15?
a) 1,800
b) 2,400
c) 3,600
d) 7,200
Answer:
c) 3,600 — The LCM of 8, 12 and 15 is 120 = 2³ × 3 × 5. To make it a perfect square, all prime exponents must be even. Multiplying by 2 × 3 × 5 = 30 gives 2⁴ × 3² × 5² = 3,600 = 60².
