Number System - Practice MCQs for CCAT

Correct Answer: B - 1275

Sum of first n natural numbers = n(n+1)/2 = 50×51/2 = 1275.

Correct Answer: B - 28 and 17

Let numbers be x and y. x-y=11 and (x+y)/5=9, so x+y=45. Solving: x=28, y=17.

Correct Answer: D - 7

7¹=7, 7²=49, 7³=343, 7⁴=2401. Pattern repeats every 4. 95÷4=23 rem 3. So unit digit is 3.

Correct Answer: D - 60

12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60.

Correct Answer: B - 12

36=2²×3², 48=2⁴×3, 60=2²×3×5. HCF = 2²×3 = 12.

Correct Answer: C - 1

2¹≡2(mod 3), 2²≡1(mod 3). Pattern: 2,1,2,1... 100 is even, so remainder is 1.

Correct Answer: A - 20

HCF × LCM = Product. So LCM = 120/6 = 20.

Correct Answer: B - 15

Primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47. Total = 15.

Correct Answer: A - 36

LCM of 12 and 18 = 36.

Correct Answer: A - 12

Since 3 and 4 are coprime, LCM(3,4) = 12. So divisible by 12.

Correct Answer: B - 2550

Sum of even numbers from 2 to 100 = 2+4+6+...+100 = 2(1+2+3+...+50) = 2×50×51/2 = 2550.

Correct Answer: C - 9968

9999 ÷ 88 = 113 remainder 55. So 9999 - 55 + 88 = 9968 or simply 88 × 113 = 9944, but 88 × 113 = 9944, and 9999 - 31 = 9968.

Correct Answer: A - 8

The powers of 3 repeat unit digits 3, 9, 7, 1, and 123 leaves remainder 3 when divided by 4, so 3¹²³ ends in 7. The powers of 7 repeat 7, 9, 3, 1, and 456 is divisible by 4, so 7⁴⁵⁶ ends in 1. The sum ends in 8.

Correct Answer: B - 14

100 ÷ 7 = 14.28. So there are 14 numbers divisible by 7 from 1 to 100 (7, 14, 21, ..., 98).

Correct Answer: B - 9 and 16

x + y = 25, xy = 144. These are roots of t² - 25t + 144 = 0. Solving: (t-9)(t-16) = 0. Numbers are 9 and 16.

Correct Answer: D - 1

17 ≡ -1 (mod 18). So 17²⁰⁰ ≡ (-1)²⁰⁰ ≡ 1 (mod 18). Remainder is 1.

Correct Answer: D - 24

Zeros = [100/5] + [100/25] + [100/125] = 20 + 4 + 0 = 24.

Correct Answer: B - 45

For 45!: [45/5] + [45/25] = 9 + 1 = 10 zeros. For 40!: [40/5] + [40/25] = 8 + 1 = 9. So n = 45.

Correct Answer: C - 84

Let digits be x and y. (10x+y) - (10y+x) = 36 → 9(x-y) = 36 → x-y = 4. x+y = 12. So x=8, y=4. Number = 84.

Correct Answer: A - 17

420 = 2² × 3 × 5 × 7. Prime factors are 2, 3, 5, 7. Sum = 2 + 3 + 5 + 7 = 17.

Correct Answer: B - 2

Powers of 2 modulo 7 repeat every 3 terms: 2, 4, 1. Since 100 leaves remainder 1 when divided by 3, 2¹⁰⁰ has the same remainder as 2¹, which is 2.

Correct Answer: C - 72

Product of numbers = HCF × LCM. So other number = (12 × 360) / 60 = 72.

Correct Answer: A - 7

Primes: 53, 59, 61, 67, 71, 73, 79 = 7 primes.

Correct Answer: B - 7

Unit digits of powers of 7 cycle: 7,9,3,1. Period=4. 245 mod 4 = 1. So unit digit = 7.

Correct Answer: A - 17

Numbers with remainder 3 when ÷7: 3,10,17,24... Check ÷5: 17÷5=3R2. So 17.

Correct Answer: A - 9944

9999 ÷ 88 = 113.625. 113 × 88 = 9944.

Correct Answer: B - 2

Numbers are 9a and 9b where a+b=5 and HCF(a,b)=1. Pairs: (1,4) and (2,3). So 2 pairs.

Correct Answer: D - 37

LCM(6,9,12) = 36. Required number = 36 + 1 = 37.

Correct Answer: C - 24

Number of factors = (3+1)(2+1)(1+1) = 4×3×2 = 24.

Correct Answer: D - 1

17 ≡ 1 (mod 16). So 17²³ ≡ 1²³ = 1 (mod 16).

Correct Answer: B - Both A and C

117 = 9×13. HCF(9,13)=1 ✓. Also 1×117, HCF=1 ✓. 3×39: HCF(3,39)=3 ✗.

Correct Answer: A - 360

120=2³×3×5. Sum=(1+2+4+8)(1+3)(1+5)=15×4×6=360.

Correct Answer: B - 13

1001 = 7 × 11 × 13. Largest prime factor is 13.

Correct Answer: C - 6

Divisible by LCM(3,5)=15. Numbers: 15,30,45,60,75,90 = 6 numbers.

Correct Answer: D - Both 3 and 9

A number whose digit sum is 9 is divisible by both 3 and 9.

Correct Answer: D - 1275

Sum = n(n+1)/2 = 50×51/2 = 1275.

Correct Answer: D - 12

Trailing zeros = floor(50/5) + floor(50/25) = 10 + 2 = 12.

Correct Answer: C - 2520

LCM(1,2,...,10) = 2520.

Correct Answer: C - 12

LCM(4,6) = 12. Number must be divisible by 12.

Correct Answer: D - 24

48=2⁴×3, 72=2³×3², 120=2³×3×5. GCD=2³×3=24.

Correct Answer: D - 128

First: 105 (7×15), Last: 994 (7×142). Count = 142-15+1 = 128.

Correct Answer: C - Cannot be determined

Numbers are 7a and 7b where a-b=2, HCF(a,b)=1. Multiple pairs possible: (3,1),(5,3),(7,5)...

Correct Answer: B - 0

Sum of digits = 1+2+3+4+5+6+7+8+9 = 45. 45÷9=5R0. Remainder is 0.

Correct Answer: A - 998001

(1000-1)² = 1000000 - 2000 + 1 = 998001.

Correct Answer: C - 11

√100=10, √400=20. Squares: 10² to 20² = 100,121,...,400. Count = 20-10+1 = 11.

Correct Answer: A - 31

LCM = 12×HCF. LCM + HCF = 403. 12H + H = 403. 13H = 403. H = 31.

Correct Answer: C - 3 and 11

2¹⁰ - 1 = 1023 = 3 x 341 = 3 x 11 x 31. It is divisible by 3 and 11, but not by 7.

Correct Answer: A - 1

Sum of cubes = [n(n+1)/2]² = [55]² = 3025. 3025 ÷ 4 = 756 R1.

Correct Answer: C - 101

101 is prime (not divisible by 2,3,5,7). Check: 101/7≈14.4, 101/11≈9.2. It is prime.

Correct Answer: A - 20

Trailing zeros in n! are counted by floor(n/5)+floor(n/25)+... . For n=20, the count is 4+0=4. For n=19, it is only 3, so the smallest such n is 20.

Correct Answer: C - 7

7¹ mod 12 = 7 and 7² mod 12 = 1. Odd powers leave remainder 7.

Correct Answer: C - 31

Trailing zeros = floor(125/5)+floor(125/25)+floor(125/125)=25+5+1=31.

Correct Answer: A - 3⁴⁸

Exponent of 3 in 100! = 33+11+3+1 = 48.

Correct Answer: A - 12

Square divisors use even exponents: for 2 choose 0,2,4 (3 ways), for 3 choose 0,2 (2 ways), for 5 choose 0,2 (2 ways). Total = 3 x 2 x 2 = 12.

Correct Answer: A - 17

7896 divided by 37 gives remainder 20. Required addition = 37 - 20 = 17.

Correct Answer: C - 7

Unit digit cycle of 3 is 3,9,7,1. 47 mod 4 = 3, so unit digit is 7.

Correct Answer: C - 7

N = 9k+4. N² mod 9 = 4² = 16 mod 9 = 7.

Correct Answer: A - 36

Divisible by 6: 30, by 10: 18, by both 30: 6. Exactly one = 30+18-2*6 = 36.

Correct Answer: C - 108

Product of numbers = HCF x LCM = 18 x 540. Other = 9720/90 = 108.

Correct Answer: A - 540

LCM = 2² x 3³ x 5 = 540.