Set B Solutions

B1. Answer: (b) 3 Unit digit of 127^173: base unit digit 7, cycle {7,9,3,1} length 4. 173÷4 = remainder 1 → 1st term = 7. Unit digit of 56^43: base unit digit 6, cycle length 1 (always 6) → unit digit 6. Sum of unit digits = 7+6 = 13 → unit digit of the total = 3.

B2. Answer: (b) 12 144 = 2⁴×3²; 180 = 2²×3²×5; 192 = 2⁶×3. HCF = 2^min(4,2,6) × 3^min(2,2,1) = 2² × 3¹ = 4×3 = 12. (5 is not common to all three, so excluded.)

B3. Answer: (b) 55 Product of the two numbers = HCF × LCM = 5 × 495 = 2475. Other number = 2475 ÷ 45 = 55.

B4. Answer: (c) 36 1800 = 2³ × 3² × 5². Number of factors = (3+1)(2+1)(2+1) = 4×3×3 = 36.

B5. Answer: (b) 2 2³ = 8 ≡ 1 (mod 7). 100 = 3×33+1, so 2^100 = (2³)^33 × 2¹ ≡ 1×2 = 2 (mod 7). Remainder = 2.

B6. Answer: (a) 01 17² = 289 → last two digits 89. 17⁴ = 89² = 7921 → last two digits 21 (an “ends in 1” pattern, tens digit 2). 400 = 4×100, so 17^400 = (17⁴)^100 = (…21)^100. Tens digit of result = (2 × 100) mod 10 = 200 mod 10 = 0 → last two digits = 01.

B7. Answer: (d) 4 Let the number = 5k+3. Its square = 25k² + 30k + 9. Both 25k² and 30k are divisible by 5, so the remainder comes only from 9 mod 5 = 4.

B8. Answer: (b) 16 Subtract the remainder from both numbers: 245−5 = 240; 1029−5 = 1024. Find HCF(240, 1024): 240 = 2⁴×3×5; 1024 = 2¹⁰. Common factor: 2^min(4,10) = 2⁴ = 16.

B9. Answer: (a) 0 Sum of first 15 natural numbers = 15×16/2 = 120. 120 ÷ 8 = 15 exactly, remainder = 0.

B10. Answer: (c) 18 N = 2³×3²×5. Total factors = (3+1)(2+1)(1+1) = 4×3×2 = 24. Odd factors (power of 2 fixed at 0) = (1)(3)(2) = 6. Even factors = Total − Odd = 24 − 6 = 18.

B11. Answer: (b) 40 LCM(6, 9, 12): 6=2×3, 9=3², 12=2²×3. LCM = 2²×3² = 36. Since the same remainder (4) is required with each divisor, and 4 < each divisor, the smallest such number = LCM + remainder = 36 + 4 = 40.

B12. Answer: (a) 1 7 mod 5 = 2. Cycle of 2’s powers mod 5: 2,4,3,1,2,4,3,1,… (length 4). 84 ÷ 4 = 21 remainder 0 → take the last (4th) term = 1.

B13. Answer: (c) 60 Let the numbers be 3x and 4x. Since 3 and 4 are co-prime, HCF of the numbers = x = 5. So the numbers are 15 and 20. LCM = product of numbers ÷ HCF = (15×20)/5 = 300/5 = 60. (Shortcut: LCM = 3×4×HCF = 3×4×5 = 60.)

B14. Answer: (c) 24 Number of trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + … For n=100: ⌊100/5⌋=20, ⌊100/25⌋=4, ⌊100/125⌋=0. Total = 20+4 = 24.

B15. Answer: (a) 1080 LCM(12, 15, 18): 12=2²×3, 15=3×5, 18=2×3². LCM = 2²×3²×5 = 180. Smallest 4-digit number: 1000 ÷ 180 = 5.55…, so the next whole multiple is the 6th multiple: 6 × 180 = 1080.

B16. Answer: (c) 8 Unit digit of 18^47: base unit digit 8, cycle {8,4,2,6} length 4. 47÷4 = 11 remainder 3 → 3rd term = 2. Unit digit of 24^36: base unit digit 4, cycle {4,6} length 2. 36÷2 = remainder 0 → last (2nd) term = 6. Sum of unit digits = 2 + 6 = 8 → unit digit = 8.

B17. Answer: (b) 21 210 = 2×3×5×7; 315 = 3²×5×7; 462 = 2×3×7×11. Common primes across all three: 3 (min power 1) and 7 (min power 1). (2 is absent from 315; 5 is absent from 462; 11 is absent from 210 and 315.) HCF = 3¹ × 7¹ = 21.

B18. Answer: (b) 450 cm This is an LCM problem: LCM(45, 50, 75). 45 = 3²×5; 50 = 2×5²; 75 = 3×5². LCM = 2¹ × 3² × 5² = 2 × 9 × 25 = 450. Check: 450÷45=10, 450÷50=9, 450÷75=6 — all exact.

B19. Answer: (c) 3 Build the cycle of powers of 5 mod 7 (period divides 6, since 7 is prime): 5¹≡5, 5²≡25≡4, 5³≡20≡6, 5⁴≡30≡2, 5⁵≡10≡3, 5⁶≡15≡1 (mod 7) — cycle {5,4,6,2,3,1}, length 6. 203 ÷ 6 = 33 remainder 5 → 5th term of the cycle = 3.

B20. Answer: (a) 07 63 ends in 3 (Case B): 63² = 3969 → last two digits 69. 63⁴ = 69² = 4761 → last two digits 61 (an “ends in 1” number, tens digit 6). 75 = 4×18 + 3, so 63^75 = (63⁴)^18 × 63³. (…61)^18: tens digit = (6×18) mod 10 = 108 mod 10 = 8 → gives “81”. 63³ = 63² × 63 → last two digits: 69 × 63 = 4347 → last two digits 47. Multiply: 81 × 47 = 3807 → last two digits = 07. (Verified independently via repeated squaring mod 100: 63²≡69, 63⁴≡61, 63⁸≡21, 63¹⁶≡41, 63³²≡81, 63⁶⁴≡61; 75=64+8+2+1, so 63⁷⁵ ≡ 61×21×69×63 ≡ 81×69×63 ≡ 89×63 ≡ 07 (mod 100) — matches.)

B21. Answer: (c) 24 2²×3³×5²×7. Odd factors exclude the factor of 2 entirely, so the exponent of 2 is fixed at 0 (1 choice). Remaining choices: exponent of 3 (0 to 3 → 4 choices), exponent of 5 (0 to 2 → 3 choices), exponent of 7 (0 to 1 → 2 choices). Odd factors = 1 × 4 × 3 × 2 = 24.

B22. Answer: (c) 8 16^0.75 = 16^(3/4) = (16^(1/4))³ = 2³ = 8 (since the 4th root of 16 is 2).

B23. Answer: (c) 64 (2^(−3) × 2⁷) ÷ 2^(−2) = 2^(−3+7) ÷ 2^(−2) = 2⁴ ÷ 2^(−2) = 2^(4−(−2)) = 2⁶ = 64. Direct check: 2^(−3)=1/8, 2⁷=128, product=16; 2^(−2)=1/4; 16 ÷ (1/4) = 64 ✓.

B24. Answer: (c) 122 37 ÷ 5 = 7 remainder 2; 7 ÷ 5 = 1 remainder 2; 1 ÷ 5 = 0 remainder 1. Reading bottom to top: 122 (base 5). Check: 1×25 + 2×5 + 2×1 = 25+10+2 = 37 ✓.

B25. Answer: (c) 62 Trailing zeros in 250! = ⌊250/5⌋ + ⌊250/25⌋ + ⌊250/125⌋ + ⌊250/625⌋ = 50 + 10 + 2 + 0 = 62.

B26. Answer: (c) 6 Number: 4,8,2,3,x,1 (from left). From the right, positions are: 1(pos1), x(pos2), 3(pos3), 2(pos4), 8(pos5), 4(pos6). Sum at odd positions (1,3,5) = 1+3+8 = 12. Sum at even positions (2,4,6) = x+2+4 = x+6. Difference = 12 − (x+6) = 6 − x, which must be 0 or a multiple of 11. Since x is a digit (0–9), 6−x ranges from −3 to 6, so only 6−x=0 works → x = 6.

B27. Answer: (c) 16 Let the numbers be 5y and 6y. Since 5 and 6 are co-prime, their HCF = y. LCM = 5 × 6 × y = 30y. Given LCM = 480 → 30y = 480 → y = 16. So HCF = 16. (Check: numbers are 80 and 96; LCM(80,96): 80=2⁴×5, 96=2⁵×3, LCM=2⁵×3×5=480 ✓; HCF=2⁴=16 ✓.)

B28. Answer: (b) 13 Subtract the remainders from the respective numbers: 70−5 = 65; 125−8 = 117. HCF(65, 117): 65 = 5×13; 117 = 3²×13. Common factor = 13. Check: 70÷13 = 5 remainder 5 ✓ (13×5=65, 70−65=5); 125÷13 = 9 remainder 8 ✓ (13×9=117, 125−117=8).

B29. Answer: (d) 64 Pattern is n²+1 for n = 1, 2, 3, …: 1²+1=2, 2²+1=5, 3²+1=10, 4²+1=17, 5²+1=26, 6²+1=37, 7²+1=50, 8²+1=65. The given series has 64 in the 8th position instead of 65 — so 64 is the wrong number.

B30. Answer: (d) 117 For any two co-prime numbers, HCF = 1. Using HCF × LCM = product: 1 × LCM = 117 → LCM = 117.

End of Chapter 1: Number System.