Set B Solutions

B1. Answer: (b) 1 Last three digits = 536; 536÷8=67 exact, so divisibility by 8 holds for any x (doesn’t involve x). For divisibility by 9: digit sum =1+x+2+5+3+6=17+x. Nearest multiple of 9 ≥17 is 18 → x=1. Number = 112536; 112536÷72=1563 exact. ✓

B2. Answer: (b) 2 Group the digits in 3s from the right: 296, 753, 4. Alternating sum =296-753+4=-453. Since -453 and 4753296 leave related remainders mod 7, reduce 453: 7×64=448, so 453 leaves remainder 5, meaning -453 leaves remainder 7-5=2 (mod 7). Direct check: 7×679042=4753294; 4753296-4753294=2. Remainder = 2.

B3. Answer: (b) Only 1,001,000 For 6,214,357: groups from right are 357, 214, 6. Alternating sum =357-214+6=149; 149÷13=11.46 → not divisible by 13 (direct check: 6214357÷13=478027.46, confirmed not divisible). For 1,001,000: this equals 1001×1000, and since 1001=7×11×13, it is divisible by 13 (in fact by 7, 11, and 13 all at once). So only 1,001,000 is divisible by 13.

B4. Answer: (d) 31 ⌊125/5⌋=25, ⌊125/25⌋=5, ⌊125/125⌋=1, ⌊125/625⌋=0. Total=25+5+1=31.

B5. Answer: (c) 41 ⌊173/5⌋=34, ⌊173/25⌋=6, ⌊173/125⌋=1, ⌊173/625⌋=0. Total=34+6+1=41.

B6. Answer: (c) 22 ⌊50/3⌋=16, ⌊50/9⌋=5, ⌊50/27⌋=1, ⌊50/81⌋=0. Total=16+5+1=22.

B7. Answer: (c) 7 Since n is odd, a^n+bn is always divisible by a+b. Here a+b=5+2=7.

B8. Answer: (b) 1 2^3=8≡17 (cycle length 3). 96=3×32 exactly → 2^(96)=(23)^(32)≡1(32)=17.

B9. Answer: (c) 100 As derived in Section 4: trailing zeros stay at 22 for n=95 to 99, then jump to 24 at n=100 (since 100 contributes two factors of 5, being 4×25). So the smallest n giving exactly 24 zeros is 100.

B10. Answer: (a) 1 5634 ÷ 7 = 804.857; 7×804=5628; 5634-5628=6 (current remainder). To reach the next multiple of 7 (7×805=5635), we need to add 5635-5634=1. Least number to add = 1.

B11. Answer: (c) 6 The last three digits are “13x”. Test each option: 130÷8 not exact; 132÷8 not exact; 134÷8 not exact; 136÷8=17 exact ✓; 138÷8 not exact. So x=6, giving last three digits 136. Check: 482136 ÷ 8 = 60267 exact. ✓

B12. Answer: (c) 24 Standard result: product of any 4 consecutive integers is divisible by 4!=24.

B13. Answer: (b) 3 With b=0: number is 63a50. Divisible by 5 automatically (ends in 0). For divisibility by 3: digit sum =6+3+a+5+0=14+a. For a=0..9, 14+a divisible by 3 when 14+a 15,18,21 → a1,4,7. That’s 3 possible values.

B14. Answer: (b) 1 7≡25. Cycle of 2^k mod 5: 2,4,3,1 (length 4). 84=4×21 exactly → 2^(84)≡24≡15 (since it completes whole cycles, matching the value at the end of one full cycle, which is 1). So 7^(84)≡15.

B15. Answer: (d) 11 Given: number =342k+47 for some integer k, and 342=18×19. So number =18(19k)+47. Now reduce 47 modulo 18: 18×2=36, so 47=18×2+11. Thus number =18(19k+2)+11, which leaves remainder 11 when divided by 18.

B16. Answer: (c) 8 72=8×9. Last three digits are “73y” =730+y. Since 730=8×91+2, we need 2+y≡08, i.e., y≡68; the only valid digit is y=6. Digit sum =5+x+4+7+3+6=25+x. Nearest multiple of 9 ≥25 is 27 → x=2. Number =524736; check 524736÷72=7288 exact. ✓ x+y=2+6=8.

B17. Answer: (c) 2 Group 8,462,573 into 3-digit blocks from the right: 573, 462, 8. Alternating sum =573-462+8=119. 119÷13=9 remainder 2 (since 13×9=117). Direct check confirms 8462573÷13 leaves remainder 2. Remainder = 2.

B18. Answer: (b) 48 6=2×3. E_2(100!): ⌊100/2⌋+⌊100/4⌋+⌊100/8⌋+⌊100/16⌋+⌊100/32⌋+⌊100/64⌋ = 50+25+12+6+3+1=97. E_3(100!): ⌊100/3⌋+⌊100/9⌋+⌊100/27⌋+⌊100/81⌋=33+11+3+1=48. Since each “6” needs one 2 and one 3, the limiting factor is (97,48)=48.

B19. Answer: (d) 30 By Wilson’s Theorem, for prime p=31: (p-1)!=30! ≡ -1 (mod 31), i.e., 30! leaves remainder 31-1=30 when divided by 31.

B20. Answer: (c) 720 General rule: the product of n consecutive integers is always divisible by n!. For n=6, 6!=720.

B21. Answer: (b) 727 The number must be 7 more than a common multiple of 12, 16, 18, 30. 12=2^2×3, 16=2^4, 18=2×3^2, 30=2×3×5. LCM=2^4×32×5=16×9×5=720. Smallest number =720+7=727. Check: 727÷12=60 remainder 7; 727÷16=45 remainder 7; 727÷18=40 remainder 7; 727÷30=24 remainder 7. ✓

B22. Answer: (c) 200 By inclusion-exclusion: (divisible by 4) + (divisible by 6) − (divisible by both, i.e., by LCM(4,6)=12). ⌊600/4⌋=150, ⌊600/6⌋=100, ⌊600/12⌋=50. Total =150+100-50=200.

B23. Answer: (a) 0 Since the digit sum (45) is itself divisible by 9, the number and its digit sum leave the same remainder mod 9 — that remainder is 0 (the number is exactly divisible by 9).

B24. Answer: (d) 30 Here a+b=17+13=30, and the exponent n=15 is odd, so by Rule 3 (Section 3), a^n+bn is always divisible by a+b=30.

B25. Answer: (c) 4 Powers of 3 mod 7 cycle: 3^1=3,32=2,3^3=6,34=4,3^5=5,36=1 (cycle length 6). 250 = 6×41+4, so 3^(250)≡34≡47. Remainder = 4.

B26. Answer: (c) 6 Number 3 x 2 8 4 7 1, digits from right: 1(1),7(2),4(3),8(4),2(5),x(6),3(7). Odd positions (1,3,5,7): 1+4+2+3=10. Even positions (2,4,6): 7+8+x=15+x. Difference =10-(15+x)=-5-x. For divisibility by 11: -5-x=0⇒ x=-5 (invalid), or -5-x=-11⇒ x=6. x=6, giving 3,628,471; check 3628471÷11=329861 exact. ✓

B27. Answer: (d) 21 2520=2^3×32×5×7. Listing all factors strictly between 10 and 100: 12, 14, 15, 18, 20, 21, 24, 28, 30, 35, 36, 40, 42, 45, 56, 60, 63, 70, 72, 84, 90 — counting gives 21 factors.

B28. Answer: (b) Only 4,58,731 458731 = 13×35287 exactly, so it is divisible by 13. Checking 458730: since 458731 is the multiple of 13, 458730 is one less, so it leaves remainder 12 on division by 13 — not divisible. Only 4,58,731 works.

B29. Answer: (c) 7 Powers of 2 mod 9 cycle: 2,4,8,7,5,1 (cycle length 6). 100=6×16+4, so 2^(100)≡24≡16≡79. Remainder = 7.

B30. Answer: (c) 365 The number must be 5 more than a common multiple of 8, 9, 10. 8=2^3, 9=3^2, 10=2×5, so LCM=2^3×32×5=360. Smallest number =360+5=365. Check: 365÷8=45 remainder 5; 365÷9=40 remainder 5; 365÷10=36 remainder 5. ✓

End of Chapter 2. Aspirants should re-attempt Set B without looking at solutions after a 24-hour gap to test retention, and revisit the Legendre’s Formula section before attempting factorial-based questions in Number System mock tests.