Free Topic-Wise General Studies MCQs
Master number system basics with Divisibility Rules and Remainder Theorem MCQs. Essential quantitative practice for the UPSC CSAT with detailed mathematical explanations.
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Explanation: Alternating sum for 4A51B2: (4+5+B) - (A+1+2) = 6+B-A. For divisibility by 11, this must be 0 or ±11. With A+B=9, B-A=5 gives A=2, B=7, making the alternating sum 11. Option 3 comes from solving A-B=6 (sign error). Option 5 comes from A+B=9 alone without divisibility. Option 7 is B's value, tempting if you solve for B instead of A.
Explanation: By Wilson's theorem, for prime p, (p-1)! ≡ -1 (mod p). Since 97 is prime, 96! ≡ -1 ≡ 96 (mod 97). Option 1 tempts those who confuse with Fermat's little theorem. Option 95 is (-2) from misremembering Wilson's as (p-2)!. Option 0 comes from thinking 97 divides 96! since 97 > 96.
Explanation: 91 = 7 × 13, so it is composite. 61, 73, and 89 are all prime. Option 89 tempts because it ends in 9 and students might test only divisibility by 2, 3, 5. Option 73 is a known prime but placed to distract. Option 61 is similarly safe. The trap is that 91 looks prime as it's not divisible by 2, 3, or 5.
Explanation: Any prime p > 3 is odd and not divisible by 3. Being odd, p² ≡ 1 (mod 8). Being not divisible by 3, p² ≡ 1 (mod 3). Thus p² - 1 is divisible by lcm(8,3) = 24. Option 6 catches those who only check divisibility by 3. Option 12 catches those who know about 3 and 4 but miss 8. Option 48 is too large and not always true (e.g., 5²-1=24).
Explanation: For 24 and 15, HCF is 3, not 1, so they are not co-prime. Their LCM would be 120, not 360. Option 8 and 45: 8=2³, 45=3²×5, LCM=360. Option 9 and 40: 9=3², 40=2³×5, LCM=360. Option 5 and 72: 5=5, 72=2³×3², LCM=360. The trap is seeing 24×15=360 and assuming they work.
Explanation: 10800 = 2⁵ × 3³ × 5². A perfect square factor must have even exponents: 2⁰,²,⁴ (3 choices), 3⁰,² (2 choices), 5⁰,² (2 choices). Total = 3×2×2 = 12. Option 8 comes from counting only square-free choices. Option 10 comes from 5+3+2=10 (adding exponents). Option 15 comes from total number of factors (6×4×3=72) divided by something arbitrary.
Explanation: n ≡ 4 (mod 7). Then 3n²+2n+5 ≡ 3(16)+8+5 = 61 ≡ 5 (mod 7). Option 0 tempts those who think n² ≡ 0. Option 3 comes from 3+2+5=10 ≡ 3 (adding coefficients blindly). Option 6 comes from 3(16)=48≡6, then forgetting to add the rest.
Explanation: 2³ = 8 ≡ 1 (mod 7). So 2^100 = 2^(99+1) = (2³)^33 × 2 ≡ 1 × 2 = 2 (mod 7). Option 1 comes from stopping at 2^99 ≡ 1. Option 4 comes from 2^100 = (2²)^50 = 4^50 and wrong cycle. Option 6 comes from thinking 2^3 ≡ -1 (mod 7) so 2^99 ≡ -1, 2^100 ≡ -2 ≡ 5, but that's not 6 either. Actually 6 comes from wrong cycle length.
Explanation: 36 = 4 × 9. For divisibility by 4, last two digits 0B must be divisible by 4, so B ∈ {4, 8} (non-zero). For B=4: digit sum = 13+A ≡ 0 (mod 9) gives A=5, so A+B=9. For B=8: digit sum = 17+A ≡ 0 (mod 9) gives A=1, so A+B=9. Option 7 comes from B=2 (but 02 is not div by 4). Option 8 comes from B=4, A=4 (wrong digit sum). Option 10 comes from B=8, A=2 (wrong digit sum).
Explanation: 7 cannot be written as sum of three odd primes because the smallest sum of three odd primes is 3+3+3 = 9. Option 9 = 3+3+3. Option 11 = 3+3+5. Option 13 = 3+3+7 or 3+5+5. The trap is thinking any odd number works, but 7 is too small.
Explanation: Solving step by step: N = 5k+3 ≡ 4 (mod 7) gives k ≡ 3 (mod 7), so N = 35m+18. Then 35m+18 ≡ 2 (mod 9) gives 8m ≡ 2 (mod 9), so m ≡ 7 (mod 9). Thus N = 315n + 263. Option 157 satisfies first two conditions but 157 mod 9 = 4, not 2. Option 298 satisfies first and third but 298 mod 7 = 4? Wait, 298 = 315-17, not in sequence. Actually 298 mod 7 = 4, yes, but 298 mod 9 = 1, not 2. Option 315 satisfies none except maybe first (315 mod 5 = 0).
Explanation: Trailing zeros require factors of 10 = 2×5. The primes between 10 and 50 are all odd and none equals 5, so the product contains no factor of 2 or 5. Option 1 tempts those who count the number of primes ending in 0 (none) or think 1 is standard. Option 2 comes from counting primes ending in 5 (only 1, but 15 isn't prime). Option 3 is arbitrary.
Explanation: Trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ... For n=85 to 89: ⌊n/5⌋ = 17, ⌊n/25⌋ = 3, total = 20. For n=90: ⌊90/5⌋ = 18, ⌊90/25⌋ = 3, total = 21. Option 85, 87, 88 all give exactly 20 zeros. Option 90 gives 21 zeros.
Explanation: 867 = 3×255 + 102; 255 = 2×102 + 51; 102 = 2×51 + 0. HCF is 51. Option 17 is a factor of 51 but not the HCF. Option 34 comes from (867-255)/18 = 34, arbitrary. Option 85 = 255/3, not related.
Explanation: Let numbers be 12a and 12b with HCF(a,b)=1. Then 144ab = 2160, so ab = 15. Factor pairs of 15 with co-prime factors: (1,15) and (3,5). These give number pairs (12,180) and (36,60). Option 1 misses one pair. Option 3 counts (5,3) separately or includes non-coprime. Option 4 counts all ordered pairs including (15,1) and (5,3) as distinct.
Explanation: 2n341 = 20341 + 1000n. Modulo 7: 20341 ≡ 6 (mod 7) and 1000 ≡ 6 (mod 7). So 6 + 6n ≡ 0 (mod 7), giving n ≡ 6 (mod 7). Thus n=6. Option 1: 21341/7 = 3048.7... Option 3: 23341/7 = 3334.4... Option 9: 29341/7 = 4191.5...
Explanation: 3^n has cycle 4: 3,9,7,1. 2025 mod 4 = 1, so unit digit is 3. 4^n has cycle 2: 4,6. 2025 is odd, so unit digit is 4. Sum = 7. Option 3 is just 3^2025. Option 5 comes from 3+2=5 (wrong cycle for 4). Option 9 comes from 3+6=9 (wrong position in 4's cycle).
Explanation: A number's remainder mod 9 equals its digit sum mod 9. Sum 1-9 = 45. Sum 10-19 = 55. Sum 20-29 = 65. Sum 30-39 = 75. Sum 40-49 = 85. Sum 50 = 5. Total = 330. 330 mod 9 = 6. Option 3 comes from 330 mod 9 = 3+3+0 = 6, but someone might do 3+3=6 then confuse. Actually 3 is just wrong. Option 5 is the last digit added. Option 8 is 330 mod 11 or similar.
Explanation: 180 = 2² × 3² × 5. Each co-prime pair must split the prime power factors completely. With 3 distinct primes, there are 2³ = 8 ordered pairs, or 2² = 4 unordered pairs: (1,180), (4,45), (9,20), (5,36). Option 2 counts only even splits. Option 3 misses one combination. Option 6 counts total factor pairs (9) or confuses with something else.
Explanation: 1001 = 7 × 11 × 13, so divisible by 13. 991 = 13×76 + 3. 997 = 13×76 + 9. 1009 = 13×77 + 8. Option 991 and 997 are primes near 1000. Option 1009 is also prime. The trap is that 1001 looks 'random' while 991 and 997 look like they could be divisible.
Explanation: By Fermat's little theorem, 5^16 ≡ 1 (mod 17). 100 = 6×16 + 4, so 5^100 ≡ 5^4 = 625 ≡ 13 (mod 17). Option 1 comes from thinking 5^16 ≡ 1 means any large power ≡ 1. Option 5 is the base. Option 16 is -1, from confusing with Wilson's or wrong exponent.
Explanation: Working backwards: q2 = 3k+1, q1 = 4(3k+1)+2 = 12k+6, N = 5(12k+6)+3 = 60k+33. Remainder is 33. Option 13 comes from adding remainders 3+2+1+7 (random). Option 23 comes from 5×4 + 3 = 23 (stopping early). Option 43 comes from 5×(4×2+2)+3 = 5×10+3 = 53... no, 43 is just wrong.
Explanation: φ(100) = 100 × (1-1/2) × (1-1/5) = 40. Option 20 is φ(100)/2 or just even numbers. Option 50 is 100/2 (numbers not divisible by 2). Option 60 is 100-40 or numbers not divisible by 5.
Explanation: 1!+2!+3!+4!+5!+6! = 1+2+6+24+120+720 = 873. 7! and above are divisible by 7. 873 mod 7: 7×124=868, remainder 5. Option 1 is 5! mod 7. Option 3 is 4! mod 7. Option 6 is 6! mod 7 or 3! mod 7.
Explanation: n³ - n = n(n-1)(n+1) is product of three consecutive integers, hence divisible by both 2 and 3, so by 6. Option n²+n = n(n+1) is always even but not always divisible by 3 (e.g., n=2 gives 6, n=3 gives 12, but n=1 gives 2). Option n³+n = n(n²+1), not always div by 3 (n=2 gives 10). Option n²-n+1: n=2 gives 3, not div by 6.
Explanation: In base 7, trailing zeros equal the exponent of 7 in the prime factorization of 50!. This is ⌊50/7⌋ + ⌊50/49⌋ = 7 + 1 = 8. Option 6 is ⌊50/7⌋ only. Option 7 is a guess. Option 10 is ⌊50/5⌋ (base 10 zeros).
Explanation: By definition, remainder must be non-negative and less than divisor. -17 = (-4)×5 + 3, so remainder is 3. Option -2 comes from -17/5 = -3.4, taking decimal part. Option 2 comes from 17 mod 5 = 2 and keeping sign. Option 4 is 5-1 or close.
Explanation: 111 = 3 × 37. 122 = 3×37 + 11. 133 = 3×37 + 22. 144 = 3×37 + 33. The pattern 111, 222, 333... are all divisible by 37 because 111 = 3×37. The trap is that 122, 133, 144 look like they follow a pattern but aren't divisible.
Explanation: Among four consecutive integers, there are two even numbers (one divisible by 4) and one multiple of 3. So the product is divisible by 4×2×3 = 24. For n=1: 1×2×3×4 = 24, which is not divisible by 48 or 120. Option 12 is too small. Option 48 fails for n=1. Option 120 = 5! is too large.
Explanation: Sum = n(n+1)/2. For divisibility by n, (n+1)/2 must be an integer, which requires n to be odd. There are 50 odd numbers from 1 to 100. Option 25 is 100/4. Option 75 is 3×25. Option 100 assumes all work.
Explanation: 1089 = 33². 1099 is between 33²=1089 and 34²=1156. 1111 is not a square (33²=1089, 34²=1156). 1127 is not a square. The trap is that 1111 looks like it could be 33.3² or has a pattern.
Explanation: Unit digit of 7^n cycles every 4: 7,9,3,1. We need 7^7 mod 4. Since 7 ≡ -1 (mod 4), 7^7 ≡ -1 ≡ 3 (mod 4). Thus unit digit is same as 7³ = 3. Option 1 comes from 7^7 mod 4 = 1 (wrong parity). Option 7 is 7^1. Option 9 is 7^2.
Explanation: For odd n, n²-1 = (n-1)(n+1). Both factors are even consecutive integers, so one is divisible by 4 and the other by 2, making the product divisible by 8. Option n²+1: n=3 gives 10, not div by 8. Option n²+n = n(n+1): n=3 gives 12, not div by 8. Option n²-n = n(n-1): n=3 gives 6, not div by 8.
Explanation: 48 = 2⁴×3, 72 = 2³×3², 120 = 2³×3×5. HCF = 2³×3 = 24. Option 12 is too small (misses a factor of 2). Option 36 = 72/2, not common to 48 or 120. Option 48 is the LCM of 48 and something, or just the first number.
Explanation: N ≡ 7 (mod 12), N ≡ 10 (mod 15), N ≡ 13 (mod 18). From first two: N = 60m + 55. Then 60m+55 ≡ 13 (mod 18) gives 6m ≡ 12 (mod 18), so m ≡ 2 (mod 3). Thus N = 180n + 175. Option 55 satisfies first two but 55 mod 18 = 1, not 13. Option 115: 115 mod 12 = 7, 115 mod 15 = 10, 115 mod 18 = 7, not 13. Option 235 = 175+60, satisfies first two but 235 mod 18 = 1, not 13.
Explanation: 720 = 2⁴×3²×5. A factor divisible by 12 = 2²×3 must have form 12×d where d|60. Since 60 = 2²×3×5 has (2+1)(1+1)(1+1) = 12 factors. Option 8 counts factors of 720/12=60 that are not multiples of 3 or 5. Option 10 is a guess. Option 15 is total factors of 720 divided by something.
Explanation: Sum = 20×21×41/6 = 2870. 2870 = 20×143 + 10, so remainder is 10. Option 0 comes from thinking n(n+1)(2n+1)/6 is always divisible by n. Option 15 comes from 20×21×41/6 mod 20 = 0×1×1/6 = 0, then wrong calculation. Option 19 is 20-1.
Explanation: Digit sum = A+6+7+9+B = A+B+22. For divisibility by 9, A+B+22 ≡ 0 (mod 9), so A+B ≡ 5 (mod 9). Smallest A is 1, giving B=4 (since B=13 is invalid). Option 2 gives A+B=3, not 5. Option 5 gives A+B=6, making sum 28, not div by 9. Option 7 gives A+B=8, making sum 30, not div by 9.
Explanation: For n ≥ 5, n! contains factors 4 and 5, so n! ≡ 0 (mod 20). Sum of 1!+2!+3!+4! = 1+2+6+24 = 33 ≡ 13 (mod 20). Option 3 is 33 mod 10. Option 7 is 1+2+4=7 (missing 3!). Option 17 is 20-3 or wrong addition.
Explanation: For any two numbers: Product = HCF × LCM. So HCF = 7425/495 = 15. Option 5 is 7425/1485 or 495/99. Option 10 is a divisor of 7425 but not correct. Option 25 is 7425/297.
Explanation: Alternating sum: (9+7+5+4+2) - (8+6+A+3+1) = 27 - (18+A) = 9-A. For divisibility by 11, 9-A = 0 or ±11. A=9 gives 0. Option 1 gives 8. Option 5 gives 4. Option 11 is invalid (not a digit).
Explanation: 96! ≡ -1 (mod 97) by Wilson's theorem. Since 96! = 96 × 95! ≡ -1 × 95! (mod 97), we get -95! ≡ -1, so 95! ≡ 1 (mod 97). Option 48 is (97-1)/2. Option 95 is -2 mod 97. Option 96 is -1 mod 97.
Explanation: N = 2×3×5×7×11 + 1. When divided by any of 2,3,5,7,11, it leaves remainder 1. Thus N is not divisible by any of them. This is the key idea in Euclid's proof of infinite primes. Option A and B are wrong because remainder is 1. Option D is absurd. Option C is the correct statement.
Explanation: 7 ≡ 2 (mod 5). Powers of 2 mod 5 cycle: 2,4,3,1 with period 4. Sum of one cycle = 10 ≡ 0 (mod 5). 50 terms = 12 complete cycles + 2 terms. Sum ≡ 2+4 = 6 ≡ 1 (mod 5). Option 0 comes from 12×0 = 0. Option 2 is the first term. Option 4 is the second term.
Explanation: 72 = 2³ × 3². For LCM(a,b) = 72, max exponents must be 3 and 2. For prime 2: 7 ordered exponent pairs give max=3. For prime 3: 5 ordered pairs give max=2. Total = 7×5 = 35. Option 15 is (3+1)(2+1) = 12, wrong formula. Option 21 is 7+5+9 or something. Option 45 is 9×5, using wrong count for 2.
Explanation: Using CRT with 15 = 3×5: 2^50 ≡ 1 (mod 3) and 2^50 ≡ 4 (mod 5). Solving x≡1(mod 3), x≡4(mod 5) gives x≡4(mod 15). Option 1 is just mod 3. Option 7 satisfies x≡1(mod 3) but 7 mod 5 = 2. Option 11 satisfies x≡1(mod 3) but 11 mod 5 = 1.
Explanation: 987654321 = 7 × 141093474 + 3. Option 1 comes from digital root or wrong step. Option 5 comes from 987654321 mod 7 = 321 mod 7 = 45 R 6... no, 321/7 = 45 R 6. Option 6 is last three digits mod 7.
Explanation: Factors of 180 divisible by 6 are 6×d where d|30. Sum of factors of 30 is 72. So sum = 6×72 = 432. Option 216 is 432/2. Option 360 is sum of all factors of 180. Option 540 is 6×90 or wrong.
Explanation: Pairing: (1-2)+(3-4)+...+(99-100) = 50×(-1) = -50. -50 ≡ 5 (mod 11) since -50 + 55 = 5. Option 0 comes from thinking pairs sum to 0. Option 6 comes from 50 mod 11 = 6 (ignoring sign). Option 10 comes from -50 ≡ -1 ≡ 10 (mod 11), but remainder is conventionally positive.
Explanation: n^5 - n = n(n-1)(n+1)(n²+1). Among any 5 consecutive integers, the product is divisible by 5! = 120. But more directly: n^5 ≡ n (mod 5) by Fermat's little theorem. Also n(n-1)(n+1) is divisible by 6. Thus n^5-n is divisible by 30. Option 10 misses factor 3. Option 15 misses factor 2. Option 60 is not always true (n=2 gives 30).
Explanation: 7^4 = 2401 ≡ 1 (mod 100). Since 2024 is divisible by 4, 7^2024 = (7^4)^506 ≡ 1 (mod 100). Last two digits are 01. Option 07 is 7^1. Option 49 is 7^2. Option 43 is 7^3 mod 100.
Explanation: 3^(3^3) = 3^27. Unit digit cycle of 3 is 4: 3,9,7,1. 27 mod 4 = 3, so unit digit is 7. Option 3 is 3^1. Option 9 is 3^2 or 3^27 mod 4 = 3, then picking position 2. Option 1 is 3^0 or 3^4.
Explanation: Product = (25!)². Zeros in 25! = ⌊25/5⌋ + ⌊25/25⌋ = 6. So (25!)² has 2×6 = 12 zeros. Option 6 is zeros in 25!. Option 10 is ⌊25/5⌋ × 2. Option 25 is just 25.
Explanation: By Fermat's little theorem, n^13 ≡ n (mod 13) for all integers n. Thus n^13 - n is always divisible by 13. Option n^12-1 is divisible by 13 only if n is not divisible by 13 (then n^12 ≡ 1). Option n^13+n ≡ 2n (mod 13), not always 0. Option n^12+1 ≡ 2 (mod 13) when n not div by 13.
Explanation: Numbers are 2x, 3x, 4x. LCM = x × LCM(2,3,4) = 12x = 240, so x = 20. HCF = x × HCF(2,3,4) = 20 × 1 = 20. Option 10 is 240/24. Option 40 is 2x. Option 60 is 3x.
Explanation: By Wilson's theorem, 22! ≡ -1 (mod 23). Since 22! = 22×21×20! ≡ (-1)(-2)×20! ≡ 2×20! (mod 23), we get 2×20! ≡ -1, so 20! ≡ -12 ≡ 11 (mod 23). Option 1 is wrong application. Option 12 is -11 or inverse confusion. Option 22 is -1 mod 23.
Explanation: a²-b² = (a-b)(a+b). For odd a,b, both a-b and a+b are even. Moreover, one of them is divisible by 4 (since (a+b)/2 and (a-b)/2 are consecutive integers). Thus product is divisible by 8. Option a²+b²: a=3,b=1 gives 10, not div by 8. Option a³+b³: a=3,b=1 gives 28, not div by 8. Option a³-b³: a=3,b=1 gives 26, not div by 8.
Explanation: 0.123... = 123/999 = 41/333. Modulo 13: 41 ≡ 2, 333 ≡ 8. Inverse of 8 mod 13 is 5. So 41/333 ≡ 2×5 = 10 (mod 13). Option 2 is just numerator. Option 8 is denominator. Option 12 is -1 mod 13.
Explanation: 360 = 2³×3²×5 has 4×3×2 = 24 factors. Factors divisible by 6 must have at least 2¹ and 3¹: 3 choices for 2's exponent, 2 for 3's, 2 for 5's = 12 factors. Not divisible by 6: 24-12 = 12. Option 8 is a guess. Option 10 is wrong subtraction. Option 16 is factors divisible by 2 or 3 but not necessarily both.
Explanation: Using mod 11: 1^5≡1, 2^5≡-1, 3^5≡1, 4^5≡1, 5^5≡1, 6^5≡-1, 7^5≡-1, 8^5≡-1, 9^5≡1, 10^5≡-1. Sum = 0. Option 1 is first term. Option 5 is number of +1 terms minus something. Option 10 is -1 mod 11.