Number System MCQs for UPSC Prelims

Write the 11-digit number as N = 3798125 x 10000 + P x 1000 + 369 = 37981250369 + 1000P. Now reduce mod 7. 37981250369 = 7 x 5425892910 - 1, so 37981250369 leaves remainder 6 (that is, -1) mod 7. Also 1000 leaves remainder 6 mod 7. So N mod 7 = 6 + 6P (mod 7). For divisibility we need 6 + 6P congruent to 0 mod 7, i.e. 6P congruent to 1 mod 7. Since 6 is -1 mod 7, this is -P congruent to 1, giving P congruent to 6 mod 7, so P = 6. Testing the other options: P = 1, 7 or 9 each leaves a non-zero remainder, so only 6 works.

35^n = 5^n x 7^n, so count factors of 5 and 7 in 7 x 343 x 385 x 1000 x 2401 x 77777. Factors of 5: 385 = 5 x 7 x 11 gives one 5, 1000 = 2^3 x 5^3 gives three 5s, total 4. Factors of 7: leading 7 (one), 343 = 7^3, 385 (one), 2401 = 7^4, 77777 = 7 x 11111 (one); total 1+3+1+4+1 = 10. The number of available 35s is limited by the scarcer prime, min(4, 10) = 4. So maximum n = 4. Answer: (b).

Option (c) is correct. If x divides 7x + 96, then since x always divides 7x, x must divide the difference, which is 96. So x can be any positive divisor of 96 = 2^5 x 3, and the number of divisors is (5 + 1)(1 + 1) = 12. Options (a) 10 and (b) 11 miscount the divisors (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96). Option (d) is wrong because the divisibility condition bounds x to divisors of 96, so the possibilities are finite.

If 1186 leaves remainder 31 on division by n, then n divides 1186 - 31 = 1155 and n must exceed the remainder, so n > 31. Factorise: 1155 = 3 x 5 x 7 x 11, which has 16 divisors: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Those greater than 31 are 33, 35, 55, 77, 105, 165, 231, 385 and 1155 - nine numbers, option (d). Options (a), (b) and (c) come from miscounting the divisor list or forgetting the divisor must be larger than 31.

We need three-digit numbers with strictly decreasing digits (hundreds > tens > units). The number 1000 has digits 1,0,0,0 and does not qualify, so the hundreds digit can only be 7, 8 or 9. For a fixed hundreds digit H, the tens and units digits must be two distinct digits both less than H, arranged in decreasing order; choosing any 2 distinct digits from {0,1,...,H-1} fixes their order, giving C(H, 2) numbers. For H = 7: C(7,2) = 21. For H = 8: C(8,2) = 28. For H = 9: C(9,2) = 36. Total = 21 + 28 + 36 = 85. Hence (c) 85.

Conditions: P = 3T with T not 0, S = Q + 4 and Q = R + 3 (so S = R + 7), and all five digits distinct. P = 3T allows only T = 1 (P = 3), T = 2 (P = 6) or T = 3 (P = 9). S = R + 7 with S a digit allows only R = 0, 1 or 2, giving (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9). Now combine and reject clashes. T = 1, P = 3: (0, 3, 7) clashes with P = 3; (1, 4, 8) clashes with T = 1; (2, 5, 9) works, giving 35291. T = 2, P = 6: (0, 3, 7) works, giving 63072; (1, 4, 8) works, giving 64182; (2, 5, 9) clashes with T = 2. T = 3, P = 9: (0, 3, 7) clashes with T = 3; (1, 4, 8) works, giving 94183; (2, 5, 9) clashes with P = 9. Valid numbers: 35291, 63072, 64182, 94183, a total of 4. The official UPSC key confirms option (b).

99% of the 400 total is 396, so the four integer scores must add to 396, i.e. the shortfalls from 100 in the four papers must add to 4 with each shortfall between 0 and 100. The number of non-negative integer solutions of d1 + d2 + d3 + d4 = 4 is C(4 + 3, 3) = C(7, 3) = 35 (the cap of 100 never binds since each shortfall is at most 4). Enumerating score patterns confirms it: (100,100,100,96) in 4 ways, (100,100,99,97) in 12, (100,100,98,98) in 6, (100,99,99,98) in 12, (99,99,99,99) in 1, totalling 35. So (d) is correct; 16, 17 and 23 come from missing some of these distributions.

This is the product of all odd numbers from 1 to 999. The product includes the factor 5 (and many multiples of 5 such as 15, 25, ...), and it is multiplied by other odd numbers. Any multiple of 5 multiplied by an odd number that is not a multiple of 5 still ends in 5 (5 x odd = ...5). Since the whole product is odd (no factor of 2) and divisible by 5, its unit digit must be 5. Official UPSC 2025 key: C.