Number system

Answer: (d) 225

Each term is the previous one multiplied by a factor that grows by 0.5 each time: ×0.5, ×1, ×1.5, ×2, ×2.5. The next factor is ×3, so X = 75 × 3 = 225.

1. Divide each term by the one before it: 10 ÷ 20 = 0.5, 10 ÷ 10 = 1, 15 ÷ 10 = 1.5, 30 ÷ 15 = 2, 75 ÷ 30 = 2.5.
2. The multiplier rises by 0.5 each time, so the next multiplier is 3.
3. X = 75 × 3 = 225.
4. Check: 225 ÷ 75 = 3 continues the pattern 0.5, 1, 1.5, 2, 2.5, 3.

Remember · When differences look irregular, try the ratio of consecutive terms; a steadily growing multiplier is a common pattern.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 8

Work through the conditions: G must be 9, E must be 5, the even digits 2, 4, 8 fill B, D, F, and the tests for 3 and 4 fix B = 4 and D = 2. The middle digits C, D, E then add up to 8 (or 14, which is not offered).

1. The digits 1, 2, 4, 5, 7, 8, 9 add up to 36, so the full number is divisible by 9 in any order.
2. ABCDEF is divisible by 6, so F is even and A + B + C + D + E + F is a multiple of 3. That sum is 36 − G, so G is a multiple of 3: G = 9.
3. ABCDE is divisible by 5 and 0 is not available, so E = 5.
4. AB divisible by 2 makes B even; ABCD divisible by 4 makes D even. So B, D, F are the even digits 2, 4, 8, and A, C are 1 and 7 in some order.
5. ABC divisible by 3: A + B + C = 1 + 7 + B = 8 + B must be a multiple of 3, so B = 4.
6. ABCD divisible by 4: the last two digits CD must be 12 or 72 (18 and 78 fail), so D = 2 and F = 8.
7. Middle three digits C + D + E = C + 2 + 5: this is 8 when C = 1 and 14 when C = 7. Only 8 is offered.
8. Check: 7412589 — 741258 ÷ 6 = 123543, 74125 ends in 5, 7412 ÷ 4 = 1853, 741 ÷ 3 = 247, 74 is even.

Remember · Divisibility chains: fix the digit forced by 5, then the even places, then use digit-sum and last-two-digit tests.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 3²¹

4¹⁸ and 8¹² both equal 2³⁶, which is smaller than 2⁴⁰. And 3²¹ = 2187³ is smaller than 2³⁶ = 4096³, so 3²¹ is the smallest of the four.

1. Write the powers of 2 in base 2: 4¹⁸ = 2³⁶ and 8¹² = 2³⁶. Among 2⁴⁰, 4¹⁸ and 8¹² the smallest is 2³⁶.
2. Compare 3²¹ with 2³⁶ using the common exponent 3: 3²¹ = (3⁷)³ = 2187³ and 2³⁶ = (2¹²)³ = 4096³.
3. 2187 < 4096, so 3²¹ < 2³⁶ < 2⁴⁰.
4. Check: 21 × log 3 ≈ 10.02 while 36 × log 2 ≈ 10.84, so 3²¹ (about 1.05 × 10¹⁰) is the smallest.

Remember · To compare powers, rewrite them with a common base or a common exponent, then compare only what differs.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 2

The first-row number N must keep 3N below 1000, so it starts with 2 or 3 and uses three of 2, 3, 7, 9. Testing the eight such numbers leaves only 273 (546, 819) and 327 (654, 981).

1. The third row is 3N, a 3-digit number, so N ≤ 333. With digits from 2, 3, 7, 9 (no repeats), N must start with 2 or 3.
2. Candidates: 237, 239, 273, 279, 293, 297, 327 and 329.
3. 237 → 474 (repeats 4); 239 → 478, 717 (repeats 7); 279 → 558 (repeats 5); 293 → 586, 879 (repeats 8 and 9); 297 → 594 (repeats 9); 329 → 658, 987 (repeats 8).
4. 273 → 546 → 819 uses each of 1–9 once; 327 → 654 → 981 also uses each of 1–9 once.
5. So 2 arrangements are possible.
6. Check: 273 × 2 = 546 and 273 × 3 = 819; 327 × 2 = 654 and 327 × 3 = 981.

Remember · Bound the search first (here 3N < 1000, so N ≤ 333); a short, systematic list beats guessing.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 294

If A gets k pens, the others get k/2, k/3, k/4, k/5 and k/6, and each must be even. So k must be a multiple of 4, 6, 8, 10 and 12; the least such k is 120, giving 120 + 60 + 40 + 30 + 24 + 20 = 294 pens.

1. Let A get k pens. Then B, C, D, E and F get k/2, k/3, k/4, k/5 and k/6.
2. Each share must be an even whole number, so k must be divisible by 4, 6, 8, 10 and 12.
3. LCM(4, 6, 8, 10, 12) = 120, so the least k is 120.
4. Shares: 120, 60, 40, 30, 24, 20 — all even. Total = 294.
5. Check: k = 60 would give 147 pens, but then D gets 15, an odd number, so 147 fails.

Remember · ‘k/n must be even’ means k is a multiple of 2n; take the LCM of all such 2n.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 1 only

The six numbers use each digit twice in every place, so x = 222 × (A + B + C). The least digit sum is 1 + 2 + 3 = 6, giving x = 1332, so x is always at least 1332 and can never be a 3-digit number.

1. The six numbers formed by A, B, C place each digit twice in the hundreds, tens and units places, so x = 2 × (A + B + C) × 111 = 222 × (A + B + C).
2. The smallest possible sum of three distinct non-zero digits is 1 + 2 + 3 = 6, so the least x is 222 × 6 = 1332, a 4-digit number. Statement 1 is correct.
3. A 3-digit x would need 222 × (A + B + C) < 1000, i.e. a digit sum of 4 or less — impossible. So x is never 3-digit, and 888 (digit sum 4) cannot occur. Statement 2 is incorrect.
4. Check: 123 + 132 + 213 + 231 + 312 + 321 = 1332.

• ✓ 1. Digit sum 6 is the minimum, giving x = 222 × 6 = 1332.
• ✗ 2. x is always a multiple of 222 with digit sum at least 6, so it is never below 1332; 888 would need digit sum 4.

Remember · Sum of all permutations of three distinct digits = 222 × (sum of the digits).

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 0

1261 = 13 × 97. The product contains 91 = 7 × 13 and also 97, so it is a multiple of 1261 and the remainder is 0.

1. Factorise the divisor: 1261 = 13 × 97.
2. The product contains 91 = 7 × 13 and the factor 97.
3. So the product is a multiple of 13 × 97 = 1261, and the remainder is 0.

Remember · Before dividing a big product, factorise the divisor and look for its factors inside the product.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 1083

A number leaving remainder 3 with each divisor is 3 more than a common multiple of them. LCM(6, 9, 12, 15, 18) = 180, and the first multiple of 180 that gives a number above 1000 is 1080, so the answer is 1083.

1. LCM(6, 9, 12, 15, 18) = 180.
2. The number must be of the form 180k + 3.
3. 180 × 5 + 3 = 903 is below 1000; 180 × 6 + 3 = 1083 is the first above it.
4. Check: 1083 − 3 = 1080 is divisible by 6, 9, 12, 15 and 18.

Remember · Same remainder r for several divisors: the number = k × LCM + r.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 9

2430 = 45 × 54, and 54 is 45 with its digits reversed. The difference between them is 54 − 45 = 9.

1. 2430 = 2 × 3⁵ × 5; look for a two-digit factor whose reverse is the other factor.
2. 2430 = 45 × 54, and 54 is 45 reversed.
3. Difference = 54 − 45 = 9.
4. Check: a two-digit number and its reverse differ by 9 × (difference of the digits) = 9 × (5 − 4) = 9.

Remember · A two-digit number minus its reverse = 9 × (difference of its digits).

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 1, 2 and 3

Each statement says ‘can be’, so one example is enough for each: 3 × 9 = 27 is odd, 6 ÷ 2 = 3 is prime, and 2 + 9 = 11 is prime. All three are correct.

1. p = 3, q = 9: p × q = 27 is odd, so statement 1 is correct.
2. p = 2, q = 6: q/p = 3 is prime, so statement 2 is correct.
3. p = 2, q = 9: p + q = 11 is prime, so statement 3 is correct.
4. All three statements are correct.

• ✓ 1. An odd prime times an odd composite is odd, e.g. 3 × 9 = 27.
• ✓ 2. e.g. 6 ÷ 2 = 3 (also 15 ÷ 5 = 3).
• ✓ 3. e.g. 2 + 9 = 11 (also 3 + 4 = 7).

Remember · ‘Can be’ statements need just one valid example; ‘must be’ statements need just one counter-example to fail.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 6

The largest power of 3 dividing 15! decides m. The multiples of 3 up to 15 give five 3s and 9 = 3² gives one more, so m can be at most 6.

1. Count the factors of 3 in 15!: the multiples of 3 up to 15 are 3, 6, 9, 12, 15 — five numbers, one 3 each.
2. 9 = 3² contributes one extra 3.
3. Total power of 3 = ⌊15/3⌋ + ⌊15/9⌋ = 5 + 1 = 6, so the maximum m is 6 (n is then a whole number).

Remember · Highest power of prime p in n! = ⌊n/p⌋ + ⌊n/p²⌋ + ….

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 252

The terms are n² × (n + 1) — that is, n³ + n² — for n = 1, 2, 3, 4, 5: 2, 12, 36, 80, 150. The next term is 6² × 7 = 252.

1. Test n²(n + 1): 1 × 2 = 2, 4 × 3 = 12, 9 × 4 = 36, 16 × 5 = 80, 25 × 6 = 150.
2. Next term: 36 × 7 = 252.
3. Check with differences: 10, 24, 44, 70, 102; second differences 14, 20, 26, 32 rise steadily by 6, as they should for a cubic pattern.

Remember · If terms grow roughly like cubes, test n³ ± n² or n³ ± n before trying differences.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Only three

Writing the integers as n − 1, n, n + 1 gives 3n = n(n² − 1), so n = 0, 2 or −2. That produces three triples: (−1, 0, 1), (1, 2, 3) and (−3, −2, −1).

1. Let the integers be n − 1, n, n + 1. Sum = 3n; product = n(n² − 1).
2. 3n = n(n² − 1) gives n = 0 or n² − 1 = 3, i.e. n = 0, 2 or −2.
3. The triples are (−1, 0, 1), (1, 2, 3) and (−3, −2, −1), with sums 0, 6, −6 equal to their products.
4. So there are exactly three possibilities. (Testing only positive integers finds just 1, 2, 3.)

Remember · ‘Integers’ includes zero and negatives; solve the algebra instead of testing only positive numbers.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error