Number system

Answer: (c) 81

111111111 × 111111111 = 12345678987654321. Its digits add up to 2 × (1 + 2 + … + 8) + 9 = 72 + 9 = 81.

1. Pattern: 11² = 121, 111² = 12321, 1111² = 1234321, and so on.
2. So 111111111² = 12345678987654321.
3. Digit sum = 2 × (1 + 2 + … + 8) + 9 = 2 × 36 + 9 = 81.
4. Check: with n ones the digit sum of the square is n², and 9² = 81.

Remember · Squares of numbers made of n ones (n up to 9) are palindromes 123…n…321 with digit sum n².

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Answer: (b) 856

From 10 to 99, each tens digit 1 to 9 appears 10 times (sum 450) and each units digit 0 to 9 appears 9 times (sum 405), which gives 855. The number 100 adds 1 more, so the total is 856.

1. Tens digits in 10–99: each of 1 to 9 appears 10 times → 10 × 45 = 450.
2. Units digits in 10–99: each of 0 to 9 appears 9 times (once in every ten) → 9 × 45 = 405.
3. Sum for 10–99 = 450 + 405 = 855.
4. Add the digits of 100: 1 + 0 + 0 = 1. Total = 856.

Remember · For digit-sum totals, count how often each digit appears in each place, and do not forget the end-points (here, 100).

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Answer: (d) 83, 89, 91, 97

In groups (a), (b) and (c) all four numbers are prime. In group (d), 91 = 7 × 13 is not prime, so that group is the odd one out.

1. (a) 17, 37, 47, 97 — all prime.
2. (b) 31, 41, 53, 67 — all prime.
3. (c) 71, 73, 79, 83 — all prime.
4. (d) 83, 89 and 97 are prime, but 91 = 7 × 13. So (d) is different.

Remember · Know the primes up to 100; 91 (7 × 13), 51 (3 × 17) and 87 (3 × 29) are the classic look-alike composites.

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Answer: (d) 9

If 1186 leaves remainder 31, the divisor divides 1186 − 31 = 1155 and must be greater than 31. 1155 = 3 × 5 × 7 × 11 has 16 divisors, of which 9 are greater than 31.

1. 1186 = n × (quotient) + 31, so n divides 1186 − 31 = 1155; also n > 31, because a remainder is always less than the divisor.
2. 1155 = 3 × 5 × 7 × 11, so it has 2 × 2 × 2 × 2 = 16 divisors.
3. Divisors greater than 31: 33, 35, 55, 77, 105, 165, 231, 385, 1155 — 9 numbers.

Remember · Remainder r from N ÷ n means n divides N − r and n > r; count divisors of N − r above r.

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Answer: (c) Both 1 and 2

pp + qq + rr = 11(p + q + r) and tt0 = 110t, so p + q + r must be 10 or 20. The triples with p < q < r are (1,2,7), (1,3,6), (1,4,5), (2,3,5), (3,8,9), (4,7,9), (5,6,9) and (5,7,8), giving 5 possible values of p and 6 of q.

1. pp = 11p, so pp + qq + rr = 11(p + q + r). Also tt0 = 100t + 10t = 110t.
2. 11(p + q + r) = 110t gives p + q + r = 10t. With digits 1 to 9 and p < q < r, the sum is at most 7 + 8 + 9 = 24, so it is 10 or 20.
3. Sum 10: (1,2,7), (1,3,6), (1,4,5), (2,3,5). Sum 20: (3,8,9), (4,7,9), (5,6,9), (5,7,8).
4. Values of p: 1, 2, 3, 4, 5 → 5. Values of q: 2, 3, 4, 6, 7, 8 → 6.
5. Both statements are correct.

• ✓ 1. p can be 1, 2, 3, 4 or 5 — five values.
• ✓ 2. q can be 2, 3, 4, 6, 7 or 8 — six values.

Remember · Repeated-digit numbers factor neatly: aa = 11a and aa0 = 110a. Turn the puzzle into a simple sum condition, then list cases.

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Answer: (b) 4:10 p.m.

Green and red last equally long, so the full cycles are 50 s, 78 s and 120 s. All three turn green together again after LCM(50, 78, 120) = 7800 s = 2 hours 10 minutes, i.e. at 4:10 p.m.

1. Each full cycle (green → red → green) is twice the green time: 2 × 25 = 50 s, 2 × 39 = 78 s, 2 × 60 = 120 s.
2. 50 = 2 × 5², 78 = 2 × 3 × 13, 120 = 2³ × 3 × 5.
3. LCM = 2³ × 3 × 5² × 13 = 7800 s = 130 minutes = 2 h 10 min.
4. 2:00 p.m. + 2 h 10 min = 4:10 p.m.
5. Check: LCM(25, 39, 60) = 3900 s, but at that moment the third signal has changed 65 times (an odd number), so it is turning red, not green.

Remember · When signals must show the same colour again, take the LCM of full cycles (both phases), not of the single-phase times.

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Answer: (b) 15

The ring moves forward 1, 2, 3, … places on successive passes, so after k passes it has moved 1 + 2 + … + k = k(k + 1)/2 places. It is back with child-1 when this is a multiple of 40, which first happens at k = 15 (120 places, i.e. 3 full rounds).

1. Child-1 → 2 → 4 → 7 → …: the jumps are 1, 2, 3, … places.
2. After k passes the ring has moved k(k + 1)/2 places round the circle of 40.
3. It is back with child-1 when k(k + 1)/2 is a multiple of 40, i.e. when k(k + 1) is a multiple of 80.
4. k = 15: 15 × 16 = 240 = 3 × 80 works; no smaller k does (for example 14 × 15 = 210 and 9 × 10 = 90).
5. So after 15 passes (120 places, 3 full rounds) the ring returns to child-1.

Remember · Growing jumps 1, 2, 3, … add up to triangular numbers k(k + 1)/2; returning to the start needs a multiple of the circle size.

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Answer: (c) Both 1 and 2

Making 78 gm as a sum of these coins needs at least 7 coins, for example 50 + 10 + 10 + 2 + 2 + 2 + 2. But on a two-pan balance coins can go on both pans: 50 + 25 + 5 against the object plus a 2 gm coin weighs 78 gm with just 4 coins.

1. Statement 1 (buying 78 gm of coins): the coin weights must add up to exactly 78.
2. 50 + 25 = 75 leaves 3, which 2 and 5 cannot make. 50 + 10 + 10 = 70 leaves 8 = 2 + 2 + 2 + 2, giving 7 coins; checking all combinations, none with 6 or fewer coins makes 78.
3. So at least 7 coins must be bought — correct.
4. Statement 2 (weighing 78 gm): put 50 + 25 + 5 = 80 gm on one pan and the object with a 2 gm coin on the other; the object weighs 80 − 2 = 78 gm.
5. That uses 4 coins, fewer than 7 — correct.

• ✓ 1. The fewest coins adding up to 78 gm is 7 (50 + 10 + 10 + 2 + 2 + 2 + 2).
• ✓ 2. Using both pans of a balance, 50 + 25 + 5 on one side and 2 with the object on the other weighs 78 gm with 4 coins.

Remember · Buying uses sums of coin values only; weighing on a balance allows coins on both pans, so differences count too.

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Answer: (d) 4

Divided by 6, the powers of 2 leave remainders 2, 4, 2, 4 … in turn: odd powers leave 2 and even powers leave 4. Since 192 is even, the remainder is 4.

1. 2¹ = 2 leaves 2; 2² = 4 leaves 4; 2³ = 8 leaves 2; 2⁴ = 16 leaves 4.
2. The remainders repeat in a cycle of 2: odd powers leave 2, even powers leave 4.
3. 192 is even, so 2¹⁹² leaves remainder 4.
4. Check: 2¹⁹² = 4⁹⁶ leaves 1 on division by 3, and it is even; of 2 and 4, only 4 is even and leaves 1 on division by 3.

Remember · For remainders of large powers, list the first few remainders, spot the cycle, then place the exponent in it.

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Answer: (a) 6

The sum D08 + GHI = 975 fixes I = 7, H = 6 and D + G = 9. The only product of two 2-digit numbers of the form D08 with CD ending in D and all digits distinct is 12 × 34 = 408, so A + B + C = 1 + 2 + 3 = 6.

1. DEF = D08. Units: 8 + I = 15, so I = 7 (carry 1). Tens: 0 + H + 1 = 7, so H = 6. Hundreds: D + G = 9.
2. AB × CD = D08, where CD ends in the digit D. Test D = 1 to 8.
3. 108: no pair of 2-digit factors. 208 = 13 × 16 and 308 = 11 × 28 = 14 × 22: no factor ends in 2 or 3 as needed.
4. 408 = 12 × 34 or 17 × 24. With CD = 34: A = 1, B = 2, C = 3, D = 4, G = 5 — all nine digits 1, 2, 3, 4, 0, 8, 5, 6, 7 are distinct. With CD = 24, AB = 17 gives B = 7 = I, a clash.
5. 508, 708 give no valid pair; 608 = 38 × 16 gives B = 8 = F, a clash; D = 8 clashes with F; D = 9 gives G = 0 = E.
6. So A + B + C = 1 + 2 + 3 = 6.
7. Check: 12 × 34 = 408 and 408 + 567 = 975.

Remember · In digit puzzles, settle the addition column by column first; it fixes most letters and leaves only a few cases to test.

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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.

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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.

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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.

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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.

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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.

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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).

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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.

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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.

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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).

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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.

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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⌋ + ….

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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.

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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.

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Answer: (c) 7

The remainder on division by 10 is simply the last digit. Last digits of powers of 3 repeat as 3, 9, 7, 1, and 2019 leaves remainder 3 on division by 4, so 3²⁰¹⁹ ends in 7.

1. The last digit of 3¹, 3², 3³, 3⁴ is 3, 9, 7, 1, and this cycle of 4 repeats.
2. 2019 = 4 × 504 + 3, so 3²⁰¹⁹ has the same last digit as 3³.
3. 3³ = 27 ends in 7.
4. The remainder on dividing by 10 equals the last digit, so the remainder is 7.
5. Check: 3⁷ = 2187 also ends in 7, and 7 = 4 + 3 — the pattern holds.

Remember · Remainder by 10 = units digit. For powers of 2, 3, 7, 8 the units digit cycles every 4, so use the exponent's remainder by 4.

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Answer: (b) 6

Because 1001 = 7 × 11 × 13, a number is divisible by 7 when the alternating sum of its three-digit groups (taken from the right) is. That sum here is 1063 − P, and 1063 leaves remainder 6 on division by 7, so P = 6.

1. Since 1001 = 7 × 11 × 13, a number is divisible by 7 if the alternating sum of its 3-digit groups, taken from the right, is divisible by 7.
2. Groups of 3798125P369 from the right: 369, 25P, 981, 37.
3. Alternating sum = 369 − 25P + 981 − 37 = 1313 − (250 + P) = 1063 − P.
4. 1063 = 7 × 151 + 6, so 1063 − P is a multiple of 7 only when P = 6.
5. Check: 37981256369 ÷ 7 = 5425893767 exactly.

Remember · For divisibility by 7, 11 or 13 in a long number, take the alternating sum of 3-digit groups from the right (1001 = 7 × 11 × 13).

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Answer: (d) 9

Fix the hundreds digit and count the ways the last two digits can make up the rest: sums of 3, 2 and 1 give 4, 3 and 2 numbers. 1000 does not qualify, so the total is 9.

1. 7ab: a + b = 3 → 703, 712, 721, 730 (4 numbers).
2. 8ab: a + b = 2 → 802, 811, 820 (3 numbers).
3. 9ab: a + b = 1 → 901, 910 (2 numbers).
4. 1000 has digit sum 1, so it does not count. Total = 4 + 3 + 2 = 9.

Remember · Two digits adding to k (k ≤ 9) can be chosen in k + 1 ways.

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Answer: (c) Both 1 and 2

With the digits 3, 6 and 9, the six numbers add up to 222 × 18 = 3996, which is 74 × 54 and 9 × 444. In general, the six arrangements of three digits add to 222 × (digit sum), and 222 = 3 × 74, so both statements hold.

1. The non-zero digits that are multiples of 3 are 3, 6 and 9; without repetition they give 3! = 6 numbers.
2. Each digit appears twice in each place, so S = 2 × (3 + 6 + 9) × 111 = 222 × 18 = 3996.
3. 3996 = 74 × 54, so S is divisible by 74.
4. 3996 = 9 × 444, so S is divisible by 9.
5. Check (general case): any three distinct digits a, b, c give six numbers summing to 222(a + b + c) = 3 × 74 × (a + b + c); when the numbers are multiples of 3, a + b + c is too, so S is a multiple of 9 × 74 = 666.

• ✓ 1. S = 222 × (digit sum) and 222 = 3 × 74, so 74 always divides S.
• ✓ 2. 222 carries one factor 3 and the digit sum (a multiple of 3) carries another, so 9 always divides S.

Remember · Sum of all arrangements of 3 distinct digits = 222 × (sum of the digits).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 202

In place-value form the four numbers add to 70 + 24P, and a units digit of 2 allows only P = 3 or P = 8. The sums 142 and 262 both keep P, Q and R different, and their mean is 202.

1. 3P + 4P + PP + PP = (30 + P) + (40 + P) + 11P + 11P = 70 + 24P.
2. The sum ends in 2, so 24P ends in 2, i.e. 4P ends in 2: P = 3 or P = 8.
3. P = 3: 70 + 72 = 142, so R = 1, Q = 4 — the digits 3, 4, 1 are different.
4. P = 8: 70 + 192 = 262, so R = 2, Q = 6 — the digits 8, 6, 2 are different.
5. Mean of the possible sums = (142 + 262) ÷ 2 = 202.

Remember · Write letter-numbers in place-value form (3P = 30 + P, PP = 11P), then shortlist with the units digit.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 2

Three times Q must end in Q, so Q is 0 or 5, and only Q = 5 works. Then 85 × 3 = 255 is the only fit, so P = 8, R = 2 and (P + R) ÷ Q = 10 ÷ 5 = 2.

1. The units digit of 3 × Q must be Q, so Q = 0 or 5.
2. Q = 0: P0 × 3 = R00 needs 30P = 100R, i.e. 3P = 10R — no digit P works with R ≠ 0.
3. Q = 5: (10P + 5) × 3 = 100R + 55 gives 30P = 100R + 40, i.e. 3P = 10R + 4, so P = 8 and R = 2.
4. 85 × 3 = 255, and (P + R) ÷ Q = (8 + 2) ÷ 5 = 2.

Remember · In digit puzzles, start with the units place — digits whose multiples end in themselves narrow things quickly.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) Both 1 and 2

Five consecutive integers add to five times the middle one, so 18 to 22 give 100. And 1, 2, 3 have product 6 and sum 6, so both statements are correct.

1. Five consecutive integers add to 5 × (middle one); 5 × 20 = 100, e.g. 18 + 19 + 20 + 21 + 22 = 100. Statement 1 is correct.
2. 1 × 2 × 3 = 6 and 1 + 2 + 3 = 6. Statement 2 is correct.
3. Both statements are correct.

• ✓ 1. 18 + 19 + 20 + 21 + 22 = 100.
• ✓ 2. 1, 2, 3: product 6 = sum 6.

Remember · A 'can be' statement is proved by one example; try the smallest numbers first.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error