Number system

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).

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: (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.

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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

Answer: (b) Wednesday

All three meet again after LCM(5, 24, 9) = 360 days. That is 51 weeks and 3 days, so the meeting falls three days after Sunday — on Wednesday.

1. They meet together again after LCM(5, 24, 9) days.
2. 24 = 2³ × 3 and 9 = 3², so LCM = 2³ × 3² × 5 = 360.
3. 360 = 7 × 51 + 3, so the weekday moves on by 3 days.
4. Sunday + 3 days = Wednesday.

Remember · Next joint meeting = LCM of the intervals; the weekday shifts by (LCM mod 7).

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 only

A 2-digit number and its reverse differ by 9 times the difference of the digits, so 54 means the digits differ by 6 — that is fixed. The digit sum could be 6, 8, 10 or 12; a known product would settle it, but so would other information, so 'only if the product is known' is not true.

1. (10a + b) − (10b + a) = 9(a − b), so 9 × (difference of digits) = 54.
2. The digits differ by 6 — Statement 2 is correct.
3. Possible numbers: 60, 71, 82, 93 (or 17, 28, 39 the other way round), with digit sums 6, 8, 10, 12.
4. Knowing the product (0, 7, 16 or 27) would fix the sum, but so would other clues, such as the larger digit — the product is not the only way, so Statement 1 is incorrect.

• ✗ 1. The product is one way to fix the sum, not the only way, so 'only if' makes the statement false.
• ✓ 2. 9 × (difference of digits) = 54 gives a difference of 6.

Remember · Number − its reverse = 9 × (difference of digits); number + its reverse = 11 × (sum of digits).

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Answer: (d) 15873

The smallest number made only of 1s that 7 divides is 111111 (six ones). 111111 ÷ 7 = 15873.

1. Divide 1, 11, 111, … by 7: the remainders are 1, 4, 6, 5, 2, 0 — the first zero comes at 111111.
2. 111111 ÷ 7 = 15873.
3. Check: 15873 × 7 = 111111.
4. Useful fact: 111111 = 3 × 7 × 11 × 13 × 37.

Remember · Divide repunits (1, 11, 111 …) by 7 until the remainder is 0; 111111 = 7 × 15873.

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