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CSAT · 140 questions

Number system

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Number system questions per year: 2016: 3, 2017: 7, 2018: 5, 2019: 13, 2020: 19, 2021: 10, 2022: 13, 2023: 21, 2024: 15, 2025: 22, 2026: 12 Asked in 11 of 11 years · most in 2025 (22)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

CSAT 2023 · Q7

Easy

What is the remainder when 85 × 87 × 89 × 91 × 95 × 96 is divided by 100?

Answer & explanation

Answer: (a) 0

85 and 95 each supply a factor 5, and 96 supplies several factors of 2, so the product contains 5 × 5 × 2 × 2 = 100. A multiple of 100 leaves remainder 0.

  1. 85 = 5 × 17 and 95 = 5 × 19, which give 5 × 5 = 25.
  2. 96 = 2⁵ × 3, which gives at least 2 × 2 = 4.
  3. So the product has 25 × 4 = 100 as a factor; the remainder on division by 100 is 0.
  4. Check: 85 × 96 × 95 = 775200, already ending in 00.

Remember · Before any remainder working, look for factors 2 and 5: enough of them make the product a multiple of 10 or 100.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q8

Easy

What is the unit digit in the expansion of (57242)^(9×7×5×3×1)?

Answer & explanation

Answer: (a) 2

Only the last digit of the base, 2, matters, and the unit digits of powers of 2 repeat every 4. The exponent 9 × 7 × 5 × 3 × 1 = 945 leaves remainder 1 on division by 4, so the unit digit is that of 2¹, which is 2.

  1. The base ends in 2 and the exponent is 9 × 7 × 5 × 3 × 1 = 945, so we need the unit digit of 2^945.
  2. Unit digits of powers of 2 run 2, 4, 8, 6 and then repeat (cycle of 4).
  3. 945 ÷ 4 leaves remainder 1.
  4. So the unit digit is the first in the cycle: 2.

Remember · Unit digit of a power: keep the base's last digit, find its cycle (2, 3, 7, 8 repeat every 4), use exponent mod 4.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q9

Medium

If ABC and DEF are both 3-digit numbers such that A, B, C, D, E and F are distinct non-zero digits such that ABC + DEF = 1111, then what is the value of A + B + C + D + E + F?

Answer & explanation

Answer: (d) 31

Add column by column. With non-zero digits the units must give C + F = 11 with a carry, then B + E + 1 = 11 and A + D + 1 = 11. So the six digits add up to 11 + 10 + 10 = 31.

  1. Units column: C + F must end in 1. C + F = 1 is impossible with non-zero digits, so C + F = 11 and 1 is carried.
  2. Tens column: B + E + 1 must end in 1, so B + E + 1 = 11, i.e. B + E = 10, and 1 is carried.
  3. Hundreds column: A + D + 1 must give 11 (the leading '11' of 1111), so A + D = 10.
  4. A + B + C + D + E + F = 10 + 10 + 11 = 31.
  5. Check: 124 + 987 = 1111, and 1 + 2 + 4 + 9 + 8 + 7 = 31.

Remember · In digit-addition puzzles, work column by column from the units, tracking carries; the digit sum then follows without finding the digits.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q10

Medium

D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred's place and the digit at the unit's place of D?

Answer & explanation

Answer: (c) 8

The ratio (number ÷ digit sum) is smallest when the hundreds digit is as small as possible and the tens and units digits are as large as possible. 199 gives 199/19 ≈ 10.47, lower than 189/18 = 10.5 or any other, so D = 199 and the difference between 9 and 1 is 8.

  1. Write D = 100a + 10b + c. Raising a adds 100 to the number but only 1 to the digit sum, so a should be 1; b and c should be as large as possible.
  2. Take a = 1, b = 9, c = 9: 199/19 ≈ 10.47.
  3. Compare the nearest rivals: 189/18 = 10.5, 179/17 ≈ 10.53, 198/18 = 11 — all larger.
  4. So D = 199: hundreds digit 1, units digit 9, difference 9 − 1 = 8.

Remember · To minimise a number divided by its digit sum, keep the high-place digit smallest and push the digit sum into lower places; test close rivals.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q14

Medium

Three of the five positive integers p, q, r, s, t are even and two of them are odd (not necessarily in order). Consider the following:

  1. 1.p + q + r − s − t is definitely even.
  2. 2.2p + q + 2r − 2s + t is definitely odd.

Which of the above statements is/are correct?

Answer & explanation

Answer: (a) 1 only

Subtracting a number changes odd/even exactly as adding it does, so p + q + r − s − t has the same parity as the sum of all five, which is even (it contains two odd numbers). In 2p + q + 2r − 2s + t the doubled terms are even, leaving the parity of q + t, which may be even.

  1. Adding or subtracting a number has the same effect on parity, so signs can be ignored.
  2. Statement 1: parity of p + q + r − s − t = parity of p + q + r + s + t. Two odds make an even and evens stay even, so the result is always even. Correct.
  3. Statement 2: 2p, 2r and 2s are even, so the parity equals that of q + t.
  4. Example: p = 1 and s = 1 are the odd ones, q = r = t = 2. Then 2 + 2 + 4 − 2 + 2 = 8, which is even. So statement 2 is not always true.
  • ✓ 1. Its parity is that of the sum of all five numbers: three evens and two odds add up to an even number, whatever the order.
  • ✗ 2. Only q and t decide the parity; if both are even (or both odd) the expression is even, so it is not definitely odd.

Remember · For parity, treat minus as plus and ignore every term multiplied by an even number.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q15

Medium

Consider the following in respect of prime number p and composite number c.

  1. 1.(p + c)/(p − c) can be even.
  2. 2.2p + c can be odd.
  3. 3.pc can be odd.

Which of the statements given above are correct?

Answer & explanation

Answer: (d) 1, 2 and 3

Each statement says 'can be', so one example proves it. p = 11, c = 9 gives (p + c)/(p − c) = 20/2 = 10, which is even; p = 2, c = 9 gives 2p + c = 13, which is odd; p = 3, c = 9 gives pc = 27, which is odd. All three statements hold.

  1. A 'can be' statement is proved by a single example.
  2. Statement 1: p = 11, c = 9 → (11 + 9)/(11 − 9) = 20/2 = 10, which is even. True.
  3. Statement 2: 2p is always even, so take an odd composite: p = 2, c = 9 → 4 + 9 = 13, odd. True.
  4. Statement 3: an odd prime times an odd composite: 3 × 9 = 27, odd. True.
  • ✓ 1. p = 11, c = 9 gives 20/2 = 10, an even number.
  • ✓ 2. 2p is even, so 2p + c is odd whenever c is an odd composite, e.g. 2 × 2 + 9 = 13.
  • ✓ 3. Both odd: p = 3, c = 9 gives 27.

Remember · 'Can be' needs one working example; 'must be' needs proof. Remember odd composites like 9, 15 and 21.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q16

Medium

A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C?

Answer & explanation

Answer: (a) 18

37DD = 3700 + 11D must be a multiple of D, so D must divide 3700; that allows D = 1, 2, 4 or 5. Only D = 4 gives a 3-digit ABC with all four digits different: 3744 ÷ 4 = 936, so A + B + C = 9 + 3 + 6 = 18.

  1. 37DD = 3700 + 11 × D. Since ABC × D = 37DD, D divides 3700 + 11D, so D divides 3700 = 2² × 5² × 37.
  2. Possible non-zero digits: D = 1, 2, 4 or 5.
  3. D = 1 gives ABC = 3711 and D = 2 gives 3722 ÷ 2 = 1861: neither is a 3-digit number.
  4. D = 5 gives 3755 ÷ 5 = 751, but then B = 5 = D, which is not allowed.
  5. D = 4 gives 3744 ÷ 4 = 936: A = 9, B = 3, C = 6, D = 4, all different. A + B + C = 18.
  6. Check: 936 × 4 = 3744.

Remember · In digit puzzles, write the unknown number algebraically (3700 + 11D), shortlist digits by divisibility, then apply the 'different digits' condition.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q17

Easy

For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by:

Answer & explanation

Answer: (d) 7, 11 and 13

XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001, and 1001 = 7 × 11 × 13. So every such number is divisible by 7, 11 and 13.

  1. XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001.
  2. 1001 = 7 × 11 × 13.
  3. So the number is always divisible by 7, 11 and 13.
  4. Check: 123123 = 123 × 1001 = 123 × 7 × 11 × 13.

Remember · A repeated 3-digit block (abcabc) is a multiple of 1001 = 7 × 11 × 13; abab is a multiple of 101.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q25

Easy

Let x be a positive integer such that 7x + 96 is divisible by x. How many values of x are possible?

Answer & explanation

Answer: (c) 12

7x is always a multiple of x, so x divides 7x + 96 exactly when x divides 96. Since 96 = 2⁵ × 3, it has (5 + 1)(1 + 1) = 12 divisors, so x can take 12 values.

  1. 7x is divisible by x, so x divides 7x + 96 only when x divides 96.
  2. 96 = 2⁵ × 3.
  3. Number of divisors = (5 + 1) × (1 + 1) = 12: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96.

Remember · If x divides kx + N, then x divides N; the number of divisors of 2^a × 3^b is (a + 1)(b + 1).

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q26

Easy

If p, q, r and s are distinct single digit positive numbers, then what is the greatest value of (p + q)(r + s)?

Answer & explanation

Answer: (b) 225

Use the four largest digits 9, 8, 7 and 6, which add up to 30. For a fixed total, the product of two parts is greatest when the parts are equal, so split 30 as 15 and 15: (9 + 6)(8 + 7) = 225.

  1. To make the product large, use the four largest distinct digits: 9, 8, 7, 6 (total 30).
  2. Split 30 into two sums as equal as possible: 9 + 6 = 15 and 8 + 7 = 15.
  3. Greatest value = 15 × 15 = 225.
  4. Check: the other splits give 17 × 13 = 221 and 16 × 14 = 224, both smaller.

Remember · With a fixed total, a product of parts is largest when the parts are as equal as possible.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q27

Medium

A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13?

Answer & explanation

Answer: (a) 11

999999 (six 9s) = 999 × 1001 is divisible by 13, so each block of six 9s leaves nothing over. 99 nines are 16 such blocks followed by 999, so N leaves the same remainder as 999, which is 11.

  1. 999999 = 999 × 1001 = 999 × 7 × 11 × 13, so it is divisible by 13.
  2. 99 = 6 × 16 + 3, so N is sixteen blocks of 999999 followed by 999.
  3. The first 96 nines form a multiple of 999999 (and so of 13), shifted three places; hence N leaves the same remainder as 999.
  4. 999 = 13 × 76 + 11, so the remainder is 11.

Remember · Six identical digits (aaaaaa) are always divisible by 7, 11 and 13; remove blocks of six and work with the digits left over.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q28

Easy

Each digit of a 9-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q29

Medium

What is the sum of all digits which appear in all the integers from 10 to 100?

Answer & explanation

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).

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q40

Easy

Choose the group which is different from the others:

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q45

Medium

How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q46

Hard

Let pp, qq and rr be 2-digit numbers where p < q < r. If pp + qq + rr = tt0, where tt0 is a 3-digit number ending with zero, consider the following statements:

  1. 1.The number of possible values of p is 5.
  2. 2.The number of possible values of q is 6.

Which of the above statements is/are correct?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q50

Medium

There are three traffic signals. Each signal changes colour from green to red and then from red to green. The first signal takes 25 seconds, the second signal takes 39 seconds and the third signal takes 60 seconds to change the colour from green to red. The durations for green and red colours are same. At 2:00 p.m, they together turn green. At what time will they change to green next, simultaneously?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q54

Medium

40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child-7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q69

Medium

There are large number of silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, 50 gm each. Consider the following statements:

  1. 1.To buy 78 gm of coins one must buy at least 7 coins.
  2. 2.To weigh 78 gm using these coins one can use less than 7 coins.

Which of the statements given above is/are correct?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q74

Easy

What is the remainder if 2¹⁹² is divided by 6?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q76

Hard

AB and CD are 2-digit numbers. Multiplying AB with CD results in a 3-digit number DEF. Adding DEF to another 3-digit number GHI results in 975. Further A, B, C, D, E, F, G, H, I are distinct digits. If E = 0, F = 8, then what is A + B + C equal to?

Answer & explanation

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.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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