Number system

Answer: (c) मात्रा-I = मात्रा-II

For a fixed sum of 10, the product p × q is largest when p = q = 5, giving 25. Allowing some negative values only adds negative products, so both maxima are 25 and the two values are equal.

1. Value-I: with p + q = 10 and both positive, p × q = p(10 − p) is largest at p = q = 5, giving 25.
2. Value-II: p ≥ −6 and q ≥ −4 allow p from −6 to 14. The product p(10 − p) still peaks at p = 5 (25).
3. The extra cases give negative products, e.g. (−6) × 16 = −96 and 14 × (−4) = −56.
4. So Value-II = 25 = Value-I.

Remember · For a fixed sum, the product is greatest when the two numbers are equal (or as close as possible).

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Answer: (c) मात्रा-I = मात्रा-II

The lowest allowed set, −5 to 5, is balanced around 0, so its average is 0. The product of 0, 1, …, 10 is 0, the smallest possible for non-negative integers. Both values are 0.

1. Value-I: the lowest possible set is −5, −4, …, 4, 5. Its average is 0; any later set has a larger average. So Value-I = 0.
2. Value-II: the set 0, 1, …, 10 has product 0; any set starting above 0 has a positive product. So Value-II = 0.
3. Value-I = Value-II.

Remember · Consecutive integers balanced around 0 average 0; a product of non-negative numbers is smallest (0) when it includes 0.

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Answer: (a) केवल I

For a number between 0 and 1, higher powers get smaller and square roots get larger. So x² > x³ is true, but x > √x is false.

1. Take x = 0.25 as a test value.
2. I: x² = 0.0625 and x³ = 0.015625, so x² > x³. In general x² − x³ = x²(1 − x) > 0 for 0 < x < 1. True.
3. II: √0.25 = 0.5, which is larger than 0.25, so x > √x is false. For 0 < x < 1, √x is always larger than x.
4. Only statement I is correct.

• ✓ I x² − x³ = x²(1 − x) is positive when 0 < x < 1, so x² > x³.
• ✗ II For a fraction between 0 and 1 the square root is larger: √0.25 = 0.5 > 0.25.

Remember · Between 0 and 1: x³ < x² < x < √x. Test with x = 0.25 to confirm quickly.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) ऐसी एक से अधिक संख्याएँ हो सकती हैं ।

Two natural numbers 10 apart have nine numbers strictly between them. These contain one multiple of 5 when both ends are multiples of 5 (5 and 15) and two otherwise (3 and 13). So there can be more than one such number, as (c) says.

1. Take 5 and 15: the only multiple of 5 strictly between them is 10 — one number.
2. Take 3 and 13: the multiples of 5 between them are 5 and 10 — two numbers.
3. The nine numbers between the two always include at least one multiple of 5, so the count is 1 or 2, never 0.
4. Only (c), 'There can be more than one such number', holds in general.

• ✗ (a) True only when both numbers are multiples of 5 (e.g. 5 and 15); 3 and 13 have two such numbers between them.
• ✗ (b) True only when neither number is a multiple of 5; 5 and 15 have just one (10) between them.
• ✓ (c) The count can be 2 (3 and 13 give 5 and 10), so more than one such number is possible.
• ✗ (d) Nine consecutive numbers always include a multiple of 5, so at least one always exists.

Remember · When a count depends on the case, test a boundary case and a general case; pick the option true in every case.

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Answer: (d) 200

The product is n raised to the power 2n for n = 1 to 25. Twos are plentiful, so count the fives: only 5, 10, 15, 20 and 25 contribute, giving 10 + 20 + 30 + 40 + 100 = 200.

1. Each zero at the end needs one 2 and one 5; there are far more 2s than 5s here, so count the 5s.
2. Only multiples of 5 carry a 5: 5¹⁰ gives 10 fives, 10²⁰ gives 20, 15³⁰ gives 30, 20⁴⁰ gives 40.
3. 25 = 5², so 25⁵⁰ = 5¹⁰⁰ gives 100 fives.
4. Total fives = 10 + 20 + 30 + 40 + 100 = 200, so there are 200 zeros at the end.
5. Check: forgetting that 25 holds two 5s gives 150, which is not even an option.

Remember · Trailing zeros = number of 5s (2s are always more). Remember 25 = 5², so its power counts double.

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Answer: (b) 8 जनवरी, 2023

He saves 1, 3, 5, 7, … rupees, so the total after n days is n². A number that is both a square and a cube is a sixth power, so n must be a perfect cube; n = 8 gives 64 = 8² = 4³.

1. Daily savings are the odd numbers 1, 3, 5, …; the sum of the first n odd numbers is n².
2. n² is also a perfect cube only when it is a sixth power, i.e. when n itself is a perfect cube.
3. The first cube after 1 is 8: total after 8 days = 64 = 8² = 4³.
4. So the date is 8th January, 2023.
5. Check: 7 days give 49 and 9 days give 81, neither a cube. (The ₹1 at the end of 1st January is trivially 1² = 1³, but that date is not among the options.)

Remember · Sum of the first n odd numbers = n². A number that is both a square and a cube is a sixth power.

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Answer: (b) 3 और 37 से किन्तु 2 से नहीं

222 = 2 × 3 × 37 and 333 = 3² × 37, so both terms, and hence their sum, are divisible by 3 and 37. But 222³³³ is even and 333²²² is odd, so the sum is odd and not divisible by 2.

1. Factorise the bases: 222 = 2 × 3 × 37 and 333 = 3 × 3 × 37.
2. Both 222³³³ and 333²²² contain the factors 3 and 37, so their sum is divisible by 3 and by 37.
3. 222³³³ is even; 333²²² is odd (a power of an odd number stays odd). Even + odd = odd.
4. So the sum is divisible by 3 and 37 but not by 2.

Remember · Factorise the bases first: a sum is divisible by any factor common to both terms. Even + odd is always odd.

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Answer: (d) 9

30³⁰ = 3³⁰ × 10³⁰, so the digit just before the zeros is the unit digit of 3³⁰. Powers of 3 end in 3, 9, 7, 1 in a cycle of 4, and 30 leaves remainder 2 on division by 4, giving 9.

1. 30³⁰ = 3³⁰ × 10³⁰; the factor 10³⁰ supplies the 30 zeros at the end.
2. The unit digits of 3¹, 3², 3³, 3⁴ are 3, 9, 7, 1, and then they repeat every 4 powers.
3. 30 = 4 × 7 + 2, so 3³⁰ ends like 3², i.e. in 9.
4. So the rightmost digit before the zeros is 9.

Remember · Split off the powers of 10 first; then use the four-step unit-digit cycle (remainder 0 means the fourth digit).

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Answer: (c) 3

If both numbers leave remainder 1, the divisor divides 420 and 426, and hence their HCF, 6. The divisors of 6 that are greater than 1 are 2, 3 and 6, so three numbers work (1 cannot leave a remainder of 1).

1. Remainder 1 means the divisor divides 421 − 1 = 420 and 427 − 1 = 426.
2. HCF(420, 426) = HCF(420, 6) = 6.
3. Divisors of 6 are 1, 2, 3 and 6. A divisor must be larger than the remainder, so 1 is ruled out.
4. Possible divisors: 2, 3 and 6, i.e. 3 numbers.
5. Check: 421 = 6 × 70 + 1 and 427 = 6 × 71 + 1.

Remember · Same remainder r: the divisor divides each number minus r. Count the HCF’s divisors that are larger than r.

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Answer: (b) 4

A bottle size must divide both 399 and 532 exactly, i.e. it must divide their HCF, 133 = 7 × 19. The number 133 has four divisors: 1, 7, 19 and 133.

1. 399 = 3 × 7 × 19 and 532 = 2² × 7 × 19.
2. HCF(399, 532) = 7 × 19 = 133.
3. Every common divisor divides 133: 1, 7, 19 and 133.
4. So 4 different bottle sizes are possible.
5. Check: 532 − 399 = 133, and 399 = 3 × 133, 532 = 4 × 133.

Remember · ‘Equal bottles for both’ means common divisors; their number equals the number of divisors of the HCF.

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Answer: (c) केवल 1 और 3

The primes below 10 are 2, 3, 5 and 7, and three distinct ones can be chosen in only four ways, giving sums 10, 12, 14 and 15. Unit digits 0 and 5 occur; 9 does not.

1. Primes less than 10: 2, 3, 5, 7. Choosing 3 distinct ones gives just 4 cases.
2. 2 + 3 + 5 = 10; 2 + 3 + 7 = 12; 2 + 5 + 7 = 14; 3 + 5 + 7 = 15.
3. Possible unit digits: 0, 2, 4, 5.
4. So statements 1 (0) and 3 (5) are correct; statement 2 (9) is not.

• ✓ 1. 2 + 3 + 5 = 10 ends in 0.
• ✗ 2. None of the four possible sums (10, 12, 14, 15) ends in 9.
• ✓ 3. 3 + 5 + 7 = 15 ends in 5.

Remember · When the set is tiny, list every case (four sums here) instead of reasoning in general.

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Answer: (d) 8

X + Y = 11 × (sum of the digits), and it must equal 99, so the digits add up to 9. Both numbers must be two-digit, so neither digit can be 0: X can be 18, 27, 36, 45, 54, 63, 72 or 81, which is 8 values.

1. Let X = 10a + b and Y = 10b + a; then X + Y = 11(a + b).
2. The greatest two-digit number is 99, so 11(a + b) = 99 and a + b = 9.
3. Y must also be a two-digit number, so b cannot be 0; this rules out X = 90 (its reverse is 09).
4. a = 1 to 8 gives X = 18, 27, 36, 45, 54, 63, 72, 81, i.e. 8 values.

Remember · A number plus its reverse is 11 × digit sum. Watch for a 0 digit that makes the reverse a one-digit number.

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Answer: (d) 1, 2 और 3

With p, q odd and r, s even, statement 1 contains the even factor qs and statement 2 the even factor s, so both are even. In statement 3, q + r and p + s are both odd, so the product is odd. All three are correct.

1. Parity rules: odd ± even = odd; any product with an even factor is even; odd × odd = odd.
2. Statement 1: (p − r)² is odd, but qs = odd × even is even, so the product is even. Correct.
3. Statement 2: s is even, so (q − s)q²s is even. Correct.
4. Statement 3: q + r is odd, so (q + r)² is odd; p + s is odd; odd × odd = odd. Correct.
5. Check with p = 1, q = 3, r = 2, s = 4: (−1)² × 12 = 12 (even); (−1) × 9 × 4 = −36 (even); 5² × 5 = 125 (odd).

• ✓ 1. qs is odd × even = even, so the whole product is even whatever (p − r)² is.
• ✓ 2. The factor s is even, so the product is even.
• ✓ 3. q + r and p + s are each odd + even = odd, and a product of odd numbers is odd.

Remember · For parity questions, plug in small numbers of the right parity and test each statement; it is faster than algebra.

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Answer: (b) 56

Count the digit 5 place by place in 1 to 260. It appears 26 times in the units place and 30 times in the tens place (50–59, 150–159, 250–259), a total of 56.

1. Units place: 5, 15, 25, …, 255, which is 26 numbers.
2. Tens place: 50–59, 150–159 and 250–259, 10 each, 30 in all.
3. Hundreds place: there is no page from 500 upwards, so 0.
4. Total fives = 26 + 30 = 56.
5. Check: 55, 155 and 255 each use two fives; they are counted once in the units count and once in the tens count, as they should be.

Remember · Count a digit place by place (units, tens, hundreds); numbers like 55 are then automatically counted twice.

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Answer: (b) 84

After the first 1, the sequence runs in blocks 1 | 2, 1 | 3, 2, 1 | 4, 3, 2, 1 | …. The first 28 terms are the lone 1, the full blocks up to 6 (21 terms, sum 56) and the first 6 terms of the block 7, 6, …, 1 (sum 27): 1 + 56 + 27 = 84.

1. Group the terms: 1 | 1 | 2, 1 | 3, 2, 1 | 4, 3, 2, 1 | 5, 4, 3, 2, 1 | …; after the first 1, block k is k, k − 1, …, 1.
2. Blocks 1 to 6 have 1 + 2 + 3 + 4 + 5 + 6 = 21 terms; with the first 1 that makes 22 terms.
3. Their sum = 1 + (1 + 3 + 6 + 10 + 15 + 21) = 1 + 56 = 57.
4. Terms 23 to 28 are the first 6 terms of block 7: 7 + 6 + 5 + 4 + 3 + 2 = 27.
5. Sum of the first 28 terms = 57 + 27 = 84.

Remember · In block sequences, count terms block by block (1, 2, 3, … terms) and handle the last, partial block separately.

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Answer: (a) 9

Two two-digit numbers add up to at most 197, so the leading 1 of 1CE is a carry. In the tens column A + C + carry = 10 + C, which forces A = 9 with a carry of 1 from the units column.

1. AB + CD ≤ 99 + 98 = 197, so the sum 1CE lies between 100 and 197.
2. Tens column: A + C + c = C + 10, where c (0 or 1) is the carry from the units column.
3. So A + c = 10. A is a single digit, so c = 1 and A = 9.
4. Check: 97 + 35 = 132 fits, with A = 9, B = 7, C = 3, D = 5, E = 2 all different.

Remember · In digit-sum puzzles, write each column as an equation with its carry; the leftmost column usually fixes a digit at once.

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Answer: (c) 10

32⁵ = (2⁵)⁵ = 2²⁵, so the sum is 2²⁵ + 2²⁷ = 2²⁵ × (1 + 4) = 5 × 2²⁵. It has the factors 2 and 5, so it is divisible by 10.

1. 32 = 2⁵, so 32⁵ = 2²⁵.
2. 32⁵ + 2²⁷ = 2²⁵ + 2²⁷ = 2²⁵ × (1 + 2²) = 5 × 2²⁵.
3. This contains 2 × 5 = 10, so it is divisible by 10.
4. Its only prime factors are 2 and 5, so it is not divisible by 3, 7 or 11.

Remember · Write every term with the same base, take out the common power, and read off the factors.

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Answer: (c) 5

For k = 3 and k = 4 there is only one pair with p < q: (1, 2) and (1, 3). For k = 5 there are two, (1, 4) and (2, 3), so 5 is the smallest k that does not fix p and q.

1. k = 3: only (1, 2). Unique.
2. k = 4: only (1, 3), since (2, 2) breaks p < q. Unique.
3. k = 5: (1, 4) and (2, 3). Not unique.
4. So the smallest such k is 5.

Remember · For a ‘smallest value’ question, test the options in increasing order and stop at the first that works.

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Answer: (b) 155

The terms are n³ + n² + n for n = 1, 2, 3, …: 3, 14, 39, 84, 155, 258. So the missing term is 125 + 25 + 5 = 155.

1. First differences: 11, 25, 45. Second differences: 14, 20, rising by 6, so the next second difference is 26.
2. Next first difference = 45 + 26 = 71, so the missing term = 84 + 71 = 155.
3. Check the last term: next second difference 32, so the next first difference is 71 + 32 = 103, and 155 + 103 = 258.
4. Formula view: n³ + n² + n gives 3, 14, 39, 84, 155, 258.

Remember · If first differences don’t settle a series, take second differences; a steady rise in them points to a cubic pattern.

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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.

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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.

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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.

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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.

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Answer: (a) केवल 1

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.

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Answer: (d) 1, 2 और 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.

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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.

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Answer: (d) 7, 11 और 13 से

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.

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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 · Report an error

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 · Report an error

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 · Report an error