Number system

Answer: (d) 14

All three touch 199′, and after that their landing marks coincide every LCM(4, 6, 5) = 60 feet. Counting 199, 259, … up to 979 gives 14 common marks in the range.

1. LCM of 4, 6 and 5 is 60, so common marks are 199 + 60k.
2. 199 − 60 = 139 is below 195, so the first common mark in range is 199′ itself.
3. 199 + 60k ≤ 1000 gives k ≤ 13.35, so k = 0, 1, …, 13.
4. Number of common marks = 14 (199′, 259′, …, 979′).

Remember · Common landing points of different step lengths repeat every LCM of the steps; count terms of the resulting sequence.

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Answer: (b) 111

Break the right side into primes: 2³⁷⁵ × 3²⁵ × 5¹¹⁴. The number of 10s is the smaller of the powers of 2 and 5, i.e., 114, and n must hold none of them, so m + 3 = 114.

1. 75²⁵ = (3 × 5²)²⁵ = 3²⁵ × 5⁵⁰.
2. 25³² = 5⁶⁴.
3. 32⁷⁵ = (2⁵)⁷⁵ = 2³⁷⁵.
4. Right side = 2³⁷⁵ × 3²⁵ × 5¹¹⁴, which contains exactly 10¹¹⁴ (limited by the 5s).
5. n = 2²⁶¹ × 3²⁵ is not divisible by 10, so 10ᵐ × 10³ = 10¹¹⁴ and m = 111.

Remember · Powers of 10 in a product = min(power of 2, power of 5). Count both, take the smaller.

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Answer: (b) II only

t + 1/t falls until t = 1 and rises after it, so Statement I breaks when x and y are on opposite sides of 1. Statement II simplifies to √(1/y² + 1) < √(1/x² + 1), which always holds because y > x > 0.

1. Statement I: try x = 1/2, y = 1. Left = 0·5 + 2 = 2·5; right = 1 + 1 = 2. 2·5 < 2 is false, so I is not always correct.
2. Statement II: divide inside the root by the square of the denominator: √(1 + y²)/y = √(1/y² + 1) and √(1 + x²)/x = √(1/x² + 1).
3. Since 0 < x < y, 1/y² < 1/x², so √(1/y² + 1) < √(1/x² + 1). II always holds.
4. Check II with numbers: x = 1, y = 1·5 gives 1·202 < 1·414.

• ✗ I Fails for x = 1/2, y = 1: 2·5 is not less than 2. The function t + 1/t decreases on (1/3, 1).
• ✓ II Equivalent to 1/y² < 1/x², true whenever 0 < x < y.

Remember · For 'always correct', hunt for one counter-example; for the other statement, simplify until the inequality is obvious.

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Answer: (b) 5

Measuring can remove water as well as add it: take 100 litres three times (300 litres) and take out 1 litre twice. That is 5 measurements, and 4 cannot work because 298 is 2 short of 300 and no single cylinder holds 2 litres.

1. 298 = 300 − 2 = 100 + 100 + 100 − 1 − 1: three fills of 100 litres and two removals of 1 litre = 5 measurements.
2. Could 4 work? With at most two 100-litre fills, the other two measures add at most 50 litres: 250 < 298.
3. So three 100-litre fills are needed, leaving one measure to remove exactly 2 litres — impossible with 1, 6, 25 or 100.
4. Minimum = 5.
5. Check: the greedy pour-in-only count 2 × 100 + 3 × 25 + 3 × 6 + 5 × 1 = 13 measures is the trap option.

Remember · For 'minimum measurements', aim just above the target with the big measure, then remove the excess.

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Answer: (c) 13

The product dominates the sum, so put the two largest values in the product for M and the two smallest for N. M = 5 × 4 + 3 = 23 and N = 2 × 3 + 4 = 10, so M − N = 13.

1. Largest: x × y = 5 × 4 = 20, z = 3 (largest value left): M = 23. (5 × 3 + 4 = 19 is smaller.)
2. Smallest: x × y = 2 × 3 = 6, z = 4 (smallest value left): N = 10. (2 × 4 + 3 = 11 is larger.)
3. M − N = 23 − 10 = 13.

Remember · To maximise or minimise xy + z, decide the product first; the added term matters less.

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Answer: (c) 8

Any power of 6 ends in 6. Powers of 7 end in 7, 9, 3, 1 in a cycle of four; 307 leaves remainder 3 on division by 4, so 7³⁰⁷ ends in 3. The product ends in the last digit of 6 × 3 = 18, i.e., 8.

1. 6¹²⁹ ends in 6.
2. 7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1; the cycle length is 4.
3. 307 = 4 × 76 + 3, so 7³⁰⁷ ends in 3.
4. 6 × 3 = 18, so the unit digit is 8.

Remember · Unit digits: 0, 1, 5, 6 never change; 2, 3, 7, 8 cycle every 4 — use the exponent's remainder on division by 4.

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Answer: (c) 3

Powers of 2 run 64, 128, 256, 512, 1024. Only 2⁷, 2⁸ and 2⁹ — 128, 256 and 512 — have three digits.

1. 2⁶ = 64 (two digits), 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, 2¹⁰ = 1024 (four digits).
2. Three-digit powers of 2: 128, 256, 512 — that is 3.

Remember · Know powers of 2 up to 2¹⁰ = 1024 by heart; they settle many CSAT items instantly.

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Answer: (b) 19

5 is the tens digit in 50–59 (10 times) and the units digit in 15, 25, …, 95 (9 times). Counting appearances, 55 contributes two, and the total is 19.

1. Tens place: 50, 51, …, 59 → 10 appearances.
2. Units place: 15, 25, 35, 45, 55, 65, 75, 85, 95 → 9 appearances.
3. Total = 10 + 9 = 19.

Remember · Count digit appearances place by place (tens, then units); a number like 55 is counted once in each place.

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Answer: (c) Both I and II

HCF × LCM equals the product of the two numbers, so a × b = a³ means b = a². The numbers are a and a²: their difference a(a − 1) is a product of consecutive integers and hence even, and a² is a perfect square.

1. For any two numbers, HCF × LCM = a × b.
2. a × b = a³ gives b = a² (a ≠ 1, as the numbers are distinct).
3. Difference = a² − a = a(a − 1), a product of two consecutive integers, so it is even.
4. a² is a perfect square.
5. Check: 3 and 9 — HCF 3, LCM 9, product 27 = 3³; difference 6, and 9 is a square.

• ✓ I a² − a = a(a − 1) is always even.
• ✓ II The second number is a², a perfect square.

Remember · HCF × LCM = product of the two numbers — the first thing to use in any HCF–LCM item.

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Answer: (d) 7

For divisibility by 11, the alternating digit sums must differ by a multiple of 11. Here that gives 7 − (x + y) ≡ 0 (mod 11), so x + y leaves remainder 7 when divided by 11.

1. Digits from the right: 0, 9, 7, y, 5, x, 4.
2. Odd places (1st, 3rd, 5th, 7th): 0 + 7 + 5 + 4 = 16.
3. Even places (2nd, 4th, 6th): 9 + y + x.
4. 16 − (9 + x + y) = 7 − (x + y) must be a multiple of 11.
5. So x + y = 7 or 18; either way the remainder on division by 11 is 7.
6. Check: x = 0, y = 7 gives 4057790 = 11 × 368890.

Remember · Divisibility by 11: (sum of digits in odd places) − (sum in even places) must be 0 or a multiple of 11.

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Answer: (b) 9

A − B = 9 × (difference of digits), so the digit difference must be a multiple of 3: 3 or 6 (9 would need a 0 as the units digit, making B a one-digit number). That gives 6 + 3 = 9 numbers.

1. A = 10a + b, B = 10b + a, so A − B = 9(a − b), with a > b.
2. 9(a − b) is a multiple of 27 when a − b is 3, 6 or 9.
3. B must be two-digit, so b ≥ 1. a − b = 9 would need b = 0 — not allowed.
4. a − b = 3: 41, 52, 63, 74, 85, 96 → 6 numbers.
5. a − b = 6: 71, 82, 93 → 3 numbers.
6. Total = 9. (Allowing b = 0 would give 12 — the trap option.)

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

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Answer: (a) 20

Every tile is 3′ wide, so the tiles lie in one row and their lengths must add up to exactly 100: 3a + 7b + 11c = 100. Using mostly 3′ tiles gives at most 32 tiles; using mostly 11′ tiles gives at least 12, so x − y = 20.

1. The strip is 3′ wide and every tile is 3′ on one side (a 7′ or 11′ side cannot fit across), so 3a + 7b + 11c = 100.
2. Maximum: 100 = 3 × 33 + 1 is not possible with 3′ tiles alone; 3 × 31 + 7 = 100 gives 32 tiles — the most.
3. Minimum: 11 × 8 + 3 × 4 = 100 gives 12 tiles (so does 11 × 4 + 7 × 8).
4. 11 tiles cannot work: with a + b + c = 11, the equation becomes 8c + 4b = 67, and the left side is even. Fewer tiles fail the same way or fall short in length.
5. x − y = 32 − 12 = 20.

Remember · Covering a strip: turn it into an equation of lengths, then push towards the smallest (max count) or largest (min count) pieces.

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Answer: (c) 8

Dividing N = p + q + r by N turns the condition into 1 = 1/a + 1/b + 1/c with three different whole numbers, and the only such solution is 1/2 + 1/3 + 1/6. So N must be a multiple of 6, and there are 8 multiples of 6 below 50.

1. Divide N = p + q + r by N: 1 = p/N + q/N + r/N = 1/a + 1/b + 1/c, where a = N/p, b = N/q, c = N/r are whole numbers.
2. Distinct factors mean a, b, c are distinct. Take a < b < c. If a ≥ 3, the sum is at most 1/3 + 1/4 + 1/5 < 1, so a = 2.
3. Then 1/b + 1/c = 1/2 with 2 < b < c: b = 3 gives c = 6; b = 4 gives c = 4 (not distinct); b ≥ 5 is too small. So (a, b, c) = (2, 3, 6).
4. Hence N must be divisible by 2, 3 and 6, i.e. a multiple of 6, and then N = N/2 + N/3 + N/6 always works.
5. Multiples of 6 below 50: 6, 12, 18, 24, 30, 36, 42, 48 — that is 8 numbers.
6. Check: 6 = 3 + 2 + 1 and 48 = 24 + 16 + 8.

Remember · When a number equals a sum of its own factors, divide through by it — the problem becomes unit fractions adding to 1.

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Answer: (d) More than 6

Why not the tempting option · UPSC's key is (d). The item says 'three prime numbers p, q and r', not three different primes, and uses 'distinct' only for the sums — so p = q = r is allowed, every prime below 20 can be the middle term and there are 8 sums. Assuming the primes must differ gives 4 sums, option (a), but that assumption is not in the item. In the exam, do not add 'distinct' where the question has not written it.

Since p − q = q − r, the three primes are in arithmetic progression and p + q + r = 3q. The item does not say the primes must be different, so p = q = r is allowed and each of the 8 primes below 20 can be q, giving 8 distinct sums — more than 6.

1. p − q = q − r gives p + r = 2q, so p + q + r = 3q. The sum depends only on the middle prime q.
2. The item does not require the primes to be different, so p = q = r is allowed (both differences are then 0).
3. So every prime q below 20 gives a value: q = 2, 3, 5, 7, 11, 13, 17, 19 gives p + q + r = 6, 9, 15, 21, 33, 39, 51, 57.
4. That is 8 distinct values, which is more than 6.
5. Trap: if one assumed the primes must be different, only (3, 5, 7), (3, 7, 11), (3, 11, 19), (5, 11, 17) and (7, 13, 19) would work, giving 4 sums — but the item says 'distinct' only of the sums, not of the primes, so equal primes count.

Remember · Equal differences mean an arithmetic progression: the sum of three terms is three times the middle one. Apply 'distinct' only where the question states it.

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Answer: (c) Three

Only three unordered triples satisfy 1/p + 1/q + 1/r = 1: (3, 3, 3), (2, 4, 4) and (2, 3, 6). Their sums 9, 10 and 11 are all different, so p + q + r can take three values.

1. Arrange so that p ≤ q ≤ r. Then 1/p is the largest of the three fractions, so 1/p ≥ 1/3, i.e. p ≤ 3; and p ≥ 2 because 1/p must be less than 1.
2. p = 3: 1/q + 1/r = 2/3 with q, r ≥ 3 forces q = r = 3. Sum = 9.
3. p = 2: 1/q + 1/r = 1/2 with q ≤ r gives (q, r) = (3, 6) or (4, 4). Sums = 11 and 10.
4. Possible values of p + q + r: 9, 10 and 11 — three values.
5. Check: 1/2 + 1/3 + 1/6 = 1, 1/2 + 1/4 + 1/4 = 1, 1/3 + 1/3 + 1/3 = 1.

Remember · For unit-fraction equations, sort the variables and bound the smallest one first; the cases then collapse quickly.

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Answer: (c) 1078

Each remainder is 2 less than its divisor, so N + 2 is a multiple of LCM(3, 5, 6, 9) = 90. The smallest 4-digit number of the form 90k − 2 is 1080 − 2 = 1078.

1. In every case the remainder is 2 short of the divisor: 3 − 1 = 5 − 3 = 6 − 4 = 9 − 7 = 2.
2. So N + 2 is divisible by 3, 5, 6 and 9, i.e. by their LCM, 90.
3. N = 90k − 2. k = 11 gives 988 (only 3 digits); k = 12 gives 1080 − 2 = 1078.
4. Check: 1078 = 3 × 359 + 1 = 5 × 215 + 3 = 6 × 179 + 4 = 9 × 119 + 7.

Remember · If divisor minus remainder is the same d for every divisor, the number is (LCM × k) − d.

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Answer: (c) 5

The product of all odd numbers up to 999 is odd and contains 5 as a factor. An odd multiple of 5 always ends in 5.

1. The product contains the factor 5, so it is a multiple of 5 and its unit digit is 0 or 5.
2. Every factor is odd, so the product is odd and cannot end in 0.
3. Hence the unit digit is 5.
4. Check: 1 × 3 × 5 = 15 already ends in 5, and multiplying any number ending in 5 by an odd number keeps the last digit 5.

Remember · Odd × 5 ends in 5; an even number times 5 ends in 0. Spot a factor of 5 before multiplying anything.

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Answer: (c) 22

Divisibility by 9 is already covered by 3, so count the numbers from 1 to 100 that are not divisible by 2, 3, 5 or 7. Below 121 these are just 1 and the primes from 11 to 97, which gives 1 + 21 = 22.

1. 9 is a multiple of 3, so only 2, 3, 5 and 7 matter.
2. Multiples of 2, 3, 5, 7 up to 100: 50 + 33 + 20 + 14 = 117.
3. Multiples of 6, 10, 14, 15, 21, 35: 16 + 10 + 7 + 6 + 4 + 2 = 45.
4. Multiples of 30, 42, 70, 105: 3 + 2 + 1 + 0 = 6; multiples of 210: 0.
5. Divisible by at least one = 117 − 45 + 6 = 78, so not divisible by any = 100 − 78 = 22.
6. Check: such numbers are 1 and the 21 primes from 11 to 97 (the smallest composite without the factors 2, 3, 5, 7 is 11² = 121), so 1 + 21 = 22.

Remember · Drop redundant divisors first (9 with 3). Up to 120, numbers free of 2, 3, 5, 7 are just 1 and primes from 11 onwards.

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Answer: (d) None of the above

The largest x + y is 8 + 7 = 15 and the smallest x − y is 4 − 7 = −3, so the ratio is −5. None of the listed values matches; 15/2 comes from wrongly taking the smallest x − y as 4 − 2.

1. Maximum of (x + y): take both at their largest, 8 + 7 = 15.
2. Minimum of (x − y): take the smallest x and the largest y, 4 − 7 = −3.
3. Ratio = 15 ÷ (−3) = −5.
4. −5 is not 6, 15/2 or −15/2, so the answer is 'None of the above'.
5. Check: the trap 15/2 uses 4 − 2 = 2, but y = 7 makes x − y smaller.

Remember · To minimise a difference, take the smallest first term and the largest second term — and keep track of the sign.

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Answer: (b) 5

If p and k were both odd, p² + k would be even and larger than 2, so one of them must be 2. With p = 2, k can be 3, 7, 13 or 19; with k = 2 (and p = 3, giving 11) k = 2 also works, so k has 5 possible values.

1. If p and k are both odd primes, p² + k is odd + odd = even and greater than 2, so it cannot be prime. Hence p = 2 or k = 2.
2. p = 2: 4 + k must be a prime below 30. k = 3, 7, 13, 19 give 7, 11, 17, 23 (prime); k = 2, 5, 11, 17, 23 give 6, 9, 15, 21, 27 (not prime).
3. k = 2: p² + 2 must be a prime below 30. p = 3 gives 11 (prime); p = 5 gives 27 (not prime). So k = 2 is possible.
4. Possible values of k: 2, 3, 7, 13, 19 — five values.

Remember · Odd + odd is even: when a sum of primes must be prime, one of them is usually 2.

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Answer: (d) More than 8

Every number in such a set is a divisor of 1001 = 7 × 11 × 13. The sets that include 1 alone number 12, already more than 8; a full count gives 32 sets of three different numbers.

1. 1001 = 7 × 11 × 13, so every number must be one of its divisors: 1, 7, 11, 13, 77, 91, 143, 1001.
2. The LCM must contain 7, 11 and 13, and the HCF must be 1, so no prime may divide all three numbers.
3. Sets containing 1 automatically have HCF 1. Those with LCM 1001: {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, and {1, x, 1001} for x = 7, 11, 13, 77, 91, 143 — 12 sets.
4. 12 is already more than 8; sets such as {7, 11, 13} and {7, 11, 143} add still more (32 in all with three different numbers).
5. So n is more than 8.

Remember · With a 'more than' option, you only need to cross the threshold: count the easiest family of cases and stop.

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Answer: (b) 3

Writing the numbers in place-value form gives 30 + R − S − T = 20P, which forces P = 1 and S + T = R + 10. With distinct digits, Q = 3 and T below 6, only (R, S) = (2, 8), (2, 7) and (4, 9) work.

1. PQR − PS = PPT means (100P + 10Q + R) − (10P + S) = 100P + 10P + T.
2. Simplify: 90P + 10Q + R − S = 110P + T, so 10Q + R − S − T = 20P. With Q = 3: 30 + R − S − T = 20P.
3. The left side is at most 30 + 9 − 1 − 1 = 37, so 20P = 20 and P = 1. Then S + T = R + 10.
4. P = 1 and Q = 3 are used, so R, S, T are different digits from 2, 4, 5, 6, 7, 8, 9, with T = 2, 4 or 5.
5. T = 2: S = R + 8 needs R = 1 (not allowed). T = 4: S = R + 6 gives R = 2, S = 8. T = 5: S = R + 5 gives (R, S) = (2, 7) or (4, 9).
6. So (R, S) has 3 possible values.
7. Check: 132 − 18 = 114, 132 − 17 = 115, 134 − 19 = 115.

Remember · For letter-digit equations, expand in place values, simplify, and use digit limits to fix the leading digit first.

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Answer: (b) 4

35ⁿ needs n factors of 5 and n factors of 7. The product has only four 5s (one from 385, three from 1000) but ten 7s, so the maximum n is 4.

1. 35 = 5 × 7, so count the 5s and the 7s in the product; n is the smaller count.
2. Fives: 385 = 5 × 7 × 11 gives one, and 1000 = 2³ × 5³ gives three; 7, 343, 2401 and 77777 have none. Total = 4.
3. Sevens: 7 (one), 343 = 7³ (three), 385 (one), 2401 = 7⁴ (four), 77777 = 7 × 11111 (one). Total = 10.
4. Each 35 needs one 5 and one 7, so n = the smaller of 4 and 10 = 4.

Remember · For divisibility by a power of a composite number, count each prime factor separately; the scarcer prime sets the limit.

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Answer: (b) 9

A perfect square with 17 digits has a square root of (17 + 1)/2 = 9 digits. Indeed 111111111² = 12345678987654321, and 111111111 has 9 digits.

1. 12345678987654321 has 17 digits.
2. A square with an odd number of digits d has a root with (d + 1)/2 digits: (17 + 1)/2 = 9.
3. Check: 111111111² = 12345678987654321, and 111111111 has 9 digits.

Remember · Digits in √N: (d + 1)/2 if N has d digits with d odd, d/2 if d is even. Also, 11…1² gives 12…n…21.

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Answer: (c) 2

1ⁿ is always 1, and 2ⁿ is a multiple of 4 once n ≥ 2. So the remainder is 3 when n = 1 and 1 for every larger n — two distinct remainders.

1. n = 1: 1 + 2 = 3, remainder 3.
2. n ≥ 2: 2ⁿ is divisible by 4 and 1ⁿ = 1, so 1ⁿ + 2ⁿ leaves remainder 1.
3. Only two remainders occur: 3 and 1.
4. Check: n = 2 gives 5 (remainder 1), n = 3 gives 9 (remainder 1).

Remember · Powers of 2 are multiples of 4 from 2² onwards; always check the first case separately.

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Answer: (c) 37

Every number of the form QQQ equals Q × 111 = Q × 3 × 37, and 481 = 13 × 37. They always share the factor 37, and 13 can never divide 3Q for a single digit Q, so the HCF is 37.

1. QQQ = Q × 111 = Q × 3 × 37.
2. 481 = 13 × 37.
3. HCF = 37 × HCF(3Q, 13). For a single digit Q, 3Q is at most 27 and is never 13 or 26, so HCF(3Q, 13) = 1.
4. So the HCF of P and 481 is 37, whatever the digit Q.
5. Check: HCF(222, 481) = 37, since 222 = 6 × 37 and 481 = 13 × 37.

Remember · Repeated-digit numbers: aaa = a × 3 × 37. Knowing 111 = 3 × 37 settles such items at sight.

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Answer: (d) 9

The numbers 1 to 99 use 9 + 180 = 189 digits. The remaining 300 digits cover exactly 100 three-digit numbers, 100 to 199, so the 489th digit is the final 9 of 199.

1. Digits from 1–9: 9. Digits from 10–99: 90 × 2 = 180. Total so far: 189.
2. Digits still needed: 489 − 189 = 300, all from three-digit numbers.
3. 300 ÷ 3 = 100 exactly, so the 489th digit is the last digit of the 100th three-digit number.
4. The 100th three-digit number is 100 + 99 = 199; its last digit is 9.
5. Check: digits 487–489 are 1, 9, 9 (from 199), and digit 490 is the 2 of 200.

Remember · Count digit blocks: 9 one-digit, 180 two-digit, 2700 three-digit. An exact division means the digit is the last one of that number.

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: (b) 4

P = 3T allows (T, P) = (1, 3), (2, 6) or (3, 9), and S = R + 7 allows (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9). Keeping all five digits different leaves 35291, 63072, 64182 and 94183 — four numbers.

1. P = 3T with T ≠ 0 and P a single digit: (T, P) = (1, 3), (2, 6) or (3, 9).
2. Q = R + 3 and S = Q + 4 = R + 7 ≤ 9, so R = 0, 1 or 2: (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9).
3. T = 1, P = 3: only (2, 5, 9) avoids repeats → 35291.
4. T = 2, P = 6: (0, 3, 7) and (1, 4, 8) work → 63072 and 64182.
5. T = 3, P = 9: only (1, 4, 8) works → 94183.
6. Total: 4 numbers.

Remember · Tie the chained conditions to one variable (here R), list the few cases for each part, then strike out repeated digits.

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) Both I and II

'There exists' needs only one example. The number 2 becomes 3 after a 50% increase and 5 after a 150% increase; all three are primes with exactly two factors, so both statements are correct.

1. Statement I: 2 has 2 factors. Increased by 50%, it becomes 3, which also has 2 factors. So I is correct.
2. Statement II: 2 increased by 150% becomes 2 × 2.5 = 5, which also has 2 factors. So II is correct.
3. Both statements are correct.
4. Check: other examples exist too, e.g. 10 → 15 (4 factors each) for I and 6 → 15 (4 factors each) for II.

• ✓ I 2 → 3: both are primes with exactly 2 factors.
• ✓ II 2 → 5: both are primes with exactly 2 factors.

Remember · 'There exists' statements need just one example — try small primes first.

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: (a) 0

Every power of 9 leaves remainder 3 when divided by 6, and there are 98 terms. 98 × 3 = 294 is a multiple of 6, so the remainder is 0.

1. 9 leaves remainder 3 when divided by 6, and 3 × 3 = 9 again leaves 3. So every power 9ᵏ leaves remainder 3.
2. Number of terms from 9³ to 9¹⁰⁰: 100 − 3 + 1 = 98.
3. Sum of remainders = 98 × 3 = 294 = 6 × 49, so the remainder is 0.
4. Check: each term is odd and a multiple of 3; 98 odd terms add to an even number, and an even multiple of 3 is divisible by 6.

Remember · For the remainder of a sum, reduce each term first; count terms carefully (3 to 100 is 98 terms).

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