Number system and basic algebra (divisibility, HCF/LCM, remainders, digits, equations): previous-year questions
192 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 95 of them came in the last five papers.
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Consider the first 100 natural numbers. How many of them are not divisible by any one of 2, 3, 5, 7 and 9?
Answer (c) & explanation
1. 9 adds nothing new: anything divisible by 9 is already divisible by 3. So we need numbers not divisible by 2, 3, 5 or 7.
2. Any number up to 100 that is not prime has a factor of 2, 3, 5 or 7 (because 11 × 11 is already more than 100).
3. So the numbers we want are just 1 and the primes bigger than 7.
4. There are 25 primes up to 100. Leave out 2, 3, 5 and 7: 21 primes. Add the number 1: 22.
9 is in the list to make you count its multiples twice.
Answer as per the official UPSC answer key.
Consider the following statements:
- I. There exists a natural number which when increased by 50% can have its number of factors unchanged.
- II. There exists a natural number which when increased by 150% can have its number of factors unchanged.
Which of the statements given above is/are correct?
Answer (c) & explanation
Both statements are correct (c). Step 1: "there exists" needs just one example. Step 2: statement I: increase by 50% means × 1.5. Take 2 → 3. Both are primes with exactly 2 factors. So the factor count is unchanged: true. Step 3: statement II: increase by 150% means × 2.5. Take 2 → 5. Again both are primes with 2 factors: true. Options (a), (b) and (d) fail.
Tip: for "there exists" statements, hunt for one simple example; for "always" statements, hunt for one counter-example. Primes are handy because they all have exactly two factors.
Answer as per the official UPSC answer key.
If 4 ≤ x ≤ 8 and 2 ≤ y ≤ 7, then what is the ratio of maximum value of (x + y) to minimum value of (x − y)?
Answer (d) & explanation
The ratio is −5, which is not listed, so the answer is (d). Step 1: the maximum of x + y uses the largest x and largest y: 8 + 7 = 15. Step 2: the minimum of x − y uses the smallest x and the largest y: 4 − 7 = −3. Step 3: ratio = 15 ÷ (−3) = −5. None of the options show −5. Option (c), −15/2, divides by −2 instead of −3. Options (a) and (b) ignore the minus sign.
Tip: to make a difference smallest, make the first number small and the subtracted number big. Always check "None of the above" before forcing a match.
to make x − y small you take y large, not small.
Answer as per the official UPSC answer key.
What is the remainder when 93 + 94 + 95 + 96 + … + 9100 is divided by 6?
Answer (a) & explanation
The remainder is 0 (a). Step 1: every power of 9 leaves remainder 3 when divided by 6. (9 = 6 + 3; 81 = 78 + 3; and so on, since 3 × 9 leaves 3 again.) Step 2: the terms run from 9³ to 9¹⁰⁰: that is 100 − 3 + 1 = 98 terms. Total remainder = 98 × 3 = 294. Step 3: 294 ÷ 6 = 49 exactly, so the remainder is 0. Options (b), (c) and (d) miscount the terms or the remainder.
Tip: count terms carefully: from a to b there are b − a + 1 terms. Then multiply by the common remainder and reduce.
Answer as per the official UPSC answer key.
Let both p and k be prime numbers such that (p2 + k) is also a prime number less than 30. What is the number of possible values of k?
Answer (b) & explanation
Five values of k are possible (b). Step 1: if p = 2, then p² = 4, and 4 + k must be a prime below 30. k = 3, 7, 13, 19 work (sums 7, 11, 17, 23). k = 2 gives 6, not prime. Step 2: if p is odd, p² is odd, so k must be 2 (the even prime) to keep the sum odd. p = 3 gives 9 + 2 = 11, prime. p = 5 gives 27, not prime. So k = 2 also works. Step 3: possible k: 2, 3, 7, 13, 19 = 5 values. Options (a), (c) and (d) miscount.
Tip: in prime questions, split into "2" and "odd" cases. Odd + odd is even, so the even prime 2 often decides things.
Answer as per the official UPSC answer key.
Let p + q = 10, where p, q are integers.
Value-I = Maximum value of p × q when p, q are positive integers.
Value-II = Maximum value of p × q when p ≥ −6, q ≥ −4.
Which one of the following is correct?
Answer (c) & explanation
The two values are equal (c). Step 1: Value-I: p + q = 10 with positive whole numbers. A product is largest when the two numbers are equal: 5 × 5 = 25. Step 2: Value-II allows p down to −6 and q down to −4, so p can range from −6 to 14. The new pairs, like (−6, 16) or (14, −4), have negative products. The best is still 5 × 5 = 25. Step 3: so Value-I = Value-II. Options (a), (b) and (d) fail.
Tip: for a fixed sum, the product is greatest when the numbers are as equal as possible. Allowing negative numbers only adds smaller (negative) products here.
Answer as per the official UPSC answer key.
There are n sets of numbers each having only three positive integers with LCM equal to 1001 and HCF equal to 1. What is the value of n?
Answer (d) & explanation
There are more than 8 such sets (d). Step 1: 1001 = 7 × 11 × 13. Each number in a set must be made from these primes (or be 1). Step 2: LCM 1001 means every prime appears somewhere; HCF 1 means no prime is in all three. Step 3: list some sets: {7, 11, 13}, {1, 77, 91}, {1, 77, 143}, {1, 91, 143}, {7, 11, 1001}, {1, 7, 143}, {1, 11, 91}, {1, 13, 77}, {77, 91, 143}. That is already 9, so n is more than 8. Options (a), (b) and (c) are too small.
Tip: when an option says "more than", you only need to find enough cases to pass it; no full count is needed. 1001 = 7 × 11 × 13 is worth remembering.
Answer as per the official UPSC answer key.
Let x be a real number between 0 and 1. Which of the following statements is/are correct?
- I. x2 > x3
- II. x > √x
Select the correct answer using the code given below:
Answer (a) & explanation
Only statement I is correct (a). Step 1: for a number between 0 and 1, multiplying by it makes things smaller. Test x = 1/4. Step 2: statement I: x² = 1/16 and x³ = 1/64. Since 1/16 > 1/64, x² > x³: true. Step 3: statement II: √x = √(1/4) = 1/2, which is bigger than 1/4. So x > √x is false. Options (b), (c) and (d) fail.
Tip: between 0 and 1, higher powers get smaller and roots get bigger: √x > x > x² > x³. Above 1 the order flips. Test with an easy fraction like 1/4.
between 0 and 1 the usual "squaring makes it bigger" instinct runs backwards.
Answer as per the official UPSC answer key.
Let PQR be a 3-digit number, PPT be a 3-digit number and PS be a 2-digit number, where P, Q, R, S, T are distinct non-zero digits. Further, PQR − PS = PPT. If Q = 3 and T < 6, then what is the number of possible values of (R, S)?
Answer (b) & explanation
1. Write it as an ordinary subtraction: PQR − PS = PPT. The answer keeps P in front, and its middle digit is also P.
2. Middle column: 3 − P (minus 1 if we borrowed) must give P. Only P = 1 with a borrow works: 3 − 1 − 1 = 1.
3. Last column, after borrowing: (R + 10) − S = T. So S + T = R + 10.
4. T can be 2, 4 or 5 (not 1 or 3, which are taken, and less than 6). Try each, using only unused digits:
5. T = 2: S = R + 8 needs R = 1, which is taken. None.
6. T = 4: S = R + 6 gives R = 2, S = 8. One pair.
7. T = 5: S = R + 5 gives (R, S) = (2, 7) or (4, 9). Two pairs.
8. Total: 3 pairs. Check one: 132 − 18 = 114.
Count pairs (R, S), not separate digits: (2, 8), (2, 7) and (4, 9).
Answer as per the official UPSC answer key.
The difference between any two natural numbers is 10. What can be said about the natural numbers which are divisible by 5 and lie between these two numbers?
Answer (c) & explanation
There can be more than one such number (c). Step 1: take two numbers 10 apart, such as 5 and 15. Strictly between them, only 10 is divisible by 5. That is one number. Step 2: now take 1 and 11. Between them, 5 and 10 are divisible by 5. That is two numbers. Step 3: the count can be one or two depending on the pair. So "only one" (a) and "only two" (b) are not always true, and (d) is false. The safe statement is (c): there can be more than one.
Tip: test both an "edge" case (ends are multiples of 5) and a general case. When answers vary, pick the option that allows for it.
Answer as per the official UPSC answer key.
What is the maximum value of n such that 7 × 343 × 385 × 1000 × 2401 × 77777 is divisible by 35n?
Answer (b) & explanation
The maximum n is 4 (b). Step 1: 35ⁿ = 5ⁿ × 7ⁿ. So we need to count both the 5s and the 7s and take the smaller count. Step 2: 7s are plentiful: 7, 343 = 7³, 2401 = 7⁴, 385 and 77777 = 7 × 11111 all add 7s. So count the 5s. Step 3: 385 = 5 × 7 × 11 gives one 5; 1000 = 2³ × 5³ gives three. The others have none. Total 5s = 4, so n = 4. Options (a), (c) and (d) miscount.
Tip: for the highest power of a composite number, split it into primes and count the scarcer prime. Here 5 is the limiting factor.
Answer as per the official UPSC answer key.
A natural number N is such that it can be expressed as N = p + q + r, where p, q and r are distinct factors of N. How many numbers below 50 have this property?
Answer (c) & explanation
Eight numbers below 50 have this property (c). Step 1: if p, q, r are factors of N, write them as N/a, N/b, N/c. Then N = N/a + N/b + N/c, so 1/a + 1/b + 1/c = 1 with a, b, c different. Step 2: the only set of different whole numbers that works is 2, 3, 6 (1/2 + 1/3 + 1/6 = 1). So N = N/2 + N/3 + N/6, and N must be divisible by 6. Step 3: multiples of 6 below 50: 6, 12, 18, 24, 30, 36, 42, 48 = 8 numbers. Example: 12 = 6 + 4 + 2. Options (a), (b) and (d) miscount.
Tip: turning factors into fractions of N often reveals a neat rule. Remember 1/2 + 1/3 + 1/6 = 1.
Answer as per the official UPSC answer key.
If N2 = 12345678987654321, then how many digits does the number N have?
Answer (b) & explanation
N has 9 digits (b). Step 1: recall the pattern: 11² = 121, 111² = 12321, 1111² = 1234321, and so on. The square rises to the number of 1s, then falls. Step 2: 12345678987654321 rises to 9, so N = 111111111, which has 9 digits. Step 3: quick check: the square has 17 digits. A number with d digits has a square of 2d − 1 or 2d digits, so d = 9. Options (a), (c) and (d) fail.
Tip: learn this palindrome pattern of repunits (numbers made of 1s). It works up to nine 1s; after that, carries spoil it.
Answer as per the official UPSC answer key.
Three prime numbers p, q and r, each less than 20, are such that p – q = q – r. How many distinct possible values can we get for (p + q + r)?
Answer (d) & explanation
More than 6 sums are possible (d). Step 1: p − q = q − r means p + r = 2q, so the sum p + q + r = 3q. Each different q gives a different sum. Step 2: the question does not say the primes must be different. So p = q = r is allowed (for example, 7 − 7 = 7 − 7). Then every prime below 20 can be q: 2, 3, 5, 7, 11, 13, 17, 19. Step 3: that gives 8 sums, which is more than 6. Options (a), (b) and (c) assume the primes must be distinct.
Tip: read the conditions closely. If a question does not say "distinct", equal values are allowed, and that often changes the answer.
read what is not said. Without "distinct", equal primes are allowed. With distinct primes you would get only 4 (q = 5, 7, 11, 13).
Answer as per the official UPSC answer key.
If n is a natural number, then what is the number of distinct remainders of (1n + 2n) when divided by 4?
Answer (c) & explanation
There are two distinct remainders (c). Step 1: try n = 1: 1 + 2 = 3, which leaves remainder 3 when divided by 4. Step 2: for n = 2 or more, 2ⁿ is a multiple of 4 (4, 8, 16, …), so it leaves remainder 0. And 1ⁿ is always 1. So 1ⁿ + 2ⁿ leaves remainder 1. Step 3: the only remainders are 3 and 1: two values. Options (a), (b) and (d) fail.
Tip: test small values of n first; patterns usually settle after one or two steps. Remember any power of 2 from 2² onward is divisible by 4.
don't stop at n ≥ 2. The single case n = 1 adds a second remainder.
Answer as per the official UPSC answer key.
On January 1st, 2023, a person saved ₹ 1. On January 2nd, 2023, he saved ₹ 2 more than that on the previous day. On January 3rd, 2023, he saved ₹ 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube?
Answer (b) & explanation
His total was a perfect square and cube on 8th January (b). Step 1: daily savings are 1, 3, 5, 7, … (odd numbers). The sum of the first n odd numbers = n². So after n days he has n² rupees. Step 2: n² is also a perfect cube only when n itself is a cube: n = 1, 8, 27, … Step 3: n = 8 gives 8² = 64 = 4³, both a square and a cube. So the 8th day, 8th January. Option (a) gives 49 (not a cube); (c) gives 81 (not a cube). Option (d) is wrong.
Tip: remember 1 + 3 + 5 + … (n terms) = n². A number that is both a square and a cube is a sixth power: 1, 64, 729.
Answer as per the official UPSC answer key.
What is the angle between the minute hand and hour hand when the clock shows 4:25 hours?
Answer (c) & explanation
The angle is 17.5° (c). Step 1: the minute hand moves 6° a minute. At 25 minutes it is 25 × 6 = 150° from 12. Step 2: the hour hand moves 30° each hour and 0.5° each minute. At 4:25 it is 4 × 30 + 25 × 0.5 = 120 + 12.5 = 132.5°. Step 3: the gap is 150 − 132.5 = 17.5°. Option (a), 12.5°, is just the hour hand's extra movement. Options (b) and (d) are wrong.
Tip: use the formula angle = |30H − 5.5M|. Here |120 − 137.5| = 17.5°. If the answer is above 180°, subtract it from 360°.
(b) 15° forgets that the hour hand also moves during the 25 minutes.
Answer as per the official UPSC answer key.
222333 + 333222 is divisible by which of the following numbers?
Answer (b) & explanation
The sum is divisible by 3 and 37 but not by 2 (b). Step 1: break the bases: 222 = 6 × 37 and 333 = 9 × 37. Both are multiples of 3 and 37, so both terms, and their sum, are divisible by 3 and 37. Step 2: check 2: 222 is even, so 222^333 is even. 333 is odd, so 333^222 is odd. Step 3: even + odd = odd, so the sum is not divisible by 2. Options (a), (c) and (d) wrongly include 2 or exclude 37 or 3.
Tip: for divisibility of a sum, check each term. Even + odd is always odd. Spotting a common factor (here 111 = 3 × 37) saves big calculations.
Answer as per the official UPSC answer key.
What is the number of fives used in numbering a 260-page book?
Answer (b) & explanation
There are 56 fives (b). Step 1: count fives in the units place: 5, 15, 25, …, 255. That is one in every 10 pages: 26 fives. Step 2: count fives in the tens place: 50–59, 150–159 and 250–259, ten each: 30 fives. (Pages 55, 155, 255 get counted in both places, which is right, since they each have two fives.) Step 3: the hundreds place has no 5, because pages stop at 260. Total = 26 + 30 = 56. Options (a), (c) and (d) miscount.
Tip: count digit by digit, place by place. Do not try to list numbers; count the units, tens and hundreds places separately.
55 and 155 have two fives each; the place-by-place count handles them without double counting.
Answer as per the official UPSC answer key.
What is the rightmost digit preceding the zeros in the value of 3030?
Answer (d) & explanation
The digit before the zeros is 9 (d). Step 1: 30^30 = 3^30 × 10^30. The 10^30 gives 30 zeros, so the digit just before them is the unit digit of 3^30. Step 2: unit digits of powers of 3 repeat in a cycle of four: 3, 9, 7, 1. Step 3: divide 30 by 4: remainder 2. So take the second digit in the cycle: 9. Options (a) 1, (b) 3 and (c) 7 are the other digits in the cycle.
Tip: for unit digits of powers, find the cycle (most digits repeat every 4). Divide the power by 4 and use the remainder; a remainder of 0 means the last digit of the cycle.
Answer as per the official UPSC answer key.