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.
How often UPSC asks this
Questions per year, counted from the official papers.
Attempt the questions
Newest papers first. Tap an option; the answer and the reason appear instantly.
How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers ?
Answer (d) & explanation
There are 9 such numbers (d). Step 1: if 1186 leaves remainder 31, then the divisor n divides 1186 − 31 = 1155. Also, n must be greater than 31 (a remainder is always smaller than the divisor). Step 2: 1155 = 3 × 5 × 7 × 11, which has 2 × 2 × 2 × 2 = 16 divisors. Step 3: those above 31 are 33, 35, 55, 77, 105, 165, 231, 385 and 1155. That is 9. Options (a), (b) and (c) miss some.
Tip: never forget the rule that the divisor must exceed the remainder. It removes the small divisors.
a remainder of 31 needs a divisor larger than 31. Counting all 16 divisors is the trap.
Answer as per the official UPSC answer key.
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. p + q + r – s – t is definitely even.
- 2. 2p + q + 2r – 2s + t is definitely odd.
Which of the above statements is/are correct ?
Answer (a) & explanation
Only statement 1 is correct (a). Step 1: subtracting a number has the same odd-even effect as adding it. So p + q + r − s − t has the same parity as the sum of all five. Step 2: the five include two odd numbers, and two odds make an even sum. So statement 1 is always even, and true. Step 3: in 2p + q + 2r − 2s + t, the doubled terms are even. The parity depends on q + t, which depends on which numbers are odd. So it is not definitely odd. Options (b), (c) and (d) fail.
Tip: for odd-even questions, ignore the signs and any doubled terms. Just count the odd numbers left.
Answer as per the official UPSC answer key.
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. The number of possible values of p is 5.
- 2. The number of possible values of q is 6.
Which of the above statements is/are correct ?
Answer (c) & explanation
Both statements are correct (c). Step 1: pp = 11p, so 11(p + q + r) = tt0 = 110t. So p + q + r = 10t: either 10 or 20. Step 2: sum 10 with p < q < r: (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). Step 3: p takes 1, 2, 3, 4, 5 (five values). q takes 2, 3, 4, 6, 7, 8 (six values). So options (a), (b) and (d) fail.
Tip: repeated-digit numbers like pp are 11 × p. Spotting this turns the puzzle into a simple sum.
Answer as per the official UPSC answer key.
Consider the following in respect of prime number p and composite number c.
- 1. (p + c)/(p – c) can be even.
- 2. 2p + c can be odd.
- 3. pc can be odd.
Which of the statements given above are correct ?
Answer (d) & explanation
All three statements can be true (d). Step 1: "can be" needs only one example each. Step 2: statement 1: p = 11, c = 9. (11 + 9) ÷ (11 − 9) = 20 ÷ 2 = 10, which is even. Statement 2: c = 9. 2p is even, and even + 9 is odd. Step 3: statement 3: p = 3, c = 9. 3 × 9 = 27, odd. So options (a), (b) and (c) fail.
Tip: remember that composite numbers can be odd (9, 15, 21). Many "can be" traps assume composites are even.
Answer as per the official UPSC answer key.
If today is Sunday, then which day is it exactly on 1010 th day ?
Answer (a) & explanation
The 10¹⁰th day is a Wednesday (a). Step 1: today (Sunday) is day 1. So the 10¹⁰th day is 10¹⁰ − 1 days later. Step 2: 10 leaves remainder 3 when divided by 7. So 10¹⁰ leaves the same remainder as 3¹⁰. Since 3⁶ leaves 1, 3¹⁰ leaves the same as 3⁴ = 81, which is 4. Step 3: so 10¹⁰ − 1 leaves 3. Sunday + 3 days = Wednesday. Option (b), Thursday, forgets that today is day 1.
Tip: "the nth day" counting today as day 1 means n − 1 days ahead. This off-by-one trap is common.
today is day 1, not day 0. Counting it as day 0 gives Thursday, which is an option.
UPSC repeats traps like this year after year. The PYQ Book gathers every trap in one place, so you see them coming. Try the free sample
Answer as per the official UPSC answer key.
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 (a) & explanation
A + B + C is 18 (a). Step 1: 37DD = 3700 + 11D. Since ABC × D = 37DD, D must divide 3700. So D is 1, 2, 4 or 5. Step 2: D = 1 or 2 makes ABC a 4-digit number. D = 5 gives 3755 ÷ 5 = 751, which repeats the digit 5. Step 3: D = 4: 3744 ÷ 4 = 936. All digits differ. A + B + C = 9 + 3 + 6 = 18. Option (d) is wrong: the data fix the answer.
Tip: split a coded number into a fixed part and a part with the unknown. Divisibility then narrows the choices quickly.
Answer as per the official UPSC answer key.
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 (b) & explanation
They turn green together again at 4:10 p.m. (b). Step 1: green and red last the same time. So each full cycle is double: 50 s, 78 s and 120 s. Step 2: LCM of 50, 78 and 120 = 2³ × 3 × 5² × 13 = 7800 s = 130 minutes. Step 3: 2:00 p.m. + 2 hours 10 minutes = 4:10 p.m. Option (a), 4:00 p.m., uses wrong cycles. Options (c) and (d) miscalculate.
Tip: always use the full cycle (green + red) for "turn green again". Using only the green time is the common trap.
LCM of 25, 39 and 60 gives the next time they all change, not the next time they all turn green.
Answer as per the official UPSC answer key.
For any choices of values of X, Y and Z, the 6-digit number of the form XYZXYZ is divisible by :
Answer (d) & explanation
XYZXYZ is always divisible by 7, 11 and 13 (d). Step 1: write it out: XYZXYZ = XYZ × 1000 + XYZ = XYZ × 1001. Step 2: 1001 = 7 × 11 × 13. Step 3: so every such number is divisible by all three. Options (a), (b) and (c) each leave one out. For example, 123123 = 123 × 7 × 11 × 13.
Tip: 1001 = 7 × 11 × 13 is worth remembering. Any 3-digit block written twice is a multiple of 1001.
Answer as per the official UPSC answer key.
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 (b) & explanation
The ring returns after 15 changes (b). Step 1: the jumps grow by one each time: 1, 2, 3 and so on. After k passes, the ring has moved 1 + 2 + ... + k = k(k + 1)/2 places. Step 2: it is back with child-1 when this is a multiple of 40. So k(k + 1) must be a multiple of 80 = 16 × 5. Step 3: the smallest such k is 15, since 15 × 16 = 240 = 80 × 3. Options (a) 14, (c) 16 and (d) 17 do not give a multiple of 80 first.
Tip: in a circle of n, you return to the start when the total steps moved is a multiple of n.
Answer as per the official UPSC answer key.
Let x be a positive integer such that 7x + 96 is divisible by x. How many values of x are possible ?
Answer (c) & explanation
12 values of x are possible (c). Step 1: 7x is always divisible by x. So 7x + 96 is divisible by x only if x divides 96. Step 2: 96 = 2⁵ × 3. Step 3: the number of divisors = (5 + 1) × (1 + 1) = 12. So x can be any of 12 values: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96. Options (a), (b) and (d) miscount.
Tip: to count divisors, write the number as a product of prime powers. Add 1 to each power and multiply.
Answer as per the official UPSC answer key.
What is the remainder if 2192 is divided by 6 ?
Answer (d) & explanation
The remainder is 4 (d). Step 1: look at remainders of powers of 2 divided by 6: 2¹ = 2 → 2; 2² = 4 → 4; 2³ = 8 → 2; 2⁴ = 16 → 4. They alternate 2, 4, 2, 4. Step 2: odd powers leave 2 and even powers leave 4. Step 3: 192 is even, so the remainder is 4. Option (c), 2, is for odd powers. Options (a) and (b) are impossible, since a power of 2 is never a multiple of 3.
Tip: for remainders of large powers, list the first few. A short repeating cycle almost always appears.
Answer as per the official UPSC answer key.
If p, q, r and s are distinct single digit positive numbers, then what is the greatest value of (p + q)(r + s)?
Answer (b) & explanation
The greatest value is 225 (b). Step 1: to make the product big, use the four largest digits: 9, 8, 7, 6. Their total is 30. Step 2: for a fixed total, a product of two parts is largest when the parts are equal. So make each bracket 15: 9 + 6 and 8 + 7. Step 3: 15 × 15 = 225. Option (a), 230, is impossible. Option (c), 224 = 16 × 14, is slightly less. Option (d), 221 = 17 × 13, is less again.
Tip: for a fixed sum, the product is greatest when the numbers are as close as possible.
Answer as per the official UPSC answer key.
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 (a) & explanation
A + B + C is 6 (a). Step 1: AB × CD = D08, a 3-digit number. So both AB and CD are small, and B × D must end in 8. Step 2: try 12 × 34 = 408. Here D = 4, E = 0 and F = 8. Then GHI = 975 − 408 = 567. Step 3: all nine digits (1, 2, 3, 4, 0, 8, 5, 6, 7) are distinct. So A + B + C = 1 + 2 + 3 = 6. Options (b), (c) and (d) do not fit.
Tip: in such puzzles, a small product means small factors. Start with the smallest two-digit numbers.
Answer as per the official UPSC answer key.
A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13 ?
Answer (a) & explanation
The remainder is 11 (a). Step 1: 999999 = 13 × 76923. So every block of six 9s is a multiple of 13. Step 2: 99 = 6 × 16 + 3. After 16 blocks, only 999 is left over. Step 3: 999 = 13 × 76 + 11. So the remainder is 11. Options (b), (c) and (d) do not match.
Tip: 999999 (six 9s) is divisible by 7, 11, 13 and 37. Use it to cut long strings of 9s into simple blocks.
Answer as per the official UPSC answer key.
What is the remainder when 85 × 87 × 89 × 91 × 95 × 96 is divided by 100 ?
Answer (a) & explanation
The remainder is 0 (a). Step 1: 100 = 4 × 25. Step 2: 85 = 5 × 17 and 95 = 5 × 19 give two 5s, so the product has 5 × 5 = 25. 96 = 32 × 3 gives plenty of 2s, so 4 also divides it. Step 3: so the product is a multiple of 4 × 25 = 100. The remainder is 0. Options (b), (c) and (d) assume something is left over.
Tip: to divide by 100, check for two 5s and two 2s among the factors. If both are there, the remainder is zero.
Answer as per the official UPSC answer key.
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 (c) & explanation
The digit sum is 81 (c). Step 1: 111111111 × 111111111 = 12345678987654321. Step 2: add the digits: 2 × (1 + 2 + ... + 8) + 9 = 2 × 36 + 9 = 81. Step 3: so the answer is (c). Option (a), 64, is 8², the result for eight ones. Options (b) and (d) are guesses.
Tip: for a number made of n ones (n up to 9), the square reads 1, 2, ..., n, ..., 2, 1. Its digit sum is simply n². Here 9² = 81.
Answer as per the official UPSC answer key.
What is the unit digit in the expansion of (57242)9×7×5×3×1 ?
Answer (a) & explanation
The unit digit is 2 (a). Step 1: only the last digit of 57242 matters, which is 2. Powers of 2 end in the cycle 2, 4, 8, 6, of length 4. Step 2: the power is 9 × 7 × 5 × 3 × 1 = 945. 945 ÷ 4 leaves a remainder of 1. Step 3: remainder 1 gives the first term of the cycle: 2. Options (b), (c) and (d) are other cycle terms.
Tip: an odd power can never be a multiple of 4. A product of odd numbers is always odd, which quickly rules out the 4th term (6).
Answer as per the official UPSC answer key.
A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
Answer (c) & explanation
The minimum is 294 pens (c). Step 1: A gets 2, 3, 4, 5 and 6 times the others. So A must be divisible by all of them: a multiple of 60. Step 2: every share must also be even. With A = 60, B = 30, C = 20, D = 15 (odd). So A must be a multiple of 120. Step 3: with A = 120: B = 60, C = 40, D = 30, E = 24, F = 20. Total = 120 + 60 + 40 + 30 + 24 + 20 = 294. Option (a), 147, is the total for A = 60, which gives odd shares.
Tip: after finding the LCM, check the extra condition (even shares). Doubling the base often fixes it.
147 (with A = 60) is the answer if you ignore "even". Then D gets 15.
Answer as per the official UPSC answer key.
Let A, B and C represent distinct non-zero digits. Suppose x is the sum of all possible 3-digit numbers formed by A, B and C without repetition.
Consider the following statements :
- 1. The 4-digit least value of x is 1332.
- 2. The 3-digit greatest value of x is 888.
Which of the above statements is/are correct?
Answer (a) & explanation
1. Six numbers can be made (ABC, ACB, BAC, BCA, CAB, CBA). Each digit sits twice in the hundreds place, twice in the tens and twice in the units.
2. So x = 222 × (A + B + C). Check with 1, 2, 3: 123 + 132 + 213 + 231 + 312 + 321 = 1332, and 222 × 6 = 1332.
3. The smallest A + B + C is 1 + 2 + 3 = 6, so the smallest x is 1332. Statement 1 is true.
4. Since x is at least 1332, it is never a 3-digit number. Statement 2 is false.
Answer as per the official UPSC answer key.
An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of original number, the resulting number is divisible by 5. After deleting three digits from the right of original number, the resulting number is divisible by 4. After deleting four digits from the right of original number, the resulting number is divisible by 3. After deleting five digits from the right of original number, the resulting number is divisible by 2. Which of the following is a possible value for the sum of the middle three digits of the number?
Answer (a) & explanation
1. Work from the left. The first 2 digits are divisible by 2, so the 2nd digit is even. The first 5 are divisible by 5, so the 5th digit is 5.
2. The first 4 digits must be divisible by 4 and the first 3 by 3. The first 6 must be divisible by 6, so the 6th digit is even. The digit sum 36 makes the whole number divisible by 9 automatically. Only two numbers pass every test: 7412589 and 1472589.
3. The middle three digits (3rd, 4th, 5th) add up to 1 + 2 + 5 = 8 in 7412589. (In the other number they give 14, which is not an option.)
"the first digit from the right" is the units digit. Read the deletions from the right carefully.
Traps like this come back in new clothes. The PYQ Book collects all of them, topic by topic, so none catch you twice. Try the free sample
Answer as per the official UPSC answer key.