Number System & Simplification: Practice Questions with Answers
HCF and LCM, divisibility, remainders, unit digits and BODMAS. Try each question first: tap an option to check it, and the explanation appears below.
Question 1Easy
Simplify: 65 + 12 × 12 − 32 ÷ 8
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✅ A. 205
Follow BODMAS: division and multiplication come before addition and subtraction. First 32 ÷ 8 = 4 and 12 × 12 = 144. Then 65 + 144 − 4 = 205. Working simply from left to right would give the wrong value, because it adds 65 and 12 before multiplying.
Question page →Question 2Medium
What is the unit digit of 4^72?
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✅ B. 6
The unit digits of powers of 4 repeat in a cycle of 2: 4, 6. Divide the power by 2: 72 ÷ 2 leaves a remainder of 0, which means the last digit of the cycle. So the unit digit of 4^72 is 6. Only the last digit of the base matters for this pattern.
Question page →Question 3Hard
What is the greatest number that divides 414, 339 and 377, leaving remainders 6, 16 and 3 respectively?
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✅ B. 17
Subtract each remainder first, so the number divides the results exactly: 414 − 6 = 408, 339 − 16 = 323 and 377 − 3 = 374. The greatest number dividing all three is their HCF: 408 = 17 × 24, 323 = 17 × 19, 374 = 17 × 22, and 24, 19, 22 have no common factor, so the HCF is 17. It must also be larger than every remainder, and 17 is.
Question page →Question 4Hard
What is the greatest number that divides 75, 334 and 195, leaving remainders 18, 11 and 5 respectively?
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✅ B. 19
Subtract each remainder first, so the number divides the results exactly: 75 − 18 = 57, 334 − 11 = 323 and 195 − 5 = 190. The greatest number dividing all three is their HCF: 57 = 19 × 3, 323 = 19 × 17, 190 = 19 × 10, and 3, 17, 10 have no common factor, so the HCF is 19. It must also be larger than every remainder, and 19 is.
Question page →Question 5Hard
What is the greatest number that divides 212, 243 and 377, leaving remainders 12, 18 and 2 respectively?
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✅ C. 25
Subtract each remainder first, so the number divides the results exactly: 212 − 12 = 200, 243 − 18 = 225 and 377 − 2 = 375. The greatest number dividing all three is their HCF: 200 = 25 × 8, 225 = 25 × 9, 375 = 25 × 15, and 8, 9, 15 have no common factor, so the HCF is 25. It must also be larger than every remainder, and 25 is.
Question page →Question 6Easy
What is the HCF (highest common factor) of 20 and 25?
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✅ C. 5
Write both numbers using a common factor: 20 = 5 × 4 and 25 = 5 × 5. The numbers 4 and 5 have no common factor other than 1, so nothing larger than 5 divides both. Therefore the HCF of 20 and 25 is 5. A quick check: the HCF must also divide the difference of the two numbers, and 5 ÷ 5 = 1.
Question page →Question 7Hard
What is the greatest number that divides 204, 666 and 419, leaving remainders 6, 6 and 1 respectively?
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✅ B. 22
Subtract each remainder first, so the number divides the results exactly: 204 − 6 = 198, 666 − 6 = 660 and 419 − 1 = 418. The greatest number dividing all three is their HCF: 198 = 22 × 9, 660 = 22 × 30, 418 = 22 × 19, and 9, 30, 19 have no common factor, so the HCF is 22. It must also be larger than every remainder, and 22 is.
Question page →Question 8Easy
Simplify: 25 + 4 × 5 − 10 ÷ 5
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✅ C. 43
Follow BODMAS: division and multiplication come before addition and subtraction. First 10 ÷ 5 = 2 and 4 × 5 = 20. Then 25 + 20 − 2 = 43. Working simply from left to right would give the wrong value, because it adds 25 and 4 before multiplying.
Question page →Question 9Medium
A number leaves a remainder of 23 when divided by 36. What remainder will it leave when divided by 4?
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✅ A. 3
The number can be written as 36n + 23 for some whole number n. Since 36 = 4 × 9, the part 36n divides exactly by 4. So only 23 matters: 23 ÷ 4 leaves a remainder of 3. This works only because 4 is a factor of 36.
Question page →Question 10Easy
Simplify: (33² − 18²) ÷ (33 − 18)
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✅ D. 51
Use the identity a² − b² = (a + b)(a − b). So (33² − 18²) ÷ (33 − 18) = (33 + 18)(33 − 18) ÷ (33 − 18) = 33 + 18 = 51. This saves squaring 33 and 18: 1089 − 324 = 765, and 765 ÷ 15 = 51 gives the same answer.
Question page →Question 11Easy
Simplify: 54 + 6 × 3 − 45 ÷ 5
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✅ D. 63
Follow BODMAS: division and multiplication come before addition and subtraction. First 45 ÷ 5 = 9 and 6 × 3 = 18. Then 54 + 18 − 9 = 63. Working simply from left to right would give the wrong value, because it adds 54 and 6 before multiplying.
Question page →Question 12Easy
What is the smallest number that must be added to 6,521 so that the result is exactly divisible by 11?
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✅ C. 2
Divide 6,521 by 11: the quotient is 592 and the remainder is 9. To reach the next multiple of 11, add 11 − 9 = 2. Check: 6,521 + 2 = 6,523, which is 11 × 593. Note that 9 is the number to subtract, not add, to get a multiple of 11.
Question page →Question 13Easy
What is 4/5 of 3/8 of 480?
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✅ B. 144
"Of" means multiply. First 3/8 of 480 = 480 × 3 ÷ 8 = 180. Then 4/5 of 180 = 144. You can also multiply the fractions first: 4/5 × 3/8 = 12/40, and 12/40 of 480 is 144.
Question page →Question 14Easy
Simplify: (48² − 6²) ÷ (48 − 6)
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✅ D. 54
Use the identity a² − b² = (a + b)(a − b). So (48² − 6²) ÷ (48 − 6) = (48 + 6)(48 − 6) ÷ (48 − 6) = 48 + 6 = 54. This saves squaring 48 and 6: 2304 − 36 = 2268, and 2268 ÷ 42 = 54 gives the same answer.
Question page →Question 15Medium
Three bells ring at intervals of 8, 10 and 15 seconds. If they ring together now, after how many seconds will they next ring together?
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✅ D. 120 seconds
The bells ring together again after a time that is a multiple of all three intervals, and the first such time is the LCM. LCM of 8 and 10 is 40, and LCM of 40 and 15 is 120. So the bells ring together again after 120 seconds. Adding the intervals or multiplying them gives a common time, but not the first one.
Question page →Question 16Hard
How many factors (divisors) does 240 have, including 1 and the number itself?
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✅ C. 20
Write the number as a product of primes: 240 = 2^4 × 3 × 5. A factor can use each prime from power 0 up to its full power, so the number of factors = (4 + 1) × (1 + 1) × (1 + 1) = 20. Multiplying (exponent + 1) for each prime is much faster than listing every factor.
Question page →Question 17Medium
Three bells ring at intervals of 9, 18 and 20 seconds. If they ring together now, after how many seconds will they next ring together?
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✅ D. 180 seconds
The bells ring together again after a time that is a multiple of all three intervals, and the first such time is the LCM. LCM of 9 and 18 is 18, and LCM of 18 and 20 is 180. So the bells ring together again after 180 seconds. Adding the intervals or multiplying them gives a common time, but not the first one.
Question page →Question 18Easy
Simplify: (18² − 13²) ÷ (18 − 13)
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✅ C. 31
Use the identity a² − b² = (a + b)(a − b). So (18² − 13²) ÷ (18 − 13) = (18 + 13)(18 − 13) ÷ (18 − 13) = 18 + 13 = 31. This saves squaring 18 and 13: 324 − 169 = 155, and 155 ÷ 5 = 31 gives the same answer.
Question page →Question 19Medium
What is the unit digit of 9^77?
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✅ D. 9
The unit digits of powers of 9 repeat in a cycle of 2: 9, 1. Divide the power by 2: 77 ÷ 2 leaves a remainder of 1, which means position 1 in the cycle. So the unit digit of 9^77 is 9. Only the last digit of the base matters for this pattern.
Question page →Question 20Hard
What is the smallest number greater than 16 which, when divided by 6, 10 and 16, leaves a remainder of 4 in each case?
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✅ A. 244
A number that leaves the same remainder 4 with each divisor is (a common multiple of 6, 10 and 16) + 4. The smallest common multiple is the LCM: LCM(6, 10) = 30 and LCM(30, 16) = 240. So the number is 240 + 4 = 244. (4 itself also leaves remainder 4, but it is not greater than 16.) Check: 244 ÷ 6 leaves 4, ÷ 10 leaves 4, and ÷ 16 leaves 4.
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