How to Use These Questions

These 34 questions mix all five topics, so a question on types of numbers can sit next to a remainder question and an HCF word problem. Before you pick a method, decide what is really being asked. Many of the word problems are HCF or LCM in disguise: equal cans, pieces or tiles need the HCF, and bells or runners meeting again need the LCM.

Cover the answer and work each one on paper first, at about 36 to 45 seconds a question. When one runs past a minute, find the method in Section 6 that would have saved the time and redo the question the next day.

The five [Tier 2] questions, such as the ones on square factors and HCF pairs, go further than Tier 1 usually does. Tier 1 students should still do them, since easier versions of these ideas turn up in Tier 1 too.

Solved Examples

Question 1

What kind of number is 0.0049×400\sqrt{0.0049} \times \sqrt{400}? (a) A natural number (b) An integer but not a natural number (c) A rational number but not an integer (d) An irrational number

Answer. 0.0049=0.07\sqrt{0.0049} = 0.07, because 0.07 × 0.07 = 0.0049. And 400=20\sqrt{400} = 20. So the product is 0.07 × 20 = 1.4. That is 14/10, a fraction that is not a whole number, so the answer is (c) a rational number but not an integer.

Question 2

A six-digit number is made by writing a three-digit number twice, like 3,86,386. Such a number is always divisible by which of these? (a) 9 (b) 17 (c) 77 (d) 111

Answer. Writing abc twice is the same as abc × 1000 + abc = abc × 1001. And 1001 = 7 × 11 × 13. So the number is always divisible by 7, 11 and 13, and by products of them like 77. The answer is (c) 77.

Question 3

Four bells ring at intervals of 9, 12, 15 and 18 minutes. They ring together at 8:00 a.m. When will they next ring together?

Answer. They ring together again after LCM(9, 12, 15, 18) minutes. 9 = 323^2, 12 = 22×32^2 \times 3, 15 = 3 × 5 and 18 = 2×322 \times 3^2. Taking the highest power of each prime: 22×32×52^2 \times 3^2 \times 5 = 180 minutes, which is 3 hours. So they next ring together at 11:00 a.m.

Question 4

What least number must be subtracted from 2,000 to make it a perfect square? What least number would have to be added instead?

Answer. 442=193644^2 = 1936 and 452=202545^2 = 2025, and 2,000 sits between them. To go down, I subtract 2,000 − 1,936 = 64. To go up, I add 2,025 − 2,000 = 25.

Question 5

A number n leaves remainder 5 when divided by 7. What is the remainder when n2+4nn^2 + 4n is divided by 7?

Answer. I only need the remainder, so I put 5 in place of n: 25 + 20 = 45. And 45 = 6 × 7 + 3, so the remainder is 3.

Question 6

Which of these is irrational? (a) 1.21\sqrt{1.21} (b) 0.9\sqrt{0.9} (c) 0.142857‾0.\overline{142857} (d) 0.0083\sqrt[3]{0.008}

Answer. 1.21=1.1\sqrt{1.21} = 1.1 and 0.0083=0.2\sqrt[3]{0.008} = 0.2, and 0.142857‾0.\overline{142857} repeats, so those three are rational. 0.9 = 9/10, and 10 is not a perfect square, so 0.9\sqrt{0.9} is irrational. The answer is (b). It is tempting to say 0.9=0.3\sqrt{0.9} = 0.3, but 0.320.3^2 is 0.09, not 0.9.

Question 7

If the seven-digit number 41x948y is divisible by 88, find x and y.

Answer. 88 = 8 × 11. For 8, the last three digits 48y must be a multiple of 8. Both 480 and 488 work, so y is 0 or 8, and I have to try both with the 11 rule.

Odd places from the right: y + 4 + x + 4. Even places: 8 + 9 + 1 = 18.

  • If y = 0, the difference is (8 + x) − 18 = x − 10. That is 0 only for x = 10 and −11 only for x = −1, so no digit works.
  • If y = 8, the difference is (16 + x) − 18 = x − 2, so x = 2.

So x = 2 and y = 8, and the number is 41,29,488.

Question 8

A floor is 8.16 m long and 3.84 m wide. What is the least number of square tiles, all the same size, that will cover it exactly?

Answer. In cm the floor is 816 by 384. The side of a tile has to fit exactly into both, so the biggest tile has side HCF(816, 384). 816 = 24×3×172^4 \times 3 \times 17 and 384 = 27×32^7 \times 3, so the HCF is 24×32^4 \times 3 = 48 cm. Number of tiles = (816 ÷ 48) × (384 ÷ 48) = 17 × 8 = 136.

Question 9

Find the unit digit of 173978×2588351739^{78} \times 2588^{35}.

Answer. Only the last digits matter, so I look at 9789^{78} and 8358^{35}. 9 to an even power ends in 1. For 8, 35 ÷ 4 leaves 3, and the cycle 8, 4, 2, 6 gives 2. Then 1 × 2 = 2, so the unit digit is 2.

Question 10

The sum 1 + 2 + 3 + … + 48 is divisible by which of these? (a) 47 (b) 48 (c) 49 (d) 96

Answer. The sum is 48×492=24×49=1176\frac{48 \times 49}{2} = 24 \times 49 = 1176. That is a multiple of 49. It is not a multiple of 48, because the 2 at the bottom took half of the 48 away (1,176 ÷ 48 = 24.5). The answer is (c) 49.

Question 11

The LCM of two numbers is 12 times their HCF, and the HCF and LCM add up to 403. If one of the numbers is 93, find the other.

Answer. Call the HCF h, so the LCM is 12h. Then h + 12h = 13h = 403, which gives h = 31 and LCM = 372. Using HCF × LCM = product, the other number is 31 × 372 ÷ 93 = 31 × 4 = 124. Check: 93 = 3 × 31 and 124 = 4 × 31, so the HCF is 31 and the LCM is 3 × 4 × 31 = 372.

Question 12

After how many decimal places does 4924×53×72\frac{49}{2^4 \times 5^3 \times 7^2} terminate?

Answer. 72=497^2 = 49, so the 49 on top cancels it, leaving 124×53\frac{1}{2^4 \times 5^3}. Only 2s and 5s are left, so the decimal terminates, and the bigger power is 4. It terminates after 4 places. Check: 24×53=20002^4 \times 5^3 = 2000, and 1/2000 = 0.0005.

Question 13

Find the smallest perfect square that is divisible by 12, 20 and 27.

Answer. It must be a multiple of LCM(12, 20, 27) = 540 = 22×33×52^2 \times 3^3 \times 5. In a square every power is even. The 2s are already fine, but 3 and 5 each need one more: 22×34×522^2 \times 3^4 \times 5^2 = 8100. The answer is 8,100, which is 90290^2.

Question 14

What least number must be added to 3,25,647 so that the result is divisible by 7, 11 and 13?

Answer. 7 × 11 × 13 = 1001, so I need the next multiple of 1001. 1001 × 325 = 3,25,325, and 3,25,647 − 3,25,325 = 322. So the remainder is 322, and I add 1001 − 322 = 679. The result is 3,26,326 = 1001 × 326.

Question 15

Find the HCF and the LCM of 6/7, 9/14 and 15/28.

Answer. For fractions, HCF = HCF of the tops ÷ LCM of the bottoms, and LCM = LCM of the tops ÷ HCF of the bottoms. The tops 6, 9, 15 have HCF 3 and LCM 90. The bottoms 7, 14, 28 have HCF 7 and LCM 28. So the HCF is 328\mathbf{\frac{3}{28}} and the LCM is 907\mathbf{\frac{90}{7}}.

Question 16

In a division sum, the divisor is 5 times the quotient and 3 times the remainder. If the remainder is 30, find the dividend.

Answer. Divisor = 3 × 30 = 90. Quotient = 90 ÷ 5 = 18. Dividend = divisor × quotient + remainder = 90 × 18 + 30 = 1,620 + 30 = 1,650.

Question 17

Find the difference between the place values of the two 8s in 3,86,41,982.

Answer. The first 8 is in the ten-lakhs place, so its place value is 80,00,000. The second 8 is in the tens place, so its place value is 80. The difference is 80,00,000 − 80 = 79,99,920.

Question 18

If the five-digit number 6x07y is divisible by 36, find the greatest possible value of x + y.

Answer. 36 = 4 × 9. For 4, the last two digits 7y must be a multiple of 4, so y = 2 (72) or y = 6 (76).

  • y = 2: the digit sum 6 + x + 0 + 7 + 2 = 15 + x must be 18, so x = 3 and x + y = 5.
  • y = 6: the digit sum 19 + x must be 27, so x = 8 and x + y = 14.

The greatest value is 14, from the number 68,076.

Question 19

Three sections of a school have 84, 126 and 196 students. Each section is split into groups of the same size, and no group mixes sections. What is the least possible number of groups?

Answer. Fewest groups means the biggest group size, and that size has to divide all three numbers. So it is HCF(84, 126, 196). 84 = 22×3×72^2 \times 3 \times 7, 126 = 2×32×72 \times 3^2 \times 7 and 196 = 22×722^2 \times 7^2, so the HCF is 2 × 7 = 14. Groups = 84/14 + 126/14 + 196/14 = 6 + 9 + 14 = 29.

Question 20

Find 413449\sqrt{413449}.

Answer. The last digit is 9, so the root ends in 3 or 7. Covering the last two digits leaves 4134. 642=409664^2 = 4096 fits and 652=422565^2 = 4225 doesn't, so the root starts with 64. 64 × 65 = 4,160, and 4,134 is less than that, so I take the smaller choice: 643.

Question 21

How many numbers from 1 to 1,000 are divisible by 7 but not by 11?

Answer. Multiples of 7 up to 1,000: 1,000 ÷ 7 gives 142. The ones that are also divisible by 11 are the multiples of 77: 1,000 ÷ 77 gives 12. So 142 − 12 = 130.

Question 22

A two-digit number is 4 times the sum of its digits. If 18 is added to it, its digits are reversed. Find the number.

Answer. Adding 18 reverses the digits, and a number and its reverse differ by 9 × (difference of the digits). So 9 × (difference) = 18, and the units digit is 2 more than the tens digit. The candidates are 13, 24, 35, 46, 57, 68 and 79. The number must be 4 times its digit sum: 24 has digit sum 6, and 4 × 6 = 24. So the number is 24. Check: 24 + 18 = 42.

Question 23

Three numbers are in the ratio 2 : 3 : 5 and their LCM is 1,800. Find their HCF.

Answer. Call the numbers 2h, 3h and 5h, where h is their HCF. Their LCM is LCM(2, 3, 5) × h = 30h. So 30h = 1800 and h = 60. The numbers are 120, 180 and 300, and their HCF is 60.

Question 24 [Tier 2]

Find the remainder when 31003^{100} is divided by 7.

Answer. 33=27=28−13^3 = 27 = 28 - 1, so 333^3 leaves −1 when divided by 7. Now 3100=(33)33×33^{100} = (3^3)^{33} \times 3, so the remainder is (−1)33×3=−3(-1)^{33} \times 3 = -3. A negative remainder means I add the divisor: −3 + 7 = 4.

Question 25

Which of these is divisible by 99? (a) 6,16,986 (b) 2,40,438 (c) 3,68,973 (d) 8,14,044

Answer. I add the digits in pairs from the right.

  • (a) 86 + 69 + 61 = 216, not a multiple of 99.
  • (b) 38 + 04 + 24 = 66, no.
  • (c) 73 + 89 + 36 = 198 = 2 × 99, yes.
  • (d) 44 + 40 + 81 = 165, no.

The answer is (c) 3,68,973. Checking 9 and 11 separately also works, but it takes longer.

Question 26

Find the greatest number that divides 631, 1,019 and 1,397 leaving remainders 7, 11 and 5 respectively.

Answer. I take away each number's own remainder: 631 − 7 = 624, 1,019 − 11 = 1,008 and 1,397 − 5 = 1,392. The number divides all three exactly, so I want HCF(624, 1008, 1392). 624 = 24×3×132^4 \times 3 \times 13, 1008 = 24×32×72^4 \times 3^2 \times 7 and 1392 = 24×3×292^4 \times 3 \times 29, so the HCF is 24×32^4 \times 3 = 48. It is bigger than all three remainders, which it has to be.

Question 27

How many zeros are there at the end of 15 × 25 × 32 × 40 × 75?

Answer. Each zero needs one 2 and one 5. 5s: 15 has one, 25 has two, 40 has one and 75 has two, so 6 in all. 2s: 32 has five and 40 has three, so 8. The 5s run out first, so there are 6 zeros. (The product is 3,60,00,000.)

Question 28 [Tier 2]

Find 0.23‾+0.23‾0.2\overline{3} + 0.\overline{23} as a fraction.

Answer. 0.23‾=23−290=2190=7300.2\overline{3} = \frac{23 - 2}{90} = \frac{21}{90} = \frac{7}{30} and 0.23‾=23990.\overline{23} = \frac{23}{99}. The LCM of 30 and 99 is 990, so 730=231990\frac{7}{30} = \frac{231}{990} and 2399=230990\frac{23}{99} = \frac{230}{990}. The sum is 461990\mathbf{\frac{461}{990}}.

Question 29

Find the largest four-digit number that leaves remainders 2, 3 and 4 when divided by 3, 4 and 5 respectively.

Answer. Divisor − remainder is 3 − 2 = 1, 4 − 3 = 1 and 5 − 4 = 1. The gap is always 1, so the number is one less than a multiple of LCM(3, 4, 5) = 60. 9,999 ÷ 60 leaves 39, so the largest four-digit multiple of 60 is 9,960. The answer is 9,960 − 1 = 9,959.

Question 30

Find the unit digit of 795−3587^{95} - 3^{58}.

Answer. 95 ÷ 4 leaves 3, so 7957^{95} ends in 3 (cycle 7, 9, 3, 1). 58 ÷ 4 leaves 2, so 3583^{58} ends in 9 (cycle 3, 9, 7, 1). 3 − 9 is negative, so I borrow 10: 13 − 9 = 4. The unit digit is 4. Borrowing is fine here because 7957^{95} is much bigger than 3583^{58}.

Question 31 [Tier 2]

Which of these divides 2924−129^{24} - 1? (a) 27 (b) 28 (c) 31 (d) 33

Answer. an−1a^n - 1 is always divisible by a − 1, so 2924−129^{24} - 1 is divisible by 29 − 1 = 28. (Because 24 is even, it is also divisible by 29 + 1 = 30.) The answer is (b) 28. Only one option can be right, so I don't need to test the other three.

Question 32 [Tier 2]

How many factors of 6,048 are perfect squares?

Answer. 6,048 = 25×33×72^5 \times 3^3 \times 7. A square factor can only use even powers. For 2 that means 202^0, 222^2 or 242^4 (3 choices), for 3 it means 303^0 or 323^2 (2 choices), and for 7 only 707^0 (1 choice). So there are 3 × 2 × 1 = 6 square factors: 1, 4, 9, 16, 36 and 144.

Question 33

How many digits are used to number the pages of a book with 1,250 pages?

Answer. Pages 1 to 9 use 9 digits. Pages 10 to 99 are 90 pages with 2 digits each, so 180. Pages 100 to 999 are 900 pages with 3 digits, so 2,700. Pages 1,000 to 1,250 are 251 pages with 4 digits, so 1,004. Total = 9 + 180 + 2,700 + 1,004 = 3,893.

Question 34 [Tier 2]

The sum of two numbers is 216 and their HCF is 27. How many such pairs of numbers are there?

Answer. I write the numbers as 27a and 27b, where a and b are co-prime (otherwise the HCF would be bigger than 27). Then 27(a + b) = 216, so a + b = 8. The pairs adding to 8 are (1, 7), (2, 6), (3, 5) and (4, 4). (2, 6) and (4, 4) share a factor, so only (1, 7) and (3, 5) count. There are 2 pairs: 27 and 189, and 81 and 135.