H.C.F. & L.C.M. Questions and Answers

Answer: C

The number is 40.

If the number leaves remainder 7 on 87 and remainder 3 on 123, it divides \(87 - 7\) and \(123 - 3\) exactly, so it is the HCF of those two numbers.

\[ \begin{aligned} 87 - 7 &= 80 = 2^{4} \times 5 \\ 123 - 3 &= 120 = 2^{3} \times 3 \times 5 \\ \text{HCF} &= 2^{3} \times 5 = 40 \end{aligned} \]

Check: \(87 = 2 \times 40 + 7\) and \(123 = 3 \times 40 + 3\). The divisor must also be larger than both remainders, which 40 is. 20 also divides 80 and 120 but is not the greatest such number, 80 does not divide 120, and 36 divides neither.

Answer: C

They will jump together at step 378.

The frog lands on multiples of 3 and the rabbit on multiples of 7, so they meet only on common multiples of 3 and 7, that is, multiples of their LCM, 21.

\[ 378 = 21 \times 18 \]

The others fail one test: 343 = 7 × 49 and 371 = 7 × 53 are multiples of 7 but not of 3 (digit sums 10 and 11), while 354 is a multiple of 3 but not of 7.

Answer: A

The H.C.F. is \((30)^{2} \times 7^{3}\).

The H.C.F. takes each common prime to the lowest power in which it appears in all three numbers:

• 2: powers 3, 2, 4, so \(2^{2}\)
• 3: powers 2, 3, 4, so \(3^{2}\)
• 5: powers 3, 4, 2, so \(5^{2}\)
• 7: powers 3, 3, 5, so \(7^{3}\)

\[ \begin{aligned} \text{H.C.F.} &= 2^{2} \times 3^{2} \times 5^{2} \times 7^{3} \\ &= (2 \times 3 \times 5)^{2} \times 7^{3} = 30^{2} \times 7^{3} \end{aligned} \]

\((30)^{3} \times 7^{3}\) uses too high a power of 2, 3 and 5, and \(30 \times 7^{5}\) takes the highest power of 7, which belongs to the L.C.M., not the H.C.F. \((15)^{3} \times 7^{4}\) leaves out 2 altogether.

Answer: B

There are 4 such numbers.

A number divisible by each of 3, 5 and 7 is divisible by their LCM, which is \(3 \times 5 \times 7 = 105\). The multiples of 105 below 500 are 105, 210, 315 and 420; the next, 525, exceeds 500.

\[ \left\lfloor \frac{500}{105} \right\rfloor = 4 \]

3, 5 and 6 come from miscounting the multiples.

Answer: D

The sum of the LCM and HCF is 660.

Let the numbers be 3k and 7k, where k is their HCF, since 3 and 7 are co-prime:

\[ \begin{aligned} \text{LCM} &= 3 \times 7 \times k = 21k = 630 \\ k &= 30 = \text{HCF} \\ \text{LCM} + \text{HCF} &= 630 + 30 = 660 \end{aligned} \]

The numbers are 90 and 210. 30 is the HCF alone, and 600 is the difference rather than the sum.

Answer: B

A number divisible by 8, 15, 16 and 21 must be divisible by their least common multiple.

\[ 8 = 2^3,\ 15 = 3 \times 5,\ 16 = 2^4,\ 21 = 3 \times 7 \]

\[ \text{LCM} = 2^4 \times 3 \times 5 \times 7 = 1680 \]

The greatest five digit number is 99999. Dividing, \(99999 = 1680 \times 59 + 879\), so the largest multiple not above it is

\[ 1680 \times 59 = 99120 \]

The number 987640 has six digits, 95780 is not a multiple of 1680 and 79685 is not even a multiple of 8. The answer is 99120, option B.

Answer: D

The LCM is 126. For two numbers, the product of the numbers equals the product of their HCF and their LCM.

\[ a \times b = \text{HCF} \times \text{LCM} \]

\[ \text{LCM} = \frac{1764}{14} = 126 \]

As a check, the LCM must be a multiple of the HCF. Now 126 = 14 x 9, which is a multiple of 14. The options 118, 120 and 124 are not multiples of 14, because 14 x 8 = 112 and 14 x 9 = 126, so none of them can be the LCM of two numbers whose HCF is 14.

Answer: B

The HCF is 120. Let the numbers be x, 2x and 3x. Their LCM is 6x, so \[ 6x = 720 \Rightarrow x = 120 \] The HCF of x, 2x and 3x is x, which is 120.

The other values do not satisfy 6x = 720.

A check: the three numbers are 120, 240 and 360, whose LCM is 720 and whose HCF is 120, so the answer follows directly.

Answer: C

The largest square tile that fits a room exactly has a side equal to the highest common factor of the length and breadth, and using it gives the least number of tiles. The HCF of 264 and 216 is 24 cm. The number of tiles is the length divided by 24 times the breadth divided by 24, which is 11 x 9 = 99. So at least 99 tiles are needed, option (C). The values 77, 143 and 63 do not equal the product of 11 and 9. The side of the tile, 24 cm, is the largest square that divides both 264 and 216, so no larger tile can cover the floor without cutting.

Answer: C

The number is 7987.

A number leaving remainder 7 on division by 20, 42 and 76 is 7 more than a common multiple of all three. The smallest such number is 7 more than their least common multiple.

\[ \begin{aligned} 20 &= 2^{2} \cdot 5,\quad 42 = 2 \cdot 3 \cdot 7,\quad 76 = 2^{2} \cdot 19 \\ \text{LCM} &= 2^{2} \cdot 3 \cdot 5 \cdot 7 \cdot 19 = 7980 \end{aligned} \]

So the number is 7980 + 7 = 7987. 3047 and 6047 do not work: 3040 and 6040 are not multiples of 42 (or of 76). 63847 does leave remainder 7, since 63840 = 8 × 7980, but it is not the smallest.