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

Answer: B

They beep together again at 3.30 pm.

They next beep together after the least common multiple of 30, 45 and 54 minutes. With \(30 = 2 \times 3 \times 5\), \(45 = 3^{2} \times 5\) and \(54 = 2 \times 3^{3}\):

\[ \text{LCM} = 2 \times 3^{3} \times 5 = 270\ \text{minutes} \]

270 minutes is 4 hours 30 minutes, so after 11 a.m. they beep together at 3.30 p.m. 4 pm and 3 pm are half an hour out, and 4.30 pm adds 5 hours 30 minutes.

Answer: B

The smallest four-digit multiple of 6, 7 and 4 is 1008.

A number divisible by all three must be a multiple of their LCM. With \(6 = 2 \times 3\), \(4 = 2^{2}\) and 7 prime:

\[ \text{LCM} = 2^{2} \times 3 \times 7 = 84 \]

The smallest four-digit number is 1000, and \(1000 \div 84 = 11\) remainder 76, so the next multiple of 84 is

\[ 84 \times 12 = 1008 \]

The other options fail: 1000 is divisible by 4 but not by 3 or 7, 1006 is not divisible by 3, 4 or 7, and 1002 is divisible by 6 but not by 4 or 7.

Answer: B

The number that cannot be the HCF is 42.

The HCF of two numbers always divides their LCM, because both numbers divide the LCM and the HCF divides both numbers. Factorise the LCM:

\[ 392 = 2^{3} \times 7^{2} \]

Now test the options: 28 = 2² × 7, 56 = 2³ × 7 and 196 = 2² × 7² all divide 392, so each could be the HCF. But 42 = 2 × 3 × 7 contains the prime 3, which 392 does not, so 42 does not divide 392 and cannot be the HCF.

Answer: D

He should buy A = 12, B = 10, C = 5 packets.

The smallest equal number of pens is the LCM of the packet sizes:

\[ \begin{aligned} 10 &= 2 \times 5,\quad 12 = 2^{2} \times 3,\quad 24 = 2^{3} \times 3 \\ \text{LCM} &= 2^{3} \times 3 \times 5 = 120 \end{aligned} \]

So he needs 120 pens of each brand: 120/10 = 12 packets of A, 120/12 = 10 of B and 120/24 = 5 of C. The other options swap the packet counts between brands; for example A = 10 would give only 100 pens of brand A.

Answer: D

The sum of the digits of x is 17.

A number leaving remainder 5 when divided by each of 8, 12, 20, 28 and 35 is 5 more than a common multiple of them. The least such number uses their LCM:

\[ \begin{aligned} 8 &= 2^{3},\ 12 = 2^{2} \cdot 3,\ 20 = 2^{2} \cdot 5 \\ 28 &= 2^{2} \cdot 7,\ 35 = 5 \cdot 7 \\ \text{LCM} &= 2^{3} \cdot 3 \cdot 5 \cdot 7 = 840 \\ x &= 840 + 5 = 845 \end{aligned} \]

The digit sum is 8 + 4 + 5 = 17. Forgetting to add the remainder gives 840, digit sum 12, which is not an option; 11, 14 and 15 come from other slips.

Answer: C

The LCM is not a multiple of 72.

Factorise each number and take the highest power of every prime:

\[ \begin{aligned} 21 &= 3 \times 7 \\ 36 &= 2^{2} \times 3^{2} \\ 66 &= 2 \times 3 \times 11 \\ \text{LCM} &= 2^{2} \times 3^{2} \times 7 \times 11 = 2772 \end{aligned} \]

72 is \(2^{3} \times 3^{2}\), but the LCM has only \(2^{2}\), so 2772 is not a multiple of 72 (\(2772 \div 72 = 38.5\)).

The others divide it exactly: \(2772 \div 42 = 66\), \(2772 \div 28 = 99\) and \(2772 \div 44 = 63\).

Answer: B

The sum is 27.

The common factors of two numbers are exactly the factors of their HCF. Since \(36 = 2^{2} \times 3^{2}\) and \(48 = 2^{4} \times 3\), the HCF is \(2^{2} \times 3 = 12\).

\[ \begin{aligned} \text{factors of } 12 &: 1, 2, 3, 4, 6, 12 \\ \text{sum excluding } 1 &= 2 + 3 + 4 + 6 + 12 \\ &= 27 \end{aligned} \]

28 is the sum with 1 included, the trap the bracket in the question warns against. 26 and 29 do not correspond to any set of common factors.

Answer: C

The value is 20.

Factorise: \(96 = 2^{5} \times 3\), \(120 = 2^{3} \times 3 \times 5\), \(160 = 2^{5} \times 5\). The LCM takes the highest power of each prime:

\[ \text{LCM} = 2^{5} \times 3 \times 5 = 480 \]

For the HCF, \(72 = 2^{3} \times 3^{2}\) and \(168 = 2^{3} \times 3 \times 7\), and the common part is

\[ \text{HCF} = 2^{3} \times 3 = 24 \]

\[ 480 \div 24 = 20 \]

Option 1 (12) would be 480 divided by 40 and option 4 (30) is 480 divided by 16, the results of mistaken HCFs.

Answer: A

The impossible LCM is 128.

The HCF of two numbers always divides their LCM, so the LCM must be a multiple of 18.

\[ \begin{aligned} 108 &= 18 \times 6, \quad 72 = 18 \times 4 \\ 144 &= 18 \times 8, \quad 128 \div 18 \approx 7.11 \end{aligned} \]

108, 72 and 144 are multiples of 18, so each could be the LCM (for example 18 and 108, 18 and 72, 18 and 144). 128 is not even divisible by 9, so it cannot be a multiple of 18 and cannot be their LCM.

Answer: D

The greatest such number is 46.

If each number leaves remainder 6, then subtracting 6 from each gives numbers the divisor divides exactly. The greatest such divisor is their HCF.

\[ \begin{aligned} 282 - 6 &= 276 \\ 696 - 6 &= 690 \\ 1616 - 6 &= 1610 \end{aligned} \]

\[ \begin{aligned} \text{HCF}(276, 690) &= 138 \\ \text{HCF}(138, 1610) &= 46 \end{aligned} \]

Check: \(276 = 46 \times 6\), \(690 = 46 \times 15\) and \(1610 = 46 \times 35\). 23 also divides all three but is not the greatest; 24 and 48 do not divide 690, so they leave different remainders.