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

Answer: B

The H.C.F. takes the lowest power of each shared prime.

\[ 2^{2} \cdot 3^{2} \cdot 7^{2} \quad\text{and}\quad 2 \cdot 3^{4} \cdot 7 \]

\[ \text{H.C.F.} = 2^{1} \cdot 3^{2} \cdot 7^{1} = 126 \]

The L.C.M. would take the highest powers instead.

Answer: B

\[ \text{H.C.F.} \times \text{L.C.M.} = \text{product} \]

\[ 13 \times 14560 = 416 \times n \]

\[ n = \frac{189{,}280}{416} = 455 \]

Answer: B

\[ 893 = 19 \times 47 \]

So \( N = 893k + 193 \) is a multiple of 47 plus 193, and only 193 matters:

\[ 193 = 47 \times 4 + 5 \]

The remainder is 5. This works because 47 divides the original divisor.

Answer: B

A number leaving the same remainder 2 under all six divisors is 2 more than a common multiple of them, so start from their lowest common multiple:

\[ \operatorname{LCM}(3,5,6,8,10,12) = 120 \quad\Longrightarrow\quad N = 120k + 2 \]

Now impose divisibility by 13. Since \( 120 = 9(13) + 3 \), we have \( 120 \equiv 3 \pmod{13} \), so

\[ 3k + 2 \equiv 0 \pmod{13} \quad\Longrightarrow\quad 3k \equiv 11 \pmod{13} \]

Multiplying by 9, the inverse of 3 modulo 13 because \( 3 \times 9 = 27 \equiv 1 \):

\[ k \equiv 99 \equiv 8 \pmod{13} \quad\Longrightarrow\quad N = 120(8) + 2 = 962 \]

Check: 962 divided by 3, 5, 6, 8, 10 and 12 leaves 2 every time, and \( 962 = 13 \times 74 \) exactly. Of the distractors, 1562 clears the remainder test but is not a multiple of 13, and 1586 is a multiple of 13 but leaves remainder 1 on division by 5.

Answer: C

Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number,
\((9999 - 399) = 9600\)

Answer: A

They meet at the start again after a whole number of laps each, so take the L.C.M. of the lap times.

\[ 27 = 3^{3}, \quad 45 = 3^{2}\cdot 5, \quad 63 = 3^{2}\cdot 7 \]

\[ \text{L.C.M.} = 3^{3} \cdot 5 \cdot 7 = 945 \text{ minutes} \]

Answer: A

Divide out the primes in turn.

\[ 252 = 2 \times 126 = 2 \times 2 \times 63 = 2^{2} \times 3^{2} \times 7 \]

Checking: \( 4 \times 9 \times 7 = 252 \).

Answer: A

For fractions the rule inverts between the two halves:

\[ \text{HCF of fractions} = \frac{\operatorname{HCF}(\text{numerators})}{\operatorname{LCM}(\text{denominators})} \]

Numerators first:

\[ \operatorname{HCF}(36,48,72) = 12 \]

Then the denominators, through their prime factors:

\[ 75 = 3 \cdot 5^{2}, \quad 150 = 2 \cdot 3 \cdot 5^{2}, \quad 135 = 3^{3} \cdot 5 \]

\[ \operatorname{LCM} = 2 \cdot 3^{3} \cdot 5^{2} = 1350 \]

\[ \therefore\; \text{HCF} = \frac{12}{1350} \]

Taking the LCM of the numerators and the HCF of the denominators instead gives the LCM of the fractions, which is the usual confusion.

Answer: C

Clear the decimals by scaling all three by 100, then rescale the answer.

\[ \text{H.C.F.}(210, 105, 63) = 21 \]

\[ \frac{21}{100} = 0.21 \]

Answer: C

Let the numbers be \(a\) and \(b\).
Then,
\(a + b = 55\) and
\(ab = 5 \times 120 = 600\)
The required sum
\(=\dfrac{1}{a} + \dfrac{1}{b}\\~\\= \dfrac{a+b}{ab}\\~\\=\dfrac{55}{600}\\~\\= \dfrac{11}{120}\)