Discussion
If \(x = 2^{3} \times 3^{2} \times 5^{3} \times 7^{3}\)
\(y = 2^{2} \times 3^{3} \times 5^{4} \times 7^{3}\), and
\(z = 2^{4} \times 3^{4} \times 5^{2} \times 7^{5}\),
then H.C.F. of x, y and z is
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.
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