Question:
If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n =
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
LCM $(a, b, c)=2^{3} \times 3^{2} \times 5 \ldots \ldots$ (I)
We have to find the value for n
Also
$a=2^{3} \times 3$
$b=2 \times 3 \times 5$
$c=3^{n} \times 5$
We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.
Therefore, by applying this rule and taking $n \geq 1$ we get the LCM as
$\operatorname{LCM}(a, b, c)=2^{3} \times 3^{n} \times 5 \ldots \ldots$ (II)
On comparing (I) and (II) sides, we get:
$2^{3} \times 3^{2} \times 5=2^{3} \times 3^{n} \times 5$
$n=2$
Hence the correct choice is (b).