Discussion
The smallest number, by which 9408 must be multiplied so that the product is a perfect square, will be
Answer: B
The smallest multiplier is 3.
Factorise 9408 into primes:
\[ \begin{aligned} 9408 &= 2^{6} \times 147 \\ &= 2^{6} \times 3 \times 7^{2} \end{aligned} \]
In a perfect square every prime has an even power. Here \(2^{6}\) and \(7^{2}\) are already even powers, and only 3 appears to an odd power, so multiplying by 3 gives \(2^{6} \times 3^{2} \times 7^{2} = 28224 = 168^{2}\).
Multiplying by 9 would leave \(3^{3}\), still odd; 4 and 2 change only the power of 2 and leave the single 3 unpaired.
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