Check whether $32^{2}$ is divisible by $4$ or $8$ or by both $4$ and $8$ ?

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Solution

$32^{2} = (4 \times 4 \times 2)^{2} = 4^{2} \times 4^{2} \times 2^{2} = 16 \times 16 \times 4 = 8 \times 2 \times 8 \times 2 \times 4$

As we can see above, The Given number can be divisible by both $4$ and $8$

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Check whether 322 is divisible by 4 or 8 or by both 4 and 8?

322=(4×4×2)2=42×42×22=16×16×4=8×2×8×2×4As we can see above, The Given number can be divisible by both 4 and 8

Check whether $32^{2}$ is divisible by $4$ or $8$ or by both $4$ and $8$ ?

$32^{2} = (4 \times 4 \times 2)^{2} = 4^{2} \times 4^{2} \times 2^{2} = 16 \times 16 \times 4 = 8 \times 2 \times 8 \times 2 \times 4$

As we can see above, The Given number can be divisible by both $4$ and $8$