Find the smallest no. which when divided by 28 and 32 leaves remainder 8 and 12.

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Solution

$28 - 8 = 20$ and $32 - 12 = 20$
$L C M$ of $28832 \Rightarrow$ Prime \factor of $28 = 2 \times 2 \times 7$
$L C M$ of $28832 \Rightarrow$ Prime factor of $32 = 2 \times 2 \times 2 \times 2 \times 2$
$L C M (28, 32) = 2 \times 2 \times 2 \times 2 \times 2 \times 7 = 224$
Therefore the required smallest NO $= 224 - 20 = 204$
Verification $\Rightarrow \frac{204}{28} = 28 \times 7 = 196 \Rightarrow 204 - 196 = 8$
$\frac{204}{32} = 36 \times 6 = 192 \Rightarrow 204 - 192 = 12$

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