Question

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

Answer 1

$28-8=20$ and $32-12=20$

$LCM$ of $28832\Rightarrow$ Prime \factor of $28=2\times 2\times 7$

$LCM$ of $28832\Rightarrow$ Prime factor of $32=2\times 2\times 2\times 2\times 2$

$LCM(28,32)=2\times 2\times 2\times 2\times 2\times 7=224$

Therefore the required smallest NO$=224-20=204$

Verification $\Rightarrow \frac{204}{28}=28\times 7=196\Rightarrow 204-196=8$

$\frac{204}{32}=36\times 6=192\Rightarrow 204-192=12$