Question

11. Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.​

Answer 1

Given that the smallest number when divided by 28&32 leaves remainder 8&12 respectively..

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So, 28-8=20 & 32-12=20 are divisible by the required numbers...

Therefore the required number will be 20 less than LCM of 28&32..

$\sf \pink{Prime \ factorization \ of \ 28=}$2×2×7

$\sf \pink{Prime \ factorization \ of \ 32=}$2×2×2×2×2

$\sf \ pink{LCM \ (28 & 32) \ =}$2×2×2×2×7=224

Therefore the required smallest number is 224-20=204

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verification:-

204/28=28×7=196

=204-198

=8

________________

204/32=32×6=192

=204-192

=12

Hence proved✔

Answer 2

Given that the smallest number when divided by 28&32 leaves remainder 8&12 respectively..

_____________________

So, 28-8=20 & 32-12=20 are divisible by the required numbers...

Therefore the required number will be 20 less than LCM of 28&32..

Prime factorization of 28=Prime factorization of 28= 2×2×7

Prime factorization of 32=Prime factorization of 32= 2×2×2×2×2

LCM (28 & 32) =2×2×2×2×7=224

Therefore the required smallest number is 224-20=204

____________________

verification:-

204/28=28×7=196

=204-198

=8

________________

204/32=32×6=192

=204-192

=12

Hence proved✔