Question

Find the smallest number which levaes a remainder 8 and 12

Answer 1

~HERE'S YOUR ANSWER~


The required number will be 20 less than the LCM of 28 and 32.


- Given the smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively


28 - 8 = 20 and. 32 - 12 = 20 are divisible by the required number .


therefore the required number will be 20 less than the LCM of 28 and 32.


prime factorization of 28 = 2*2*7

prime factorization of 32 = 2*2*2*2*2

LCM ( 28, 32 ) = 2*2*2*2*2*7

= 224.

therefore the required smallest number
= 224 - 20

= 204.


Verification : - 204/ 28 = 28*7 = 196.

= 204 - 198

= 8.


204 / 32 = 32 * 6 = 192 = 204 - 192

= 12 .

HOPE IT HELP YOU !!

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