find the smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively
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Solution:-
Given,
The smallest number when divided by 28 and 2 leaves remainder 8 and 12 respectively.
28 - 8 = 20
32 - 12 = 20
These numbers are divisible by the required Numbers.
Hence,
The smallest number be less than 20.
Now,
We should find the L.C.M. of 28 and 2
Prime Factorization of 28 = 2 * 2 * 7
Prime Factorization of 32 = 2 * 2 * 2 * 2 * 2
L.C.M. of 28 and 32 = 2 * 2 * 2 * 2 * 2 * 7
= 224
∴ The smallest number = 224 - 20
= 204.
Verification :
204/28 = 28 x 7 = 196.
= 204 - 196
= 8
204/32 = 32 x 6 = 192
= 204 - 192
= 12.
∴ The smallest Number = 204.
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