1.

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

Answer» sankey

sanket
Ayush
sanket


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sanket
28-8=20 & 32-12=20 So the reqd no will be 20 less than the L. C. M. of 28 and 32 . L. C. M. of 28 & 32 is = 224 Reqd smallest no =224-20 = 204.


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