Question

find the smallest number which when devides by 28 and 32 leave remainder 8 and 12 respectively​

Answer 1

Answer:

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Step-by-step explanation:

let the no. be x

hence, 28-8=20

32-12=20

thus the smallest no. is HCF(20,20)=1

Answer 2

Question

find the smallest number which when devides by 28 and 32 leave remainder 8 and 12 respectively

Answer

Given that the smallest no.

When divided by 28 & 32 leaves remainder 8 & 12.

Therefore,

28–8=20 & 32–12=20 are divisible by

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

Hence,

the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 is 204

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