Question

5. What is the smallest number that when divided by 35,56 and 91 leaves remainder 7 in each case​

Answer 1

Answer:

First, let’s find the smallest number which is exactly divisible by all 35, 56 and 91.

Which is simply just the LCM of the three numbers.  

By prime factorisation,

we get  35 = 5 × 7  56 = 23 × 7  91 = 13 × 7  

∴ L.C.M (35, 56 and 91) = 23 × 7 × 5 × 13 = 3640  

Hence,

3640 is the smallest number which is exactly divisible 28, 42 and 84 i.e.

We will get a remainder of 0 in each case.

But, we need the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.

So that is found by,  

3640 + 7 = 3647  

∴ 3647 should be the smallest number that when divided by 35, 56 and 91 leaves the remainder of 7 in each case.

Hope it Help's you

Drop some Thanks &

Mark as Brainliest

ƪ(˘‿˘)ʃ