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What is the Smallest Number That, When Divided by 35, 56 and 91 Leaves Remainders of 7 in Each Case? - Mathematics

Numerical

What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

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Solution

TO FIND: Smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case

L.C.M OF 35, 56 and 91

`35= 5xx7`

`56=2^2xx7`

`91=13xx7`

L.C.M of 35,56 and 91 = `2^2xx5xx7xx13`

=3640

Hence 84 is the least number which exactly divides 28, 42 and 84 i.e. we will get a remainder of 0 in this case. But we need the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case

Therefore

= 3640 +7 

= 3640 

Hence 3640  is smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case.

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 1 Real Numbers
Exercise 1.4 | Q 10 | Page 40
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