What is the smallest number that when divided by 35 56 and 91 leaves remainder of 17 each case?

Question 1 Real Numbers Exercise 1.4

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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 factorization, 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.

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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.