# RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.4 in Chapter 1 - Real Numbers

Question 1 Real Numbers Exercise 1.4

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

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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