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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?

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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