 # RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.2 in Chapter 1 - Real Numbers

If the HCF of 657 and 963 is expressible in the form 657x + 963 x – 15, find x.

Firstly, the HCF of 657 and 963 is to be found. By applying Euclid’s division lemma, we get 963 = 657 x 1+ 306.

Here, the remainder ≠ 0 and so we apply Euclid’s division lemma on divisor 657 and remainder

306

657 = 306 x 2 + 45.

Now, continue applying division lemma till the remainder becomes 0. 306 = 45 x 6 + 36.

Again, the remainder ≠ 0

45 = 36 x 1 + 9.

Again, the remainder ≠ 0

36 = 9 x 4 + 0.

Now, the remainder = 0.

Hence, the last divisor is the H.C.F of 657 and 963 i.e., 9

So, this HCF is expressed as a linear combination which given as, 9 = 657x + 936 (-15).

Solving for x, we get 9 = 657x —14445

9 + 14445 = 657x

14454 = 657x

⇒ x = 14454 / 657

∴ x = 22

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