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

Find the HCF of the following pair of integers and express it as a linear combination of them,

(i) 963 and 657

(ii) 592 and 252

(iii) 506 and 1155

(iv) 1288 and 575

By applying Euclid’s division lemma on 963 and 657, we get 963 = 657 x 1 + 306… ......... (1)

As the remainder ≠ 0, apply division lemma on divisor 657 and remainder 306 657 = 306 x 2 + 45…............ (2)

As the remainder ≠ 0, apply division lemma on divisor 306 and remainder 45 306 = 45 x 6 + 36… ............ (3)

As the remainder ≠ 0, apply division lemma on divisor 45 and remainder 36

45 = 36 x 1 + 9… ............... (4)

As the remainder ≠ 0, apply division lemma on divisor 36 and remainder 9

36 = 9 x 4 + 0… ................ (5)

Thus, we can conclude the H.C.F. = 9.

Now, in order to express the found HCF as a linear combination of 963 and 657, we perform 9 = 45 – 36 x 1 [from (5)]

= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6 [from (3)]

= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1 [from (2)]

= 657 x 7 – 306 x 14 – 306 x 1

= 657 x 7 – 306 x 15

= 657 x 7 – [963 – 657 x 1] x 15 [from (1)]

= 657 x 7 – 963 x 15 + 657 x 15

= 657 x 22 – 963 x 15.

(ii) 592 and 252

By applying Euclid’s division lemma on 592 and 252, we get

592 = 252 x 2 + 88…....... (1)

As the remainder ≠ 0, apply division lemma on divisor 252 and remainder 88 252 = 88 x 2 + 76… ........ (2)

As the remainder ≠ 0, apply division lemma on divisor 88 and remainder 76 88 = 76 x 1 + 12… ........... (3)

As the remainder ≠ 0, apply division lemma on divisor 76 and remainder 12

76 = 12 x 6 + 4… ............. (4)

Since the remainder ≠ 0, apply division lemma on divisor 12 and remainder 4 12 = 4 x 3 + 0… ................ (5)

Thus, we can conclude the H.C.F. = 4.

Now, in order to express the found HCF as a linear combination of 592 and 252, we perform 4 = 76 – 12 x 6 [from (4)]

= 76 – [88 – 76 x 1] x 6 [from (3)]

= 76 – 88 x 6 + 76 x 6

= 76 x 7 – 88 x 6

= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]

= 252 x 7- 88 x 14- 88 x 6

= 252 x 7- 88 x 20

= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]

= 252 x 7 – 592 x 20 + 252 x 40

= 252 x 47 – 592 x 20

= 252 x 47 + 592 x (-20)

(iii) 506 and 1155

By applying Euclid’s division lemma on 506 and 1155, we get 1155 = 506 x 2 + 143… ............. (1)

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143 506 = 143 x 3 + 77…................. (2)

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77 143 = 77 x 1 + 66… .................. (3)

Since the remainder ≠ 0, apply division lemma on divisor 77 and remainder 66

77 = 66 x 1 + 11… ..................... (4)

As the remainder ≠ 0, apply division lemma on divisor 66 and remainder 11 66 = 11 x 6 + 0… ...................... (5)

Thus, we can conclude the H.C.F. = 11.

Now, in order to express the found HCF as a linear combination of 506 and 1155, we perform 11 = 77 – 66 x 1 [from (4)]

= 77 – [143 – 77 x 1] x 1 [from (3)]

= 77 – 143 x 1 + 77 x 1

= 77 x 2 – 143 x 1

= [506 – 143 x 3] x 2 – 143 x 1 [from (2)]

= 506 x 2 – 143 x 6 – 143 x 1

= 506 x 2 – 143 x 7

= 506 x 2 – [1155 – 506 x 2] x 7 [from (1)]

= 506 x 2 – 1155 x 7+ 506 x 14

= 506 x 16 – 1155 x 7

(iv) 1288 and 575

By applying Euclid’s division lemma on 1288 and 575, we get

1288 = 575 x 2+ 138 ............... (1)

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143

575 = 138 x 4 + 23 .................... (2)

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77

138 = 23 x 6 + 0 ......................... (3)

Thus, we can conclude the H.C.F. = 23.

Now, in order to express the found HCF as a linear combination of 1288 and 575, we perform 23 = 575 – 138 x 4 [from (2)]

= 575 – [1288 – 575 x 2] x 4 [from (1)]

= 575 – 1288 x 4 + 575 x 8

= 575 x 9 – 1288 x 4

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