Question

Express the HCF of 52 and 117 as 52x and 117y where x and y are integers

Answer 1

EXPLANATION.

The H.C.F of 52 and 117 as 52x and 117y

where x and y are integers.

According,to the Euclid division lemma.

if we have two positive integers = a and b

then there exist unique integers = x and y

satisfied the conditions = a = bx + y

where, 0 ≤ r ≤ b and x is quotient and

y be the remainder.

=> 117 = 52 X 2 + 13

=> 52 = 13 X 4 + 0

therefore,

remainder is 0 so H.C.F is 13

=> 13 = ( 117 X 1 ) - ( 52 X 2 )

=> 13 = - ( 52 X 2 ) + ( 117 X 1 )

=> 13 = 52x + 117y

Therefore,

=> x = -2 and y = 1

Answer 2

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.

.

.☞ By Euclid division lemma.☜

.

Let us find HCF of 52 and 117 by Euclid's division algorithm.

Since 117>52

117 = 52 × 2 + 13

52 = 13 × 4 + 0

Therefore HCF (52, 117) = 13

Now, 13 = 52×(-2) + 117×1

13 = 52x + 117y where

X = -2

y = 1