Question

11. Find the HCF of 52 and 117 and express it in the form 52x + 117y.

Answer 1

Solution:-
By Euclid's Division Lemma 117 > 52
117 = (52 × 2) + 13 (52 is the divisor)
52 = 13 × 4 + 0 ; The division process ends here, as remainder is 0. So, HCF is 13 (Here, 13 is divisor)
13 can also be expressed as 52x + 117y i.e. as 52 (-2) + 117 (1)


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Answer 2

${Answer :-}$

${Easy \:method :}$

Here, we have,

Factors of 52 = 1,2,4,13,26 and 52.
Factors of 117 = 1,3,9,13,39 and 117.

Common factors = 1 and 13.
Highest Common Factor = 13.

Thus, HCF = 13.

${Division \:method:}$

On dividing 117 by 52, we have
Quotient = 2 and Remainder = 13.
117 = ( 52*2) + 13............... (1)

Again, dividing 52 by 13,
Quotient = 4 and remainder = 0.
52 = (13*4) + 0.

Thus, HCF is 13 since Remainder is 0 and thus division stops.

Now, we have from (1),
Dividend = (divisor*quotient)+remainder
Or, 117 = ( 52*2) + 13.
Or, 117 - (52*2) = 13.
Or, 13 = 117(1) + 52(-2)
Or, 13 = 52(-2) + 117(1)

Thus, the HCF of 117 and 52 is 13, which can be expressed in the form 52x + 117y as 52(-2) + 117(1) where x = (-2) and y = 1.

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