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

Tutor

From Euclid's division lemma,we know that a=bq+r,Since 117>52 ,we can take a=117 & b=52 Now,117=52*2+13(52 is divisor) so,52=13*4+0,the division process stops here,as the remainder becomes 0. Hence HCF =13. Since 13 is divisor ,so 13 can also be written as 52*(-2)+117*1 Where,* means multiplica...

The HCF of 52 and 117 is 13. And it can be expressed as 52(-2) + 117(1) = 13

Maths Tutor

Hcf 13

Math Tutor

Hcf is 13. Explanation is given in the pic.

Maths Science Expert

Between 65 and 117; 117 is greater than 65 Division lemma of 117 and 65: 117 = 65 x 1 + 52 65 = 52 x 1 + 13 52 = 4 x 13 + 0 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 ... read more
Between 65 and 117; 117 is greater than 65 Division lemma of 117 and 65: ⇒ 117 = 65 x 1 + 52 ⇒ 65 = 52 x 1 + 13 ⇒ 52 = 4 x 13 + 0 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 Now, 13 = 65 – 52 x 1 ⇒ 52 = 117 – 65 x 1 ⇒ 13 = 65 – (117 – 65 x 1) x 1 ⇒ 13 = 65 x 2 – 117 ⇒ 13 = 65 x 2 + 117 x (–1) Therefore H.C.F. of 65 and 117 is of the form 65m + 117n, where m = 2 and n = –1. read less

Hcf will be 13 52*13+117*13=2197

Tutor

hcf is 13 we have to divide 117/52,then we get quotient 2,remainder is 13 but this equation be calculte on eculid's divison lamma 52(-2)+117(-1)=13

Trainer

Euclid's Division Lemma : a = bq + r Here ,a= 117, b=52 117 = 52 x 2 + 13 52 = 13 x 4 + 0 H.C.F = 13 117x + 52 y = 13 x = 1 , y = -2 117 1 + 52 - 2 = 117 + -104 = 13 Thank you.

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13

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