Question

Find the largest divisor which divides the following numbers to give the required remainder in each case . The sum is 47 , 80 , & 124 remainder :3

Answer 1

Answer:

What is the largest number which divides 62, 132, 237 to leave same remainder in each case?

Assume that ‘d’ is the largest divisor , which divides 62,132,& 237 leaving remainder ‘ r' in each case.

By Euclid's division lemma, we state that

a= d*q + r , where 0 < r < d

dividend = divisor * quotient + remainder

62 = d * q1 + r

132 = d * q2 + r

237 = d * q3 + r

OR

62 -r = d *q1

132 -r = d * q2

237 -r = d* q3

This concludes that (62-r), (132-r) & (237-r) are exactly divisible by d as remainder now = 0. Or we can say that d is gcd of all these three numbers .

As we know that , if d divides a & b. Then d divides (a- b) too

So, d divides (132-r) - (62-r)

=> d divides 132 -r -62 +r

=> d divides 70 ………..(1)

Similarly d divides (237 - r) - (62 -r)

=> d divides 237 -r -62 +r

=> d divides 175 ……….(2)

=> By (1) & (2)

d is gcd of 70 & 175

70 = 2*5*7

175 = 5*5*7

So, gcd = 5*7 = 35

ANS: largest divisor is 35

Answer 2

Answer:

A number when divided by 342 leaves a remainder of 47. ... divides 62, 132, and 237 to leave the same remainder in each case is 5. ... The required number is 35 ... View 124 Upvoters.

Step-by-step explanation:

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