Question

Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case.

Answer 1

Hey User✋

Let's Try To Find Out The Number☺️
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Let's, The Largest Number = n

$62 \: mod \: n \: \\ \: \: \: \: \: \: \: = 132 \: mod \: n \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 237 \: mod \: n$

So (132-62 ) mod n = 0. Or 70 mod n = 0. So n divides 70 evenly. 

[ If two numbers have the same remainder, then the difference of their remainders is 0 ]

So Maybe 70.

$237 \: mod \: 70 = 27$

But,

$162 \: mod \: 70 = 62$

Ok, Let's Try, 35

$62 \: mod \: 35 = 27 \\ 132 \: mod \: 35 = 27 \\ 237 \: mod \: 35 = 27$

$so \: 35 \: is \: the \: answer$
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Answer 2

Thank you for asking this question:

We will assume the largest number as x which will give as the same remainder which will be divided by 62,132 and 237.

Take 62=ax+r......(1)

       132=bx+r.. .  (2)

       237=cx+r......(3)

(2)-(1),(3)-(2) and (3)-(1)

=>70=px,105=qx and 175=rx

Where p=b-a,q=c-b and r=c-a

Now x=H.C.F of 70,105,175

= 35

So 35 is the answer to this question.

If there is any confusion please leave a comment below.