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Question

Question 4 : Find the largest number which divides 64, 136 and 238 to leave the same remainder in each case.
Solution : To find the required number, we need to find the HCF of (136-64), (238-136) and (238-64), i.e., HCF (72, 102, 174).
72 = 2³ x 3²
102 = 2 x 3 x 17
174 = 2 x 3 x 29
Therefore, HCF (72, 102, 174) = 2 x 3 = 6
hence, 6 is the required number.

why we find hcf of difference between pairs of number

Comments

Basics of HCF :-

HCF(a,b,c) = HCF(|a-b|,|b-c|,|c-a|).

 

for getting lower value, they use that formula that's it,

Answer 1

Consider the same remainder to be r. So, $64-r, 136-r, 238-r$ is divisible by the largest number say d.

We know that if $n$ and $m$ are divisible by $k$ then $ n-m$ is also divisible by $k$.

So $136-r-(64-r)$ i.e. $72$ and $238-r-(136-r)$ i.e. $102$ are also divisible by d which can be found out by finding the $HCF(72,102)$