Question

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

Answer 1

Hint: In the given question, we are required to find the largest number which divides the three numbers provided to us in the problem itself to leave the same remainder in each case. In such questions, we first assume the remainder left as a constant. Then, we find the numbers which are completely divisible by subtracting the remainder from the three numbers. Now, we find the HCF of these numbers in order to get to the final answer.

Complete step by step answer:
The given numbers are: \[62\], $132$ and $237$. Now, we have to find the largest number which divides the three numbers.Let us assume that the remainder left in each case be $m$.Then, we have to find the largest number which divides all the three numbers \[\left( {62 - m} \right)\], $\left( {132 - m} \right)$ and $\left( {237 - m} \right)$.

Now, if these obtained numbers are divisible by the required number, then the difference of pairs of \[\left( {62 - m} \right)\], $\left( {132 - m} \right)$ and $\left( {237 - m} \right)$ will also be divisible by the required number. So, subtracting \[\left( {62 - m} \right)\] from $\left( {132 - m} \right)$, we get,
$\left( {132 - m} \right) - \left( {62 - m} \right)$
Opening the brackets and carrying out the calculations, we get,
$ \Rightarrow 70$
So, subtracting $\left( {132 - m} \right)$ from $\left( {237 - m} \right)$, we get $\left( {237 - m} \right) - \left( {132 - m} \right)$.

Opening the brackets and carrying out the calculations, we get,
$ \Rightarrow 105$
Now, finding the HCF of $70$ and $105$,
$70 = 2 \times 5 \times 7$
$\Rightarrow 105 = 3 \times 5 \times 7$
In order to find the HCF using prime factorization method, we first represent the given two numbers as a product of their prime factors and then find the product of the lowest powers of all the common factors.So, the highest common factor of $70$ and $105$$ = 5 \times 7 = 35$.

Hence, the largest number which divides \[62\], $132$ and $237$ to leave the same remainder in each case is $35$.

Note: Highest common factor is the greatest number that divides both the given numbers. Similarly, the highest common factor can also be found by using the prime factorization method as well as using Euclid’s division lemma. Highest common divisor is just a product of common factors with lowest power. In such questions, innovative and analytical things are important.