Question

Find the largest number which divides $245$ and $1029$ leaving remainder $5$ in each case.

Answer 1

Hint: To find the largest divisible number leaving remainder $5$ each. Find the required number subtracting $5$ from each. And then find H.C.F. of obtained numbers and the result we get.

Complete step-by-step answer:
Subtract 5 from each number we get the required number.
The required number divides $\left(245-5\right)=240$ and
Another required number is $\left(1037-5\right)=1032$
Step2: Now, find the H.C.F. of the obtained new numbers.
We know that H.C.F is the highest common factor. The greatest number which divides each of the two or more numbers is known as the highest common factor.
Find H.C.F of $\left(240,1032\right)$
Factors of $240=\left(2^4\times 3\times 5\right)$
Or $\left(2\times 2\times 2\times 2\times 3\times 5\right)$
 and
Factors of $1032=\left(2^3\times 3\times 43\right)$
Or $\left(2\times 2\times 2\times 3\times 43\right)$
Find common factors in both numbers i.e. the number appears in both number’s factor list.
Highest Common factor H.C.F $\left(240,1032\right)=\left(2^3\times 3\right)=24$
Hence, highest common factor of required number is $=24$
Therefore, the required number is also $24$

Note: The highest common factor is found by multiplying all factors which appear in both lists of factors of given two numbers.
In this problem it is required to find the largest number which divides two given number but leaving a remainder, in this case to find exact divisible number i.e. there should not be any remainder, we need to subtract remainder from number so that no remainder is left the number is divisible .Hence we have subtract $5$ from each. Similarly if we need to find the smallest number divisible by a given number we will find the lowest common factor. Common factors play a vital role in finding divisible numbers.