Question

Find the smallest number which when increased by $17$ is exactly divisible by both $520$ and $468$.

Answer 1

Hint: The smallest number which is exactly divisible by the given two or three numbers is obtained by finding the LCM (lowest common multiple) of the numbers. Here, we have to find the smallest number which when increased by $17$ then is exactly divisible by the numbers. So, to find the required smallest number we have to subtract $17$ from the LCM (lowest common multiples) of the number.

Complete step-by-step answer:
Here, we find the smallest number which when increased by $17$ is exactly divisible by both $520$ and $468$.
Firstly, we have to calculate the LCM (lowest common multiple) of $520$ and $468$. We can find LCM by prime factorization method.
The prime factor of $468 = 2 \times 2 \times 3 \times 3 \times 13$.
The prime factors of $520 = 2 \times 2 \times 2 \times 5 \times 13$.
LCM of two numbers is written by multiplying the factors only once if it is common in both otherwise multiplying all the factors of both numbers.
So, LCM of $520$ and $468$ is $2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 13 = 4680$.
We have to find the smallest number which when increased by $17$ is divisible by two numbers.
So, the required number is decreased by $17$ from $4680$.

Thus, the required number is $4680 - 17 = 4663$.

Note:
The LCM (lowest common multiples) of the given two numbers can also be calculated by division method which is a basic method.
Similar concept is applied when we have to find the largest $5$ or $n$ digits numbers which is exactly divisible by the given few numbers. for this also firstly find the LCM of the given numbers and then divide the largest $n$ digits number by the LCM and then subtract the remainder from the largest $n$ digits number to get the required results.