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

Given: 520 and 468.

To find: Here we have to find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Solution:

LCM of two numbers is the smallest number which is exactly divisible by both numbers.

First, we need to find the LCM of 520 and 468.

Now, calculating the LCM of 520 and 468 by the prime factorization method:

Writing the numbers as a product of their prime factors:

Prime factorisation of 520:

• $2\ \times\ 2\ \times\ 2\ \times\ 5\ \times\ 13\ =\ 2^3\ \times\ 5^1\ \times\ 13^1$

Prime factorisation of 468:

• $2\ \times\ 2\ \times\ 3\ \times\ 3\ \times\ 13\ =\ 2^2\ \times\ 3^2\ \times\ 13^1$

Multiplying the highest power of each prime number:

• $2^3\ \times\ 5^1\ \times\ 13^1\ \times\ 3^2\ =\ 4680$

LCM(520, 468)  $=$  4680

But we have to find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. So,

The required number  $=$  LCM(520, 468)  $-$  17

The required number  $=$  4680  $-$  17

The required number  $=$  4663

So, the smallest number which when increased by 17 is exactly divisible by both 520 and 468 is 4663.

Updated on: 10-Oct-2022

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