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# 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.

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