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## HCF of 2472 and 1284 and a third number 'n' is 12. If their LCM is 2^{3}×3^{2}×5×103×107. Find the number 'n'.

The largest possible number which divides the given numbers completely is called their highest common factor (HCF)

## Answer: The number 'n' is 180 ( 2^{2} × 3^{2} × 5)

Let us see how to find the value of n.

**Explanation:**

The highest common factor (HCF) of any set of numbers is the largest possible number which divides the given numbers exactly without any remainder.

Given that, LCM (2472, 1284, and n) is 2^{3 }× 3^{2 }× 5 × 103 × 107

Let us express the numbers 2472, and 1284 as a product of prime numbers.

2472 = 2 × 2 × 2 × 3 × 103 (2^{3} × 3 × 103)

1284 = 2 × 2 × 3 × 107 (2^{2 }× 3 × 107)

HCF (2472, 1284)= 2 × 2 × 3 (2^{2 }× 3)

'n' should also have one of the factors as HCF, and another factor as the missing element of other numbers from the LCM (i.e., 3 × 5)

Therefore,

n = 2^{2} × 3 × (3 × 5)

n = 2^{2} × 3^{2} × 5

n = 4 × 9 × 5

n = 180

### Therefore, if the HCF of 2472 and 1284 and a third number 'n' is 12 and if their LCM is 2^{3} × 3^{2 }× 5 × 103 × 107, then the number 'n' is 180 (2^{2} × 3^{2} × 5)

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