Show that the product of any 5 consecutive natural numbers is always divisible by 120.
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Let the five consecutive integers be x, x+1, x+2, x+3, x+4, where x can be any integers
So, sum of five consecutive integers is:
x + x + 1 + x + 2 + x + 3 + x + 4
= 5x + 10
= 5 (x + 2)
Since the sum of any integers is 5(x+2) which is a multiple of 5. So, the sum is always divisible by 5.
So, sum of five consecutive integers is:
x + x + 1 + x + 2 + x + 3 + x + 4
= 5x + 10
= 5 (x + 2)
Since the sum of any integers is 5(x+2) which is a multiple of 5. So, the sum is always divisible by 5.
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