Question

The product of four consecutive natural
numbers is 840. Find the numbers.
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Answer 1

Let the consecutive numbers be n, (n+1) , (n + 2 ) (n+ 3).

Given that their product = 840

then n(n+1)(n + 2 )(n+ 3) = 840.

this can also written as n(n+3)⋅(n+1)(n+2) = 840

Add and subtarct 1 to the first number of the product.

((n^2+3n+1)−1)((n2+3n+1)+1) = 840

(n^2+3n+1)^2−1 = 840

(n^2+3n+1)^2 = 841

(n^2+3n+1)^2 = 29 ^2

(n^2+3n+1) = 29 implies n^2+3n+1–29=0 = n^2 + 3n - 28

(n + 7) (n - 4) = 0

if n = 4 we have 4, 5, 6, 7 n = -7 is not possible -7 does not belongs to the natural number.