Question

find 4 consecutive integers whose product is 840

Answer 1

4,5,6,7

Step-by-step explanation:

If n is the smallest of the four consecutive positive integers, then

840=n(n+3)⋅[(n+1)(n+2) ]

=[(n2+3n+1)−1]x[(n2+3n+1)+1)]

=(n2+3n+1)2−1.

Hence, n2+3n+1=29, so that n=4 (we may neglect n=−7, which is negative). We may verify that 4x5x6x7=840