Prove that the product of two consecutive multiples of 5 is always an even number.?

Question

Prove that the product of two consecutive multiples of 5 is always an even number.

Answer 1

5n and 5(n+1)
their product is:
5n5(n+1)
=25n(n+1),
then either n is even or n+1 is even, in either case
the product is also even

Answer 2

Multiples of 5 will end in 5 or 0. If you have two consecutive multiples of 5, one will end in 5 and the other will end in 0. Their product will end in 0, making it an even number.

Answer 3

Two consecutive multiples of 5 have the form 5k and 5(k+1).

(5k)*(5(k+1)) = 25k(k+1).

If k is even, then k+1 is odd. The product of two odd numbers is odd, so 25(k+1) is odd. The product of an odd number and an even number is even, so k*(25(k+1)) is even.

If k is odd, then k+1 is even. 25k is odd, therefore 25k(k+1) is even.

Answer 4

This is tru for to consecutive multiples of any integer.
let m= that integer & n= the number by whisc we will multiply it.

mn*m(n+) is always even because
m^2*n(n+1)
2 possibilities:
1) n is even, in which case the product n(n+1) will be even.
2) n is odd, in which case n+1 will be even so n*(n+1) will be even.

since m is an integer, m^2 will also be an integer. any integer multiplied by an even number is even.

Answer 5

5n[5(n + 1)] = 5n[5n + 5) = 25n^2 + 25n = 25n(n + 1)

If n is odd, 25n ends in 5 (is odd), n + 1 is even, result is even.

If n is even, 25n ends in 0 (is even), n + 1 is odd, result is even.