Prove that the sum of two consecutive multiples of 5 is always an odd number?

Question

Prove that the sum of two consecutive multiples of 5 is always an odd number. Please give full workings, thanks!

Answer 1

5 and 10
5+10=15
15= odd number

10 and 15
10+15=25
25= odd number

etc

Answer 2

Let the first of the two consecutive multiples be 5k, where k is some integer. The next multiple of 5 will be 5(k+1). Their sum is then: 5k + 5(k+1) = 5k + 5k + 5 = 10k + 5 Since k is an integer, 10k is an even number, and an even number + 5 is an odd number. Is that the sort of thing you're looking for?

Answer 3

The multiples of 5 are x, x+5, x+10, x+15,...

Take any two consecutive ones and you will get 2x + and odd number. 2x is even. An even and an odd always is an odd. the proof is that all odd numbers can be written as an even plus 1.

I will take (x+10)+(x+15)

2x+25
2x+24+1
2(x+12) +1

Hope I've helped

Answer 4

Any two consecutive multiples of 5 can be defined as 5x (where x is any integer) and 5x+5. Adding the two together yields 10x+5. 10x will always end in a 0 and adding 5 two that makes the number end in a 5 (odd number).

Answer 5

Well any two consecutive multiples of five would include one odd number and one even number. Any odd number plus any even number will always yield an odd number.

Hope this helps. Good Luck.

Answer 6

because any even # multiplied by 5 is going to end in a 0 and any odd # multplied by 5 will end in a 5... and 2 consecutive means even and odd...

3*5=15
4*5=20
15+20=35....

Answer 7

Actually, this is true for every odd number. If p is an odd number, then each of its multiples is given by n*p, n integer. The next is given by (n+1)*p. Therefore, their sum is s= n*p + (n+1)*p = (2n+1)*p= 2n*p + p. Since 2n*p is always even and p is odd, it follows s is odd.

Answer 8

One of the consecutive numbers will always be odd and one will always be even, therefore, the sum of the two will always be odd.

Answer 9

let 5k be the first multiple of 5
then 5k + 5 is the next m=ultiple of 5

adding the two gives 10k + 5. this is mod 2 equal to 1 thus odd.

Answer 10

5n + 5(n + 1) = 5n + 5n + 5 = 10n + 5

10n is always even (divisible by 2)

An even # plus an odd # is odd