Prove that given any 17 integers, there exist nine of them whose sum is divisible by 9.

(it might to use the fact that given any five integer you can find three whose sum is divisible by 3)

Step-by-step explanation:

let x,x+1,x+2,x+3,x+4,x+5,x+6,x+7,x+8

be 9 integers

x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8=9x+36=9(x+4)

which is divisible by 9

the correct answer is 1: 2

step-by-step explanation:

hope this your welcome : )

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)