I would do this by first listing the multiples of 6 until I start to see a pattern with the one's digit.
6x0=0
6x1=6
6x2=12
6x3=18
6x4=24
6x5=30
6x6=36
6x7=42
6x8=48
...
The digits in bold are the one's digits so those are the only ones we really care about. If you list just them it looks like: 0,6,2,8,4,0,6,2,8
Notice how the first set of 5 numbers seems as though it repeats in the 6th, 7th, and 8th numbers. This probably means the pattern continues infinitely so the first 5 numbers are all the one's digits that can come from multiples of 6. Thus your answer is: 0,6,2,8,or 4

7 0

3 years ago

(2y^3+5y^2-y+15) ÷ (y-4)

Sveta_85 [38]

ANSWER: 2y^2 + 13y +51 r(219/y-4)

picture explanation too: