SOLUTION: Hi, I'm stuck with this problem... I don't really know how to tackle it. So if you can help me that'd be much appreciated. Let a be a non-negative integer, written in standard d

Algebra ->  Divisibility and Prime Numbers  -> Lessons -> SOLUTION: Hi, I'm stuck with this problem... I don't really know how to tackle it. So if you can help me that'd be much appreciated. Let a be a non-negative integer, written in standard d      Log On


   

Question 792475: Hi, I'm stuck with this problem... I don't really know how to tackle it. So if you can help me that'd be much appreciated.
Let a be a non-negative integer, written in standard decimal notation as
akak−1 . . . a2a1a0, where 0 ≤ ai ≤ 9 for i = 0,1, . . . , k. Prove that a is
divisible by 8 if, and only if, 4a2 + 2a1 +a0 is divisible by 8.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The k digits of the number a are, in order
a%5Bk%5D, a%5Bk-1%5D, ....a%5B2%5D, a%5B1%5D, and a%5B0%5D.
Let A be the non-negative integer written in standard decimal notation by the first k-3 digits.
The value of a is
a=1000A%2B100a%5B2%5D%2B10a%5B1%5D%2Ba%5B0%5D
a=8%2A125%2AA%2B%2896%2B4%29a%5B2%5D%2B%288%2B2%29a%5B1%5D%2Ba%5B0%5D
a=8%2A125%2AA%2B96a%5B2%5D%2B4a%5B2%5D%2B8a%5B1%5D%2B2a%5B1%5D%2Ba%5B0%5D
a=8%2A125%2AA%2B96a%5B2%5D%2B8a%5B1%5D%2B%284a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D%29
a=8%2A125%2AA%2B8%2A12a%5B2%5D%2B8a%5B1%5D%2B%284a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D%29
a=8%2A%28125%2AA%2B12a%5B2%5D%2Ba%5B1%5D%29%2B%284a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D%29

125%2AA%2B12a%5B2%5D%2Ba%5B1%5D=M is a non-negative integer

If 4a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D=8n for some non-negative integer n,
a=8M%2B8n=8%28M%2Bn%29 is a multiple of 8.

If 4a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D is not divisible by 8, then
4a%5B2%5D%2B2a%5B1%5D%2Ba%5B0%5D=8n%2Br for some non-negative integers n and r with 1%3C=r%3C=7, and
a=8M%2B8n%2Br=8%28M%2Bn%29%2Br is not a multiple of 8.