Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was asked to find divisibility tests for 2,3, and 4.

I could do this for 2 and 3, but for 4.

I could come only as far as:

let $a_na_{n-1}\cdots a_1a_0$ be the $n$ digit number.

Now from the hundredth digit onwards, the number is divisible by 4 when we express it as sum of digits.

So, the only part of the proof that's left is to prove that $10a_1+a_0$ is divisible by 4.

So if we show that this happens only when the number $a_1a_0$ is divisible by 4, the proof is complete.

So the best way to show it is by just taking all combinations of $a_1,a_0$ or is there a better way?

share|improve this question
    
If $a_1$ is even, then the number is divisible by $4$ if and only if $a_0$ is divisible by $4$ ($a_0=0$, $4$, or $8$). If $a_1$ is odd, then the number is divisible by $4$ if and only if $a_0$ is even and not divisible by $4$ ($a_0=2$ or $a_2=6$). This because $20$ is a multiple of $4$, and $10$ leaves a remainder of $2$. –  Arturo Magidin Nov 13 '11 at 8:57
    
$10a_1 + a_0$ is by definition the number represented by the decimal string "$a_1a_0$", so I don't see any problem. Indeed, a number is divisible by 4 iff the number represented by the last 2 digits is. –  Henno Brandsma Nov 13 '11 at 8:59
    
@kira. $a_na_{n-1}\cdots a_1a_0$ is an $(n+1)$-digit number (not $n$-digit). –  matt Nov 13 '11 at 9:02
    
Why not just say $a_1a_0$ is divisible by 4? This would be an elementary "test". You could also use $a_1a_0$ is div. by 2 and the quotient is div. by 2. –  David Mitra Nov 13 '11 at 9:14
    
@matt ya sorry for the mistake –  Bach Nov 13 '11 at 9:28

1 Answer 1

up vote 1 down vote accepted

Note that $10a_1+a_0\equiv2a_1+a_0$ (mod 4). So for divisibility by 4, $a_0$ must be even and in this case $2a_1+a_0=2(a_1+\frac{a_0}{2})$. So, $a_1$ and $\frac{a0}{2}$ must be of same parity (means both are either even, or odd).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.