show 3|n iff 3|s

**CoCo_RoAcH** · March 31st 2009, 03:21 AM

hey! the question is as follows:

Show that 3|n if and only if 3|s, where s is the sum of the digits of n (expressed as usual in base 10).

i was wondering if some one could please give me a hint as how to start off or give me a push in the right direction as i don't have any idea how to start(go about the question).
However, i do know that i need to show that 3|s and if this is true then 3|n will hold true as 3|n <=> 3|s.

plus i know that

n = ( $a_k$ ) * ( $10^k$ ) + (a_(k-1)) * (10^(k-1)) + ... + a * 10 + ( $a_0$ ) * $10^0$

and

s= ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ )

oh and i was thinking that i may need to use the definition of divisibility where x|y <=> y = xn as

3 | ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ ) <=> ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ ) = 3 m
but then once again im lost in what to do

thankyou very much,
from an extremely stressed CoCo_RoAcH

**CaptainBlack** · March 31st 2009, 08:03 AM

Originally Posted by CoCo_RoAcH

hey! the question is as follows:

Show that 3|n if and only if 3|s, where s is the sum of the digits of n (expressed as usual in base 10).

i was wondering if some one could please give me a hint as how to start off or give me a push in the right direction as i don't have any idea how to start(go about the question).
However, i do know that i need to show that 3|s and if this is true then 3|n will hold true as 3|n <=> 3|s.

plus i know that

n = ( $a_k$ ) * ( $10^k$ ) + (a_(k-1)) * (10^(k-1)) + ... + a * 10 + ( $a_0$ ) * $10^0$

and

s= ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ )

oh and i was thinking that i may need to use the definition of divisibility where x|y <=> y = xn as

3 | ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ ) <=> ( $a_k$ ) + (a_(k-1)) + ... + a + ( $a_0$ ) = 3 m
but then once again im lost in what to do

thankyou very much,
from an extremely stressed CoCo_RoAcH

So:

$n \text{ mod } 3=\sum (a_i \text{ mod } 3) (10^i \text{ mod } 3)=\sum (a_i \text{ mod } 3)=\left(\sum a_i\right) \text{ mod } 3$

CB

**CoCo_RoAcH** · March 31st 2009, 09:47 PM

Originally Posted by CaptainBlack

So:

$n \text{ mod } 3=\sum (a_i \text{ mod } 3) (10^i \text{ mod } 3)=\sum (a_i \text{ mod } 3)=\left(\sum a_i\right) \text{ mod } 3$

CB

Hey CaptainBlack
thankyou very much for that formula but i won't be able to use it as we haven't learnt it.

is there another way to start this question?

thankyou, CoCo_RoAcH

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