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So the problem is as follows:

Prove that if the sum of digits of a decimal $n$ is three's multiple, then n is three's multiple by direct proof. For example, $11234567$ is 3's multiple because $1+1+2+3+4+5+6+7=24$, and in fact, $11234567$ is $3744856 * 3$. Use $10^n \equiv 1 \pmod3$ if necessary.

I am having a lot of problems with this question seeing as it is a rather difficult problem. Could someone please help me solve this? Thank you!

Rol
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Roger
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2 Answers2

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In decimal basis, your number $n$ can be written as $n = d_0 + d_1 10 + \cdots + d_k 10^k \equiv d_0 + d_1 + \cdots + d_k \pmod3$.

I used $10^m \equiv 1 \pmod 3$ for all natural $m$.

Moreover, it follows that the sum of the digits of $n$ is the remainder of $n$ divided by 3.

Rol
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  • right but could you please hep me solve the rest of the problem? Thank you Rol! – Roger Apr 17 '15 at 03:45
  • If $d_0 + d_1 + \cdots + d_n \equiv 0 \pmod3$ then $n \equiv 0 \pmod3$ (i.e. 3 divides n) and viceversa. You can plug in the numbers from the example you gave to convince yourself. – Rol Apr 17 '15 at 03:51
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Maybe you could first prove that the sum of the digits subtracted from the number is a multiple of 9.

For example, a 4 digit decimal number such as $1234=1\times1000+2\times100+3\times10+4$

But $1000=999+1$, $100=99+1$ and $10=9+1$.

So $1234=1\times\left(999+1\right)+2\times\left(99+1\right)+3\times\left(9+1\right)+4$.

And finally, $1234=1\times999+2\times99+3\times9+\left(1+2+3+4\right)$, or $1234-\left(1+2+3+4\right)=999+2\times99+3\times9$.

I'm sure you can come up with the rest of the proof. All you have to do is show that every power of $10$ is $1$ more than a multiple of $9$. And remember, $9$ is a power of $3$ so the difference between the digital sum and the number is also a multiple of 3.

Okay - here is a proof that any power of 10 is 1 more than a multiple of 9, by induction:

$10^0=1=0\times9+1$

$10^1=10=9\times1+1$

$10^2=100=9\times11+1$

Clearly our hypothesis is true for $n=0,1,2$.

Assume it is true for some arbitrary $n$, so $10^n$ is $1$ more than a multiple of $9$.

More formally: $10^n=9\times m+1$, for some integer $m$.

Then $10^{n+1}=10^n\times10\\ =\left(9\times m + 1\right)\times10\\ =9\times10\times m+10\\ =9\times10\times m+9+1\\ =9\times\left(10\times m+1\right)+1$

We are given $m$ is an integer, so $10\times m$ is an integer and so is $10\times m+1$ (multiplication and addition are closed on the integers). Then $9\times\left(10\times m+1\right)$ is $9$ times an integer or in other words, a multiple of $9$. So clearly, when $10^n$ is $1$ more than a multiple of $9$ so is $10^{n+1}$.

Paul Hanson
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  • I feel stupid saying this, but I actually can't haha. I'm just really looking for someone to do this problem and explain step by step what's going on. That's always been the best way I learn. – Roger Apr 17 '15 at 03:55
  • I added the proof that $10^n\equiv1(mod 9)$ which also proves that $10^n\equiv1(mod 3)$. However, I didn't notice you were already given that fact. Based on that, go with Rol's answer! Remember, $(a\times b) (mod 3)\equiv a (mod 3)\times b(mod 3)$ – Paul Hanson Apr 17 '15 at 05:14
  • One more hint, is the sum of 2 multiples of 3 also a multiple of 3? – Paul Hanson Apr 17 '15 at 05:22
  • Hey Paul thank you so much for taking your time to help me. You're efforts have not gone unnoticed and I thoroughly appreciate it!! Is this the whole proof? – Roger Apr 17 '15 at 06:34