Iff a number $n$ is divisible by $3$, then the sum of its digits is also divisible by $3$.

You are watching: A number is divisible by 3 if the sum of its digits is divisible by 3

Proof:

We know $n \mod 3 = 0$. By the basis representation theorem, $n$ can be re-written as $n_k10^k + n_{k-1}10^{k-1} + \cdots + 10n_1 + n_0 \equiv 0 \mod 3$. By the modular arithmetic addition rule, we can take $n_k10^k \mod 3 + n_{k-1}10^{k-1} \mod 3 + \cdots + 10n_1 \mod 3 + n_0 \mod 3$. But then we just get $n_k + n_{k-1} + \cdots + n_1 + n_0$, since $10 \mod 3 \equiv 1$.

Now suppose that the sum of the digits was divisible by 3. How can I prove that the representation in base $10$ is divisible by $3$?


elementary-number-theory
Share
Cite
Follow
asked Jan 17 "15 at 23:44
*

Don LarynxDon Larynx
4,38533 gold badges3434 silver badges5757 bronze badges
$\endgroup$
4
Add a comment |

2 Answers 2


Active Oldest Votes
3
$\begingroup$
The key point which you seem to have noted is that $10\equiv 1\pmod{3}$. This means that a number is congruent to the sum of its digits mod 3 because $10^k\equiv 1\pmod{3}$, which you also seem to have noted. Thus $n$ is divisible by $3$ (congruent to $0$ mod 3) if and only if the sum of its digits is. The same is true for other remainders modulo 3; a number has a remainder of $1$ when divided by $3$ if and only if the sum of its digits does, etc.

Your observation that $10^k\equiv 1\pmod{3}$ is the reason why it has been commented that you have already proved the result.


Share
Cite
Follow
answered Jan 17 "15 at 23:56
*

Matt SamuelMatt Samuel
55k1010 gold badges6161 silver badges9292 bronze badges
$\endgroup$
Add a comment |
2
$\begingroup$
I hope this will suffice:$$n= (d_{m-1}\cdots d_0)_{10} = \sum_{k=0}^{m-1} d_k 10^k= \sum_{k=0}^{m-1} d_k (10^k - 1) + d_k = \sum_{k=0}^{m-1} d_k (10^k - 1) + S(n)$$with $S(n)$ being the digit sum in base $10$ and$$10^k - 1 = 9 \frac{10^{k} - 1}{10-1} = 9 \sum_{i=0}^{k-1} 10^i = \sum_{i=0}^{k-1} 9 \, 10^i= (\underbrace{9\cdots 9}_{k \times})_{10} $$is divisible by $3$, it depends on the divisibility of $S(n)$.

The same statement should hold for division by $9$.


Share
Cite
Follow
edited Jan 18 "15 at 0:12
answered Jan 17 "15 at 23:55
*

mvwmvw
33.4k22 gold badges2929 silver badges6161 bronze badges
$\endgroup$
Add a comment |

Your Answer


Thanks for contributing an answer to mmsanotherstage2019.comematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid

Asking for help, clarification, or responding to other answers.Making statements based on opinion; back them up with references or personal experience.

Use mmsanotherstage2019.comJax to format equations. mmsanotherstage2019.comJax reference.

To learn more, see our tips on writing great answers.

See more: How Many Feet Are 10 Meters Is Equal To How Many Feet Are 10 Meters


Draft saved
Draft discarded

Sign up or log in


Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Submit

Post as a guest


Name
Email Required, but never shown


Post as a guest


Name
Email

Required, but never shown


Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy


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


Featured on Meta
Visit chat
Linked
0
Why is can number be divided by 3(ofc I mean we get an integer result) if its sum of digits can be divided by 3
1
Proof of x = 0 modulo 3 only if the sum of its digits 0 modulo 3
Related
0
Particular number is divisible by 11
2
Comparing two statements of Chinese Remainder Theorem (Sun-Ze Theorem)
0
Divisibility by 11, How I get this part?
1
Proof that number ending in $66$ or $06$ is congruent to $2 \text{ mod } 4$
2
Digits sums and sum of power
Hot Network Questions more hot questions

Question feed
Subscribe to RSS
Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader.


*

mmsanotherstage2019.comematics
Company
Stack Exchange Network
site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. rev2021.9.17.40233


mmsanotherstage2019.comematics Stack Exchange works best with JavaScript enabled
*

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.