How to quickly tell if an "unknown" number is divisible by 3?

1. Every third number is divisible by three.
2. Every number divisible by three has a digit sum divisible by 3.
3. Every third number has a digit sum divisible by 3.
4. In between these, every third number has a digit sum congruent to 1 and then 2 mod 3.

Take a look:

n    digit sum mod 3
0    0
1    1
2    2
3    0
4    1
5    2
6    0
...
10   1
11   2
12   0
...
19   1
20   2
21   0
...

Say we have a string of digits constructed as you describe, and the number we just added was divisible mod 3. When we append the next number's digits, we are appending digits whose sum is congruent to 1 mod 3, and when added to those in our number, we will get a combined digit sum congruent to 1 mod 3, so our answer for the next one will be "no". The next one will add a number with digit sum congruent to 2 mod 3, and this causes the total to become congruent to 0 again, so the answer here is "yes". Finally, adding the next number which must be divisible by 3 keeps the digit sum congruent to 0.

The takeaway?

• if n is congruent to 0 modulo 3, then the answer is "yes"
• if n is congruent to 1 modulo 3, then the answer is "no"
• if n is congruent to 2 modulo 3, then the answer is "yes"

In particular, your example for n=15 is wrong; the digit string obtained represents a number that should be divisible by 3, and indeed it is (try it on a big enough calculator to verify).

All that is left is to find an implementation that is fast enough and handles all the required cases. If n is guaranteed to be under ~2 billion, then you are probably safe with something like

return (n % 3) != 1;

If n can be an arbitrarily large number, never fear; you can check whether the digit sum is congruent to 0 modulo 3 by adding up the digits in linear time. If not, you can add 1 from the number by coding addition like you do it by hand on paper and then check the result of that for divisibility by 3, again in linear time. So something like:

if (digit_sum_mod_3(n) == 0) return true;
else if (digit_sum_mod_3(add_one(n)) == 0) return false;
else return true;

Then you would have something like

digit_sum_mod_3(n[1...m])
    sum = 0
    for k = 1 to m do
        sum = sum + n[k]
        // keep sum from getting too big
        if sum >= 18 then
            sum = sum - 18
    return sum % 3

add_one(n[1...m])
    // work from right to left, assume big-endian
    for k = m to 1 do
        if n[k] < 9 then // don't need to carry
            n[k] = n[k] + 1
            break
        else then // need to carry
            n[k] = 0
    if n[1] = 0 then // carried all the way to the front
        n[1] = 1
        n[m+1] = 0
    return n

Any three consecutive numbers sum up to 0 == a + a + 1 + a + 2 mod 3. The answer reduces to k%3 == 0, or 2k-1 % 3 == 0. The latter is equivalent to k%3 == 2, which leaves out k%3==1 which then simplifies further to k%3 != 1.

It is a known trick in mathematics that a number is divisible by three if the sum of its individual decimal digits is divisible by three.

Example:

2271

2+2+7+1 = 12

12 is divisible by 3, therefore so is 2271

Additionally, the sum of any three consecutive integers must be divisible by three. This is because:

((n)+(n+1)+(n+2))/3 = (3n+3)/3 = n+1 = integer

Therefore:

If k mod 3 == 0, then concatenation of 1 to k is divisible by three.

If k mod 3 == 1, then concatenation of 1 to k is not divisible by three.

If k mod 3 == 2, then it is a bit trickier. In this case, concatenation of 1 to k is divisible by three if the sum of k and the number before k (which evaluates to (k)+(k-1), which is 2k-1) is divisible by three.

Therefore, the final condition is:

(k mod 3 == 0) || ((k mod 3 == 2) && (2k-1 mod 3 == 0))

However, this can be even further simplified.

It turns out that k mod 3 can only equal 2 whenever 2k-1 mod 3 equals 0 and vice versa.

See simple graph below that shows cyclic pattern of this behavior.

Therefore, the formula can be further simplified just to:

(k mod 3 == 0) || (k mod 3 == 2) 

Or, even more simply:

(k mod 3 != 1)

I realize answerer already provided this answer so I don't expect this to be the accepted answer, just giving a more thorough mathematical explanation.

A number is divisible by three if the sum of its digits is divisible by three (see here). Therefore, there is no need to "construct" your number, you need simply add the digits of the individual numbers. Thus for your 15 case, you do not need to "construct" 123456789101112131415, you just need to sum all of the digits in [1, 2, 3, 4, ... 14, 15].

This is simpler than it sounds because the problem only needs to check numbers of a very specific format: 12345789101112131415…k. You can use Gauss's method to quickly get the sum of the numbers 1 to k and then check if that sum is divisible by three using the usual methods. The code for that is:

'NO' if (k*(k+1)/2)%3 else 'YES'

If you look at the pattern that occurs as k increases (NO, YES, YES, NO, YES, YES, ...), you don't even need the multiplication or division. In short, all you need is:

'YES' if (k-1)%3 else 'NO'

Here is Python code which reads integers from a file and, if it wouldn't take too long also checks the answer the hard way so you can see that it is right. (Python numbers can be infinitely long, so you don't need to worry about overflow):

#!/usr/bin/python3
# Read integers from stdin, convert each int to a triangular number
# and output YES (or NO) if it is divisible by 3.

def sumgauss(x):
    '''Return the sum from 1 to x using Gauss's shortcut'''
    return (x*(x+1)/2)

def triangle(n):
    '''Given an integer n, return a string with all the integers 
       from 1 to n concatenated. E.g., 15 -> 123456789101112131415'''
    result=""
    for t in range(1, k+1):
        result+=str(t)
    return result

import sys
for k in sys.stdin.readlines():
    k=int(k)
    print ( 'YES' if (k-1)%3 else 'NO', end='')

    # If it wouldn't take too long, double check by trying it the hard way
    if k<100000:
        kstr=triangle(k)
        print("\t// %s modulo 3 is %d" % (kstr, int(kstr)%3))
    else:
        print('\t// 123456789101112131415...%d%d%d modulo 3 is %d' %
              tuple([k-2, k-1, k, sumgauss(k)%3]))

Speaking of Gauss's shortcut for summation, this problem seems a lot like a homework assignment. (Gauss invented it as a student when a teacher was trying to get the class out of his hair for a while by making them add up the numbers from 1 to 100.) If this is indeed a class assignment, please make sure the teacher knows to give the A to me and stackoverflow. Thanks!

Sample output:

$ cat data
2
6
15
130000000
130000001

$ ./k3.py < data
YES // 12 modulo 3 is 0
YES // 123456 modulo 3 is 0
YES // 123456789101112131415 modulo 3 is 0
NO  // 123456789101112131415...129999998129999999130000000 modulo 3 is 1
YES // 123456789101112131415...129999999130000000130000001 modulo 3 is 0

The first 32 triangular numbers:

$ seq 32 | ./k3.py
NO  // 1 modulo 3 is 1
YES // 12 modulo 3 is 0
YES // 123 modulo 3 is 0
NO  // 1234 modulo 3 is 1
YES // 12345 modulo 3 is 0
YES // 123456 modulo 3 is 0
NO  // 1234567 modulo 3 is 1
YES // 12345678 modulo 3 is 0
YES // 123456789 modulo 3 is 0
NO  // 12345678910 modulo 3 is 1
YES // 1234567891011 modulo 3 is 0
YES // 123456789101112 modulo 3 is 0
NO  // 12345678910111213 modulo 3 is 1
YES // 1234567891011121314 modulo 3 is 0
YES // 123456789101112131415 modulo 3 is 0
NO  // 12345678910111213141516 modulo 3 is 1
YES // 1234567891011121314151617 modulo 3 is 0
YES // 123456789101112131415161718 modulo 3 is 0
NO  // 12345678910111213141516171819 modulo 3 is 1
YES // 1234567891011121314151617181920 modulo 3 is 0
YES // 123456789101112131415161718192021 modulo 3 is 0
NO  // 12345678910111213141516171819202122 modulo 3 is 1
YES // 1234567891011121314151617181920212223 modulo 3 is 0
YES // 123456789101112131415161718192021222324 modulo 3 is 0
NO  // 12345678910111213141516171819202122232425 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526 modulo 3 is 0
YES // 123456789101112131415161718192021222324252627 modulo 3 is 0
NO  // 12345678910111213141516171819202122232425262728 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526272829 modulo 3 is 0
YES // 123456789101112131415161718192021222324252627282930 modulo 3 is 0
NO  // 12345678910111213141516171819202122232425262728293031 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526272829303132 modulo 3 is 0 

Actually the answer is pretty straight forward, if the sum of the digits divisible by three then the number is also divisible by 3.

string ans=(((1 + num) * num) / 2) % 3 == 0 ? "YES" : "NO";

according to the problem sum of digit can be considered as sum of numbers from 1 to n, sum=(n*(n+1))/2 *Make sure you divide the whole thing by 2

Another approach: string ans=n % 3 !=1 ? "YES" : "NO";

You can prove that if n or n-2 is divisible by 3, then the sum up to n is divisible by 3 (e.g., in your case sum(1...8), sum(1..9), sum(1..11), etc.).