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You must be referring to the quick way to determine whether a given integer is divisible by 3; that is, 3 divides an integer [tex]x[/tex] whenever the digits of [tex]x[/tex] sum to a multiple of 3.

Suppose [tex]x[/tex] has [tex]n\ge1[/tex] digits. We can expand it as a sum of the numbers in any given digits place by the corresponding power of 10. For example, if [tex]x=2148[/tex], we can write

[tex]2148=2\times10^3+1\times10^2+4\times10^1+8\times10^0[/tex]

More generally, if

[tex]x=d_{n-1}d_{n-2}\ldots d_1d_0[/tex]

(where [tex]d_i[/tex] denotes the numeral in the [tex]10^i[/tex]-th's place), then we have the expansion

[tex]x=d_{n-1}10^{n-1}+d_{n-2}10^{n-2}+\cdots+d_110^1+d_0[/tex]

Notice that for any integer [tex]k[/tex], we have

[tex]10^k-1=\underbrace{99\ldots99}_{k\text{ 9s}}[/tex]

which is clearly divisible by 3. So from each power of 10 in the expansion of [tex]x[/tex], we can add and subtract 1, then rearrange the terms of the sum:

[tex]x=d_{n-1}10^{n-1}+\cdots+d_110^1+d_0[/tex]
[tex]x=d_{n-1}(10^{n-1}-1+1)+\cdots+d_1(10^1-1+1)+d_0[/tex]
[tex]x=d_{n-1}(10^{n-1}-1)+\cdots+d_1(10^1-1)+(d_{n-1}+\cdots+d_1+d_0)[/tex]

We know [tex]10^k-1[/tex] is divisible by 3, which means the remainder upon dividing [tex]x[/tex] by 3 is just the sum of the digits of [tex]x[/tex]. If this remainder is divisible by 3, then so must be the original number, [tex]x[/tex].

Back to our previous example: if [tex]x=2148[/tex], then we have the expansion

[tex]2148=2\times10^3+10^2+4\times10+8[/tex]
[tex]2148=2(999+1)+(99+1)+4(9+1)+8[/tex]
[tex]2148=2\times999+99+4\times9+(2+1+4+8)[/tex]

Dividing through by 3, we get a remainder of [tex]2+1+4+8=15[/tex], which is divisible by 3, and so 2148 must also be a multiple of 3.

In case for some reason you're not convinced 15 is a multiple of 3, you can apply the same trick:

[tex]15=10+5[/tex]
[tex]15=9+1+5[/tex]

Dividing through by 3 leaves a remainder of [tex]1+5=6[/tex], which is also a multiple of 3, so that 15 must be, too.

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