Divisibility by 3

Number Theory Level 2

We know that the number 1234567 is not divisible by 3 because the sum of its digits is not divisible by 3. Is $1234567^{3}-1234567$ divisible by 3?

Yes No

3 solutions

Sam Bealing
May 16, 2016

$n^3-n=n(n^2-1)=n(n+1)(n-1)$

These are three consecutive numbers so at least one will be divisible by 3.

Nice work. Sam, even if you wrote something that does not make sense, I was going to agree with you because my son's name is Sam.

Hana Wehbi - 5 years, 1 month ago

Please respect other members. How come this does not make sense?

展豪張 - 5 years ago

I did not say it does not make sense. I said "even if". I was just trying to be nice to him because I have a son whose name is Sam, nothing more or less.

Hana Wehbi - 5 years ago

@Hana Wehbi – Sorry I didn't see the words 'even if'. But for me it's still weird to see something like this......

展豪張 - 5 years ago

Hana Wehbi
May 16, 2016

For any positive integer $n$ , $n^{3}-n$ is divisible by $3$ .

Mateo Matijasevick
May 18, 2016

Applying Fermat's Little Theorem, as $3$ is prime and $(3,1234567)=1$ , then ${ 1234567 }^{ 3 }\equiv 1234567(mod\quad 3)$

Nice, never thought of it like that.

Hana Wehbi - 5 years ago

Divisibility by 3

3 solutions

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