Proof for statement: It's impossible to find a pair of consecutive natural numbers whom digit sums would divide without reminder by 3

Question

I am searching for a mathematical proof of this statement: It's impossible to find a pair of consecutive natural numbers whom digit sums would divide without reminder by 3.

I have tried:

To make a system of linear equations where $\overline{xy} = 10x + y$

$x + y = 3k$ (A number should be divisible by 3)

$x + y + 1 = 3k + 3$ (A number that goes after it should be divisible by 3)

However, this proof is wrong, when considered two numbers $29$ and $30$ $x = 2$ $y = 9$ because $2 + 9 + 1 = 12$, however, $3 + 0 = 3$.

The pair $1,2$ has digit sum $1+2=3$. What exactly means "whom digit sums" ? — Dietrich Burde, May 01 '16 at 12:52
Well, assume by contradiction that there is such pair, then both elements in the pair are divisible by $3$, which is impossible. — barak manos, May 01 '16 at 12:52
@DietrichBurde I meant that 1 and 2 digits should be divisible by 3. — Zyberg, May 01 '16 at 12:59
@barakmanos I meant that the sum of each numbers digits should be divisible by 3. — Zyberg, May 01 '16 at 13:02
If the sum of digits is divisible by $3$, then so is the number. — barak manos, May 01 '16 at 13:11
@barakmanos That was quite a silly mistake of mine to overlook the divisibility rule. Thank you very much! :) — Zyberg, May 01 '16 at 13:22

score 1 · Accepted Answer · answered May 01 '16 at 13:23

The classic divisibility test for divisibility by $3$ is that a number is divisible by $3$ iff its sum of digits is divisible by $3$. Since two successive numbers cannot both be divisible by $3$, neither can their digit sums. In your example, though the digit sum of $30$ is not one more than the digit sum of $29$, it is so $\bmod 3$, because $2+9=11 \equiv 2 \pmod 3, 2+9+1=12 \equiv 0 \pmod 3$

Proof for statement: It's impossible to find a pair of consecutive natural numbers whom digit sums would divide without reminder by 3

1 Answers1