New divisible rule?

Number Theory Level 3

True or false?

If a natural number is divisible by $99$ , then the sum of its digits must be divisible by $18$ .

For example: For $1485=15*99$ the statement is true, because $1+4+8+5=18$ is divisible by $18$ .

True False

2 solutions

H K
Jul 1, 2017

It is quite normal to conjecture that this is false, and with some coding, the first counterexample turns out to be 10989 = 99*111.

Marta Reece
Jul 1, 2017

To actually get divisibility by 99, you need to get the sum of odd digits, say $D$ , and sum of even digits, say (E). Then $D+E$ needs to be divisible by 9, and $D-E$ by 11.

The first counterexample could is obtained by setting $D+E=27$ (so that it would not be 18, but divisible by 9) and $D-E=11$ (the lowest number divisible by 11. Zero will not work, since it's not odd)

This gives us $D=19$ and $E=8$ .

The lowest number with these properties is one with $1, 9, 9$ in odd places, and $0, 8$ in even ones.

The counterexample is $10989$ .

But many others can be made using the same process, for example $53559$ .

New divisible rule?

2 solutions

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