Track / Stage 6 / 154 of 400 #1154 of 2000

Problem 1154

National olympiad, first round

Number theory Difficulty 6.2 Prove it

$3+$ [ Divisibility rules for 3 and 9 ]

In the following multi-digit numbers, the digits have been replaced by letters (the same digits by the same letters, and different digits by different letters). It turned out that DEVYANOSTO is divisible by 90, and DEVYATKA is divisible by 9. Can SOTKA be divisible by 9?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Since ДЕВЯНОСТО is divisible by 90, then $\mathrm{O}=0$ , so the sum Д + Е + В + Я + (Н + С) + Т is divisible by 9. Since ДЕВЯТКА is divisible by 9, the sum Д + Е + В + Я + Т + (К + А) is also divisible by 9. Note that these two sums contain nine different letters, meaning all digits from 1 to 9 are used. The sum of all digits from 1 to 9 is 45, so Д + Е + В + Я + Т + (К + А) + (Н + С) = 45, which is also divisible by 9. Therefore, К + А is divisible by 9 and Н + С is divisible by 9.

Assume that СОТКА is divisible by 9, meaning the sum С + Т + К + А is divisible by 9. Since К + А is divisible by 9, then С + Т is divisible by 9. Then, considering that Н + С is divisible by 9, we get: (Н + С) - (С + Т) = Н - Т is divisible by 9. This is only possible if $\mathrm{H}=\mathrm{T}$ . This contradiction implies that the assumption is incorrect, meaning СОТКА is not divisible by 9.

## Answer

It cannot.

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