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Problem 439

AMC 12 late, AIME early

Number theory Difficulty 4.8 Find the answer

Task 1. On the board, in a row, the digits are written as follows:

$\begin{array}{lllllllll} 1 & 1 & 1 & 2 & 2 & 2 & 5 & 5 & 5 . \end{array}$

Between them, several plus signs can be placed so that the resulting sum ends in the digit one:

$1+1+12+2+25+5+5=51$

How can several plus signs be placed so that the resulting sum ends in zero? An example is sufficient.

Your answer

Official solution

Answer: $111+22+2+555=690$ .

Note. There are many other correct examples.

## Criteria

## 4 p. A correct example is provided.

(Examples with only one plus sign are also accepted.)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.