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CEMC Gauss (Grade 7), 2016

Number theory Difficulty 2.4 Junior Find the answer Canada

In the sum shown, PP and QQ each represent a digit.

Figure 0

The value of P+QP+Q is

Pick one

Solution

The sum of the units column is Q+Q+Q=3QQ+Q+Q=3Q.

Since QQ is a single digit, and 3Q3Q ends in a 6, then the only possibility is Q=2Q=2.

Then 3Q=3×2=63Q=3\times2=6, and thus there is no carry over to the tens column.

The sum of the tens column becomes 2+P+2=P+42+P+2=P+4, since Q=2Q=2.

Since PP is a single digit, and P+4P+4 ends in a 7, then the only possibility is P=3P=3.

Then P+4=3+4=7P+4=3+4=7, and thus there is no carry over to the hundreds column.

We may verify that the sum of the hundreds column is 3+3+2=83+3+2=8, since P=3P=3 and Q=2Q=2.

The value of P+QP+Q is 3+2=53+2=5, and the final sum is shown.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.