# Problem of the Week Problem A and Solution What Number Am I?

## Problem

I am a five-digit number made of the digits $$0$$, $$3$$, $$4$$, $$6$$, and $$9$$. In my number, the following are true.

• The digit $$4$$ is in a position whose place value is $$10$$ times the place value of the position of the digit $$9$$ and $$100$$ times the place value of the position of the digit $$0$$.

• The digit $$3$$ is in a position whose place value is $$100$$ times the place value of the position of the digit $$9$$.

• The digit $$0$$ in a position whose place value is $$10$$ times the place value of the position of the digit $$6$$.

What number am I?

## Solution

The positions in a five-digit number are: ones (units), tens, hundreds, thousands, and ten thousands.

Since the $$4$$ is in a position whose place value is $$10$$ and $$100$$ times the place value of other positions, the $$4$$ cannot be in the ones or tens position.

Since the $$3$$ is in a position whose place value is $$100$$ times the place value of another position, the $$3$$ cannot be in the ones or tens position.

Since the $$0$$ is in a position whose place value is $$10$$ times the place value of another position, the $$0$$ cannot be in the ones position.

So the ones position must contain either the $$9$$ or the $$6$$.

Letâ€™s assume that the $$9$$ is in the ones position. Based on the first clue, the $$4$$ must be in the tens position since $$10 \times 1 = 10$$. However, we have already stated that the $$4$$ cannot be in the ones or tens position. So we cannot put the $$9$$ in the ones position.

Therefore, we can conclude that the $$6$$ is in the ones position. Based on the third clue, the $$0$$ must be in the tens position since $$10 \times 1 = 10$$. Knowing that the $$0$$ is in the tens position, we can use the first clue to conclude that the $$4$$ must be in the thousands position, since $$100 \times 10 = 1000$$. Also, the $$9$$ must be in the hundreds position, since $$10 \times 100 = 1000$$. Now the only position that is left, the ten thousands position, must contain the only digit left which is the $$3$$. This is consistent with the second clue since $$100 \times 100 = 10\,000$$.

Therefore, the number must be $$34\,906$$.