Maths Olympiad Prep

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, 2022

Number theory Difficulty 3.5 AMC 10/12 Find the answer Canada

A positive integer aa is
input into a machine. If aa is odd,
the output is a+3a+3. If aa is even, the output is a+5a+5. This process can be repeated using
each successive output as the next input. For example, if the input is
a=1a=1 and the machine is used three
times, the final output is 12. If the input is a=15a=15 and the machine is used 51 times,
the final output is

Pick one

Solution

If aa is odd, the output is
a+3a+3, which is even because it is
the sum of two odd integers.

If aa is even, the output is a+5a+5, which is odd, because it is the sum
of an even integer and an odd integer.

Starting with a=15a=15 and using the
machine 2 times, we obtain $15 \to 15 + 3 =
18 \to 18 + 5 = 23$.

Starting with 23 and using the machine 2 times, we obtain 2323+3=2626+5=3123 \to 23 + 3 = 26 \to 26 + 5 = 31.

Starting with an odd integer and using the machine 2 times, the net
result is adding 8 to the input, because the odd input generates a first
output that is 3 larger (and so even) and a second output that is 5
larger than the first output.

This generates a net result that is 3+53+5 larger than the input.

Therefore, using the machine 46 more times (that is, repeating the 2
steps a total of 23 more times), we add 8 a total of 23 more times to
obtain the output $31 + 23 \cdot 8 =
215$.

To this point, the machine has been used 50 times.

Using the machine for the 51st time, $215 \to
215 + 3 = 218$ and so the final output is 218.

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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.