Library / /86 of 213

, 2026

Number theory Difficulty 1.4 Junior Find the answer Canada

Fatima writes the list of positive integers in order, $1, 2, 3, 4, \ldots$ and so on. The 15th
positive even integer in the list is subtracted from the 25th
positive odd integer in the list. The result is

Pick one

Solution

The first three even integers in the list are $2=2\times1$ , $4=2\times 2$ , and $6=2\times3$ .

Each even integer in the list is of the form $2\times n$ where $n$ is a positive integer.

Thus, the 15th even integer is $\$2\times$
15=30$.

The first three odd integers in the list are each one less than the
first three even integers in the list, respectively. These are $1=2\times1-1$ , $3=2\times 2-1$ , and $5=2\times3-1$ .

Each odd integer in the list is of the form $2\times n-1$ where $n$ is a positive integer.

Thus, the 25th odd integer is $\$2\times$
25-1=49$.

The result of subtracting the 15th even integer from the 25th odd
integer is $49-30=19$ .

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

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