Maths Olympiad Prep

Library / CEMC Fermat /349 of 566

CEMC Fermat, 2026

Number theory Difficulty 2.4 Junior Find the answer Canada

How many positive integers less than 100100 can be written as the sum of three
consecutive positive integers?

Pick one

Solution

The smallest positive integer less than 100100 that can be written as the sum of
three consecutive integers is 1+2+3=61+2+3=6.

The second smallest positive integer less than 100100 that can be written as the sum of
three consecutive integers is 2+3+4=92+3+4=9, which is an increase of 33 from the previous sum.

In general, for positive integers aa, the sum of three consecutive positive
integers is

a+(a+1)+(a+2)=3a+3a+(a+1)+(a+2)=3a+3.

The next largest sum of three consecutive positive integers is (a+1)+(a+2)+(a+3)=3a+6(a+1)+(a+2)+(a+3)=3a+6 which is (3a+6)−(3a+3)=3(3a+6)-(3a+3)=3 more than the previous
sum.

That is, successive integers in the list of such sums will continue to
increase by 33.

The largest positive integer less than 100100 that can be written as the sum of
three consecutive integers is 32+33+34=9932+33+34=99. (We note that 33+34+35=102 gt;100\text{33+34+35=102 gt;100}.)

Thus the positive integers less than 100100 that can be written as the sum of
three consecutive positive integers are of the form 3a+33a+3 for positive integers 1≤a≤321\leq a\leq32, and so there are 3232 such numbers.

Want a route through all this instead of an archive? The track puts 2,604 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.