2020 CIME II Problems/Problem 5

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Problem

A positive integer $n$ is said to be $k$-consecutive if it can be written as the sum of $k$ consecutive positive integers. Find the number of positive integers less than $1000$ that are either $9$-consecutive or $11$-consecutive (or both), but not $10$-consecutive.

Solution

The smallest $9$-consecutive positive integer is $1+2+3+4+5+6+7+8+9=45$, and every multiple of $9$ greater than $45$ is also $9$-consecutive, with the last one less than $1000$ being $999$. There are $\frac{999-45}{9}+1=107$ $9$-consecutive positive integers less than $1000$. The smallest $11$-consecutive positive integer is $1+2+3+4+5+6+7+8+9+10+11=66$, and the largest one less than $1000$ is $990$. There are $\frac{990-66}{11}+1=85$ of these. However, we counted $99,198,297,396,495,594,693,792,891,990$ twice and we only wanted to count them once, so we subtract $10$ from our total, giving us a total of $107+85-10=182$ that are either $9$- or $11$-consecutive. The ones that are also $10$-consecutive are \[135,165,225,275,315,385,405,495,585,605,675,715,765,825,855,935,945\] for a total of $17$ integers to be removed. The answer is $182-17=\boxed{165}$.

See also

2020 CIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

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