2020 CIME II Problems/Problem 5

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 $182-17=\boxed{165}$ .

2020 CIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2020 CIME II Problems/Problem 5

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