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Answer: $201$
Sum of 2 consecutive numbers is $n + (n+1)=2n+1$
Sum of 5 consecutive numbers is $m + (m+1)+(m+2)+(m+3)+(m+4) = 5m+10$
($m$ and $n$ different numbers because the sums have different starting points)
$2n+1=5m+10$
Can find $n$ from any $m$, as long as $5m+10$ is odd
$5m+10$ is odd whenever $m$ is odd
And $5m+10 \lt 2017$
$\therefore 5m \lt 2007\\
\therefore m \le 401$
There are $201$ odd numbers up to $401$
Doubly Consecutive Sums
Age 11 to 14
ShortChallenge Level 
Answer: $201$
Sum of 2 consecutive numbers is $n + (n+1)=2n+1$
Sum of 5 consecutive numbers is $m + (m+1)+(m+2)+(m+3)+(m+4) = 5m+10$
($m$ and $n$ different numbers because the sums have different starting points)
$2n+1=5m+10$
Can find $n$ from any $m$, as long as $5m+10$ is odd
$5m+10$ is odd whenever $m$ is odd
And $5m+10 \lt 2017$
$\therefore 5m \lt 2007\\
\therefore m \le 401$
There are $201$ odd numbers up to $401$
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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