Copyright © University of Cambridge. All rights reserved.

'Nine in a Line' printed from https://nrich.maths.org/

Show menu


Answer: $227$


Algebra from the first number
Let the first number be $n$, so the nine numbers are:
$n$, $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, $n+6$, $n+7$, $n+8$.

We are told that they all add up to $2007$, so
$n+n$$+$$1+n$$+$$2+n$$+$$3+n$$+$$4+n$$+$$5+n$$+$$6+n$$+$$7+n$$+$$8=2007$,
$9n+36=2007$,
$9n=1971$,
$n=219$.

If the smallest number is $219$, the largest will be $219+8=227$.


Algebra from the middle number
Let the middle number be $n$, so the nine numbers are:
$n-4$, $n-3$, $n-2$, $n-1$, $n$, $n+1$, $n+2$, $n+3$, $n+4$.

Summing these gives $9n=2007$ so $n=223$.

Therefore the largest number, $n+4$, is $227$.


Averages
The average of the nine numbers is $2007\div9=223$, so they are $219$, $220$,$...$, $226$, $227$.

So the last number is $227$.


 

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

 

 

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.