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Answer: 0
$$\underbrace{\_\_\_\_\quad\_\_\_\_}_{1004}\quad \_\_\_\_\qquad \therefore \quad \underbrace{\_\_\_\_\quad\_\_\_\_}_{1004}\quad \underline{1005}\\
\quad\\
\_\_\_\_\quad \underbrace{\_\_\_\_\quad \underline{1005}}_{1005}\qquad \therefore \quad \underline{1004}\quad\underbrace{\underline{\ \ \ \ 0\ \ \ \ }\quad \underline{1005}}_{1005}\\
\quad\\
\qquad\qquad\qquad\qquad\qquad\qquad 1004\times0\times1005=0$$
Alternatively:
The sum of the first two numbers and the last two numbers is 1004 + 1005 = 2009.
This counts the middle number twice.
But the sum of all three numbers is 2009, so the middle number is 0.
Hence the product of all three numbers is 0.
Using algebra
Let the three numbers be a, b and c.
We have
In Sum-mary
Age 11 to 14
ShortChallenge Level 
Answer: 0
$$\underbrace{\_\_\_\_\quad\_\_\_\_}_{1004}\quad \_\_\_\_\qquad \therefore \quad \underbrace{\_\_\_\_\quad\_\_\_\_}_{1004}\quad \underline{1005}\\
\quad\\
\_\_\_\_\quad \underbrace{\_\_\_\_\quad \underline{1005}}_{1005}\qquad \therefore \quad \underline{1004}\quad\underbrace{\underline{\ \ \ \ 0\ \ \ \ }\quad \underline{1005}}_{1005}\\
\quad\\
\qquad\qquad\qquad\qquad\qquad\qquad 1004\times0\times1005=0$$
Alternatively:
The sum of the first two numbers and the last two numbers is 1004 + 1005 = 2009.
This counts the middle number twice.
But the sum of all three numbers is 2009, so the middle number is 0.
Hence the product of all three numbers is 0.
Using algebra
Let the three numbers be a, b and c.
We have
$a+b=1004$
$b+c=1005$
and $a+b+c=2009$
Adding the first two equations gives
$a+2b+c=2009$
and subtracting the third equationg from this gives
$b=0$
Thus the product $abc=0$.
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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