Question: [Countdown](http://en.wikipedia.org/wiki/Countdown_%28game_show%29) is a British game show in which people try and make the longest possible word out of nine given letters, reach a target number using six given numbers and basic arithmetic, and then solve a nine-letter anagram, each within 30 seconds.

In the numbers round, the players choose numbers from groups of "large" and "small" numbers. The large numbers contain one instance each of $25$, $50$, $75$, and $100$, while the small numbers contain two instances of each number from $1$ to $10$. They then use addition, subtraction, multiplication, division and bracketing with these numbers to try and reach a target number which can be anywhere between $100$ and $999$ inclusive.

[This video](https://www.youtube.com/watch?v=pfa3MHLLSWI) is famous for involving numbers as large as $23850$ in the solution process, where the starting numbers are $3,6,25,50,75,100$ and the target is: $$952 = \dfrac{(100+6)\cdot 3\cdot 75-50}{25} = \dfrac{23850-50}{25}$$
But is there a set of given numbers and a target number that requires even larger numbers as part of the solution? In particular, what set of numbers requires the largest intermediate number in the solution process?

Note that ideally the solution should be the **only** solution.

Answer: 
Concerning the question "is there a set of given numbers and a target number that requires even larger numbers as part of the solution?"

Yes, there is. Here we get $62550$ as an intermediate result:

$834 = \frac{25\cdot 5\cdot 5\cdot 100+50}{75} = \frac{62550}{75}$

According to *Lopsy* there is no other possible way to write $834$ using these numbers (see the first comment).
With numbers 3, 3, 25, 50, 75 & 100,

$ 996 = \dfrac{((50+3)\*25+3)\*75}{100} $

which has an intermediate result of $99600$. The technique of tweaking

$ \dfrac{25\*50\*75}{100} = 937 \dfrac12 $

using the small numbers needs a total divisible by 3. But $999$ is liable to be reachable as $1000-1$.