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For puzzles about forming numbers using other numbers and mathematical operations.

4
votes
$300,000$ is such a
answered Oct 5 '16 by Rosie F
2
votes
Some of these become possible with decimal point: $$\begin{matrix} 11=.2^{0-1}+6 \\ 30=.2^{0-1}*6 \\ 35=\dfrac{20+1}{.6} \\ 50=\dfrac{20}{1-.6} \end{matrix}$$
answered Sep 11 '16 by Rosie F
0
votes
In case anyone's still interested in the earlier target of $30,000$, I found an expression that seems more interesting than mine for $300,000$.
answered Oct 5 '16 by Rosie F
1
vote
Further to existing answers, 38 is also
answered Nov 4 '17 by Rosie F
74
votes
Here's an answer which
answered Oct 4 '16 by Rosie F
3
votes
$300,000$ can be reached using
answered Oct 5 '16 by Rosie F
0
votes
If you're allowed an increasing monadic operator (e.g. !) by which you can make arbitrarily large integers, a decreasing monadic operator (e.g. sqrt) by which you can turn an arbitrarily large number …
answered Oct 5 '17 by Rosie F
3
votes
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 t …
answered Jun 3 '16 by Rosie F
14
votes
If we use LESS to indicate subtraction, then we have -1 = ONE LESS ONE LESS ONE 0 = ONE LESS ONE 1 = ONE 2 = ONE PLUS ONE etc., using only seven letters ELNOPSU.
answered Sep 10 '16 by Rosie F
4
votes
If the "floor" function is acceptable:
answered Feb 14 '18 by Rosie F