new puzzle

**samsum** · October 8th 2007, 07:42 PM

You are equipped with two 2's, three 3's, and the ability to combine them using addition, subtraction, multiplication, division, and exponentiation.

Create all of the integers from 37 to 67 (inclusive).

You may use any number of parentheses to control the order of operations and need not use all five numbers each time.

For example: 11 = 2^3 + 3
where 2^3 means 2 to the 3rd power

**Soroban** · October 10th 2007, 04:10 PM

Hello,samsum!

This wasn't too bad . . .

You are equipped with two 2's, three 3's, and the ability to combine them
using addition, subtraction, multiplication, division, and exponentiation.

Create all of the integers from 37 to 67 (inclusive).

You may use any number of parentheses to control the order of operations
and need not use all five numbers each time.

For example: . $11 \:= \:2^3 + 3$

$\begin{array}{ccccccc} 37 & = & (2\cdot3)^2 + \frac{3}{3} & \quad & 53 & = & 2\cdot3^3-3+2 \\ 38 & = & (3+3)^2 + 2 & \quad & 54 &= & 2\cdot3^3\\ 39 & = & (2\cdot3)^2 + 3 & \quad & 55 &= &2\cdot3^3 + 3 - 2 \\ 40 &= &(2+3)(2+3+3) & \quad &56 &= &2\cdot3^2 + 2 \\ 41 & = & (3+3)^2+3+2 & \quad &57 &= &2\cdot3^3 + 3 \\ \end{array}$
$\begin{array}{ccccccc} 42 & = & (2\cdot3)^2\!+\!3\!+\!3 &\qquad\; &58 &= &2(3^3+2) \\ 43 &= &3^3 + 2\cdot2^3 &\quad &59 &= &2\cdot3^3 + 2 + 3 \\ 44 & = & (3+3)^2 + 2^3 &\quad &60 &= &2\cdot3^3 + 2\cdot3 \\ 45 & = & (2\cdot3)^2 + 3\cdot3 &\quad &61 &= &(2^3)^2 - 3 \\ 46 & = & (3\cdot3-2)^2-3 &\quad &62 &= &2\cdot3^3 + 2^3 \\ \end{array}$
$\begin{array}{ccccccc}47 &= &3\cdot3(2 + 3) + 2 &\qquad\; &63 &= &(2^3)^2-\frac{3}{3} \\ 48 & = & 2^3(3+3) &\quad &64 &= &(2^3)^2\\ 49 & = & (3\cdot3-2)^2 &\quad &65 &= &(2^3)^2 + \frac{3}{3} \\ 50 & = & 2^3(3+3) + 2 &\quad &66 &= &2(3^3+2\cdot3) \\ 51 & = & 3(2^3+3^2) &\quad &67 &= &(2^3)^2 + 3\\ 52 & = & (2+3)^2+3^3 & & & &\end{array}$

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