Results 1 to 2 of 2

Math Help - new puzzle

  1. #1
    Junior Member samsum's Avatar
    Joined
    Sep 2006
    Posts
    25

    new puzzle

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    768
    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}<br />
37 & = & (2\cdot3)^2 + \frac{3}{3} & \quad & 53 & = & 2\cdot3^3-3+2 \\<br />
38 & = & (3+3)^2 + 2 & \quad & 54 &= & 2\cdot3^3\\<br />
39 & = & (2\cdot3)^2 + 3 & \quad & 55 &= &2\cdot3^3 + 3 - 2 \\<br />
40 &= &(2+3)(2+3+3) & \quad &56 &= &2\cdot3^2 + 2 \\<br />
41 & = & (3+3)^2+3+2 & \quad &57 &= &2\cdot3^3 + 3 \\<br />
\end{array}
    \begin{array}{ccccccc}<br />
42 & = & (2\cdot3)^2\!+\!3\!+\!3 &\qquad\; &58 &= &2(3^3+2) \\<br />
43 &= &3^3 + 2\cdot2^3 &\quad &59 &= &2\cdot3^3 + 2 + 3 \\<br />
44 & = & (3+3)^2 + 2^3 &\quad &60 &= &2\cdot3^3 + 2\cdot3 \\<br />
45 & = & (2\cdot3)^2 + 3\cdot3 &\quad &61 &= &(2^3)^2 - 3 \\<br />
46 & = & (3\cdot3-2)^2-3 &\quad &62 &= &2\cdot3^3 + 2^3 \\<br />
\end{array}
    \begin{array}{ccccccc}47 &= &3\cdot3(2 + 3) + 2 &\qquad\; &63 &= &(2^3)^2-\frac{3}{3} \\<br />
48 & = & 2^3(3+3) &\quad &64 &= &(2^3)^2\\<br />
49 & = & (3\cdot3-2)^2 &\quad &65 &= &(2^3)^2 + \frac{3}{3} \\<br />
50 & = & 2^3(3+3) + 2 &\quad &66 &= &2(3^3+2\cdot3) \\<br />
51 & = & 3(2^3+3^2) &\quad &67 &= &(2^3)^2 + 3\\<br />
52 & = & (2+3)^2+3^3 & & & &\end{array}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A new puzzle?
    Posted in the Math Puzzles Forum
    Replies: 1
    Last Post: October 23rd 2010, 01:29 PM
  2. Set Puzzle
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 6th 2008, 08:46 AM
  3. Puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: December 22nd 2006, 04:38 PM
  4. A puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: December 9th 2006, 08:40 PM
  5. more puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 15
    Last Post: October 11th 2006, 09:26 PM

Search Tags


/mathhelpforum @mathhelpforum