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Laws of Exponents

Directions: Using the digits 1 to 20, at most one time each, fill in the boxes to create equivalent expressions.

Hint

Hint

What types of numbers are easy to arrange into products?

Answer

Answer

There are many possible solutions. For example,

(2^3)^4 = ((2^5)^12)/((2^6)^8) = 2^10 x 2^2 = (2^19)/(2^7)

(2^2)^5 = ((2^6)^11)/((2^7)^8) = 2^1 x 2^9 = (2^20)/(2^10)

More solutions are given in the comments

Source: Shaun Errichiello

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9 comments

  1. This problem was fantastic! My awesomely skilled sixth graders found seven more solutions to this problem! I wonder if we found them all? Thank you!

  2. I see that some students have found more solutions. I thought I might share a solution found by one of my 8th graders.

    (2^2)^5 = (2^6)^11/(2^7)^8 = 2^3 x 2^7 = 2^20/2^10

  3. I think I found a ton that all equal 32. First I rewrote it:

    (2^a)^b = (2^c)/(2^d) = 2^g * 2^h = 2^j / 2^k

    Then {a,b,c,d,e,f,g,h,j,k} can be

    1,5,4,17,7,9,2,3,11,6
    1,5,4,17,7,9,2,3,13,8
    1,5,4,17,7,9,2,3,15,10
    1,5,4,17,7,9,2,3,16,11
    1,5,4,17,7,9,2,3,18,13
    1,5,4,17,7,9,2,3,19,14
    1,5,4,17,7,9,2,3,20,15
    1,5,7,11,4,18,2,3,15,10
    1,5,7,11,4,18,2,3,17,12
    1,5,7,11,4,18,2,3,19,14
    1,5,7,11,4,18,2,3,20,15

    or any permutation of any of these that switches the numbers in the a,b and/or c,d and/or e,f and/or g,h pairs.

    I suspect there are a lot of other solutions too!

  4. I got a different answer than the one provided. I got:
    (2^2)^3 = (2^15)^6/(2^7)^12= 2^1 x 2^5 = 2^10/2^4

    Did anyone else get this answer? Is this correct?

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