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

### Hint

What types of numbers are easy to arrange into products?

### 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

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!

Good on them! But not sure they did find them all Mr Breadstick (see Julie’s comment below)

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

We found a mistake ( that is after I double checked. ugh!) change 2^3 x 2^7 to 2^1 x 2^9

Well done to your 8th grader Holly. I’ve added their solution to the answers! 🙂

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!

there are 102400 possible combonations

i think

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?

There are lots of possible solutions Amy. Yours is correct – well done!

Any chance we could change the word “digits” to numbers since 11-20 are not digits?

one solution I have is:

(2^2)^9 = {(2^6)^5}/{(2^3)^4} = (2^11) x (2^7) = (2^19)/(2^1)

I have this same problem, but the directions say I can only use digits 0-9. Can anyone come up with a solution?

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

I just spent more time than I should have finding solutions where all of the exponents simplify to 5, 6, 7, all the way to 18. I know I didn’t find all of the solutions, but I think those are all of the possible exponents. You can’t simplify to anything less than 5 because you would have to use two 1s, and you can’t simplify to 19 for the same reason. Thanks for a great problem