Prime Factorization 2

Directions: Using the digits 0 to 9 at most one time each, fill in the boxes to make the greatest possible product.

Hint

What digits would make good/bad exponents?
What digits would make good/bad bases?

Answer

The greatest product that’s been found so far is 80 = 5^1 * 2^4.

Source: Robert Kaplinsky

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Greatest Common Factor

Directions: Using the digits 0 to 9 at most one time each, place a digit …

42 comments

  1. i did _____________=9^8×7^6

  2. i did 96=2^5+3^1

  3. Danitza Marenco

    That´s exactly what i thought 96=2^5+3^1

  4. I think it would be 80 = 5^1*2^4

  5. 5.0644e12 =

  6. 96=2^5 x 3^1

  7. 80=5^1 * 2^4

  8. You did a terrific job! This post seem extremely great.

  9. I am not very sure but i am guessing that it would be 93= 2^5* 3^2 This is kinda confusing to me so i think i might be a bit off

  10. I did 9^2*3*2=

  11. I got 96 = 6^1 x 4^2

  12. I did 64=1^0*8^2 because anything to the power of one equals zero.

  13. 50 = 2^1 * 5^2
    96 = 2^5 * 3^1

  14. up i did 8⁹*6⁷ and got 37,572,373,905,408 i think i did something wrong oops

  15. I love that one box represents only one digit. This is also a great exercise in order of operations. Thank you! It was also great that I realized the larger digits had to be the product AND the product had to be less than 100.
    4^2 x 5^1 = 16 x 5 = 80
    but even larger
    2^5 x 3^1 = 32 x 3 = 96

  16. I got 96=2^5*3^1 which is 32×3

  17. 2^5*3^1 = 96

  18. we did addition in sted of multiplication so I got 4^3+2^5

  19. I don’t think anyone has 40 = 2^3 x 5^1 which uses all of the numbers from 0-5!

  20. Rudolf Österreicher

    Here are all the possibilities that make a true statement (except for switching the factors):

    48 = 2^3 * 6^1
    49 = 1^3 * 7^2
    49 = 1^5 * 7^2
    49 = 1^6 * 7^2
    49 = 1^8 * 7^2
    54 = 3^2 * 6^1
    56 = 2^3 * 7^1
    64 = 1^3 * 8^2
    64 = 1^5 * 8^2
    64 = 1^7 * 8^2
    64 = 1^9 * 8^2
    64 = 2^3 * 8^1
    96 = 2^5 * 3^1

    As you can see, 96 is the greatest.

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