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?

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

Source: Robert Kaplinsky

Check Also

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

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