March 2012
Problem of the Month

Divide and Conquer

What is the smallest number that is divisible evenly by all of the digits 1 through 9?

Solution to the Problem:

The answer is 2520.

To be divisible by 2, you need a 2 in the answer.
To be divisible by 3, you need a 3.
To be divisible by 4, you need an additional 2 (You don't need a 4 because there is already a 2 earlier).
To be divisible by 5, you need a 5.
To be divisible by 6, you don't need any additional numbers (there is already a 2 and a 3).
To be divisible by 7, you need a 7.
To be divisible by 8, you need an additional 2.
To be divisible by 9, you need an additional 3.
Now multiply 2 x 3 x 2 x 5 x 7 x 2 x 3 = 2,520.

However, delahoydep@kpmath.com and Dakota Rees sent in the answer of 0, and 0 is divisible by each of the numbers 2, 3, 4, ..., 9!!!

Correctly solved by:

1. Chad Fore	Gate City, Virginia
2. Alp Aribal	Istanbul, Turkey
3. Mariah Stoddard	Mountain View High School, Mountain View, Wyoming
4. James Alarie	Flint, Michigan
5. delahoydep@kpmath.com	Mountain View High School, Mountain View, Wyoming
6. Blace Martin	Mountain View High School, Mountain View, Wyoming
7. Dakota Rees	Mountain View High School, Mountain View, Wyoming
8. Chelsea Anglen	Mountain View High School, Mountain View, Wyoming
9. McKinna Salsbury	Mountain View High School, Mountain View, Wyoming
10. Halee Salsbury	Mountain View High School, Mountain View, Wyoming
11. Cami Micheli	Mountain View High School, Mountain View, Wyoming
12. Felicia Lopez	Mountain View High School, Mountain View, Wyoming
13. Tyler Cantrell	Mountain View High School, Mountain View, Wyoming
14. Karsten Hauf	Mountain View High School, Mountain View, Wyoming
15. Lisa Merwyn	Taylor, Michigan

Send any comments or questions to: David Pleacher