Difference between revisions of "2019 AMC 12B Problems/Problem 3"
(→Problem) |
(→Solution) |
||
| Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
| + | n=19 | ||
| + | sum is 10 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}} | {{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:48, 14 February 2019
Problem
If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?
Solution
n=19 sum is 10
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
