Difference between revisions of "2019 AIME I Problems/Problem 9"
m (AIME needs to be capitilized) |
(→Problem 9) |
||
| Line 2: | Line 2: | ||
==Problem 9== | ==Problem 9== | ||
| + | Let <math>\tau (n)</math> denote the number of positive integer divisors of <math>n</math>. Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>. | ||
| + | |||
==Solution== | ==Solution== | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=8|num-a=10}} | {{AIME box|year=2019|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:36, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 9
Let 
denote the number of positive integer divisors of 
. Find the sum of the six least positive integers that are solutions to 
.
Solution
See Also
| 2019 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
