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 $\tau (n)$ denote the number of positive integer divisors of $n$. Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$.

Solution

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS