Art of Problem Solving

2006 UNCO Math Contest II Problems/Problem 6: Difference between revisions

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==Solution==
==Solution==
{{solution}}
(a) <math>\frac{(2^{n+1}-1)(3^{n+1}-1)}{2}</math>
 
(b) <math>\frac{(2^{2n+1}-1)(3^{n+1}-1)}{2}</math>


==See Also==
==See Also==

Latest revision as of 01:33, 13 January 2019

Problem

The sum of all of the positive integer divisors of $6^2=36$ is $1+2+3+4+6+9+12+18+36=91$

(a) Determine a nice closed formula (i.e. without dots or the summation symbol) for the sum of all positive divisors of $6^n$.

(b) Repeat for $12^n$.

(c) Generalize.

Solution

(a) $\frac{(2^{n+1}-1)(3^{n+1}-1)}{2}$

(b) $\frac{(2^{2n+1}-1)(3^{n+1}-1)}{2}$

See Also

2006 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions