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

Latest revision as of 01:33, 13 January 2019

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.

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

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

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 ==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==

2006 UNCO Math Contest II (Problems • Answer Key • Resources)
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