Difference between revisions of "2008 AMC 12B Problems/Problem 23"
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<cmath>= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5) </cmath> | <cmath>= \frac{n(n+1)^2}{2} \log(2) + \frac{n(n+1)^2}{2} \log(5) </cmath> | ||
<cmath> = \frac{n(n+1)^2}{2} </cmath> | <cmath> = \frac{n(n+1)^2}{2} </cmath> | ||
− | + | Trying for answer choices we get <math>n=11</math> | |
==See Also== | ==See Also== |
Revision as of 20:50, 15 September 2019
Problem 23
The sum of the base- logarithms of the divisors of is . What is ?
Contents
Solutions
Solution 1
Every factor of will be of the form . Using the logarithmic property , it suffices to count the total number of 2's and 5's running through all possible . For every factor , there will be another , so it suffices to count the total number of 2's occurring in all factors (because of this symmetry, the number of 5's will be equal). And since , the final sum will be the total number of 2's occurring in all factors of .
There are choices for the exponent of 5 in each factor, and for each of those choices, there are factors (each corresponding to a different exponent of 2), yielding total 2's. The total number of 2's is therefore . Plugging in our answer choices into this formula yields 11 (answer choice ) as the correct answer.
Solution 2
We are given The property now gives The product of the divisors is (from elementary number theory) where is the number of divisors. Note that , so . Substituting these values with in our equation above, we get , from whence we immediately obtain as the correct answer.
Solution 3
For every divisor of , , we have . There are divisors of that are . After casework on the parity of , we find that the answer is given by .
Solution 4
The sum is Trying for answer choices we get
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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