2008 AMC 12B Problems/Problem 23
Contents
Problem
The sum of the base- logarithms of the divisors of is . What is ?
Solutions
Solution 1
Every factor of will be of the form . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property . For any factor , there will be another factor . Note this is not true if is a perfect square. When these are multiplied, they equal . so the number of factors divided by 2 times n equals the sum of all the factors, 792.
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 factors. . We then plug in answer choices and arrive at the 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
Solution 5
Let integer be the number of divisors has. Then, we set up pairs of divisors such that each pair satisfies -- ex. , etc. Then the sum of the base- logarithms is We can use the property that only perfect squares have an odd number of factors, as for perfect square , we have ordered pair that works. For even , then, can be multiplied by itself to get , so is odd. But, in our summation, does not exist for even as is then odd, so must be even. And, since = , We want to find a for our odd options such that = . For , works, and integer can not be found for other odd . So, we get
Alternative thinking
After arriving at the equation , notice that all of the answer choices are in the form , where is . We notice that the ones digit of is , and it is dependent on the ones digit of the answer choices. Trying for , we see that only yields a ones digit of , so our answer is .
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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