2008 AMC 12B Problems/Problem 23
The sum of the base- logarithms of the divisors of is . What is ?
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
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.
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 .
The sum is Trying for answer choices we get
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
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 .
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