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 . 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.
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
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 4, and that it is dependent on the ones digit of the answer choices. Trying for , we see that only yields a ones digit of .
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