Difference between revisions of "2008 AMC 12B Problems/Problem 23"
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| − | Every factor of <math>10^n</math> will be of the form <math>2^a * 5^b , a\leq n , b\leq n</math>. Logarithmically, addition and multiplication are interchangeable (i.e. <math>log(a*b) = log(a)+log(b)</math>), so we need only count the number of 2's and 5's occurring in total. For every factor <math>2^a * 5^b</math>, there will be another <math>2^b * 5^a</math>, 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 <math>log(2)+log(5) = log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>. | + | Every factor of <math>10^n</math> will be of the form <math>2^a * 5^b , a\leq n , b\leq n</math>. Logarithmically, addition and multiplication are interchangeable (i.e. <math>\log(a*b) = \log(a)+\log(b)</math>), so we need only count the number of 2's and 5's occurring in total. For every factor <math>2^a * 5^b</math>, there will be another <math>2^b * 5^a</math>, 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 <math>\log(2)+\log(5) = \log(10) = 1</math>, the final sum will be the total number of 2's occurring in all factors of <math>10^n</math>. |
There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer. | There are <math>n+1</math> choices for the exponent of 5 in each factor, and for each of those choices, there are <math>n+1</math> factors (each corresponding to a different exponent of 2), yielding <math>0+1+2+3...+n = \frac{n(n+1)}{2}</math> total 2's. The total number of 2's is therefore <math>\frac{n*(n+1)^2}{2} = \frac{n^3+2n^2+n}{2}</math>. Plugging in our answer choices into this formula yields 11 (answer choice A) as the correct answer. | ||
Revision as of 12:42, 1 December 2008
Problem 23
The sum of the base-
logarithms of the divisors of 
is 
. What is 
?
Solution
Every factor of will be of the form 
. Logarithmically, addition and multiplication are interchangeable (i.e. 
), so we need only count the number of 2's and 5's occurring in total. 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 A) as the correct answer.
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 | |
