Difference between revisions of "2008 AMC 12B Problems/Problem 23"
(→Solution 3) |
m (→Solution 1) |
||
(18 intermediate revisions by 7 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | == Problem == |
The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>? | The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>? | ||
<math>\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math> | <math>\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15</math> | ||
− | + | == Solutions == | |
− | ==Solutions== | ||
=== Solution 1 === | === Solution 1 === | ||
− | Every factor of <math>10^n</math> will be of the form <math>2^a \times 5^b , a\leq n , b\leq n</math>. | + | Every factor of <math>10^n</math> will be of the form <math>2^a \times 5^b , a\leq n , b\leq n</math>. 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 <math>\log(a \times b) = \log(a)+\log(b)</math>. For any factor <math>2^a \times 5^b</math>, there will be another factor <math>2^{n-a} \times 5^{n-b}</math>. Note this is not true if <math>10^n</math> is a perfect square. When these are added, they equal <math>2^{a+n-a} \times 5^{b+n-b} = 10^n</math>. <math>\log 10^n=n</math> so the number of factors divided by 2 times n equals the sum of all the factors, 792. |
− | + | ||
− | 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> | + | 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>(n+1)^2</math> total factors. <math>\frac{(n+1)^2}{2}*n = 792</math>. We then plug in answer choices and arrive at the answer <math>\boxed {11}</math> |
− | |||
− | |||
− | |||
+ | === Solution 2 === | ||
We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property <math>\log(ab) = \log(a)+\log(b)</math> now gives <cmath> \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 </cmath> The product of the divisors is (from elementary number theory) <math>a^{d(n)/2}</math> where <math>d(n)</math> is the number of divisors. Note that <math>10^n = 2^n\cdot 5^n</math>, so <math>d(n) = (n + 1)^2</math>. Substituting these values with <math>a = 10^n</math> in our equation above, we get <math>n(n + 1)^2 = 1584</math>, from whence we immediately obtain <math>\framebox[1.2\width]{(A)}</math> as the correct answer. | We are given <cmath> \log_{10}d_1 + \log_{10}d_2 + \ldots + \log_{10}d_k = 792 </cmath> The property <math>\log(ab) = \log(a)+\log(b)</math> now gives <cmath> \log_{10}(d_1 d_2\cdot\ldots d_k) = 792 </cmath> The product of the divisors is (from elementary number theory) <math>a^{d(n)/2}</math> where <math>d(n)</math> is the number of divisors. Note that <math>10^n = 2^n\cdot 5^n</math>, so <math>d(n) = (n + 1)^2</math>. Substituting these values with <math>a = 10^n</math> in our equation above, we get <math>n(n + 1)^2 = 1584</math>, from whence we immediately obtain <math>\framebox[1.2\width]{(A)}</math> as the correct answer. | ||
Line 25: | Line 22: | ||
<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> | |
+ | |||
+ | == Alternative thinking == | ||
+ | After arriving at the equation <math>n(n+1)^2 = 1584</math>, notice that all of the answer choices are in the form <math>10+k</math>, where <math>k</math> is <math>1-5</math>. We notice that the ones digit of <math>n(n+1)^2</math> is <math>4</math>, and it is dependent on the ones digit of the answer choices. Trying <math>1-5</math> for <math>n</math>, we see that only <math>1</math> yields a ones digit of <math>4</math>, so our answer is <math>11</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 17:57, 17 February 2021
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 added, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.