Difference between revisions of "2008 AMC 12B Problems/Problem 23"
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− | == 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>? | ||
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== 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>. 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^ | + | 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 multiplied, 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> |
− | We then plug in answer choices and arrive at the answer | ||
=== Solution 2 === | === Solution 2 === | ||
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<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> | Trying for answer choices we get <math>n=11</math> | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Let integer <math>d</math> be the number of divisors <math>10^n</math> has. Then, we set up <math>\frac{d}{2}</math> pairs of divisors such that each pair <math>(a,b)</math> satisfies <math>ab = 10^n</math> -- ex. <math>(1, 10^n), (2, 5*10^{n-1})</math>, etc. Then the sum of the base-<math>10</math> logarithms is <math>\sum_{j=1}^{\frac{d}{2}} \log_{10} a_j + \log_{10} b_j = \sum_{j=1}^{\frac{d}{2}} \log_{10} a_j b_j = \sum_{j=1}^{\frac{d}{2}} \log_{10} 10^n = \sum_{j=1}^{\frac{d}{2}} n = \frac{nd}{2}</math> | ||
+ | We can use the property that only perfect squares have an odd number of factors, as for perfect square <math>p^2</math>, we have ordered pair <math>(p,p)</math> that works. For even <math>n</math>, then, <math>10^{\frac{n}{2}}</math> can be multiplied by itself to get <math>10^n</math>, so <math>d</math> is odd. But, in our summation, <math>\frac{d}{2}</math> does not exist for even <math>n</math> as <math>d</math> is then odd, so <math>d</math> must be even. And, since <math>\frac{nd}{2}</math> = <math>792</math>, We want to find a <math>d</math> for our odd options <math>n</math> such that <math>nd</math> = <math>1584</math>. For <math>n=11</math>, <math>d=144</math> works, and integer <math>d</math> can not be found for other odd <math>n</math>. So, we get <math>\framebox[1.2\width]{(A) 11}</math> | ||
== Alternative thinking == | == Alternative thinking == |
Latest revision as of 18:02, 17 January 2023
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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