Difference between revisions of "1986 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | Let <math>\displaystyle S</math> be the sum of the base <math>\displaystyle 10</math> | + | Let <math>\displaystyle S</math> be the sum of the base <math>\displaystyle 10</math> [[logarithm]]s of all the [[proper divisor]]s (all [[divisor]]s of a number excluding itself) of <math>\displaystyle 1000000</math>. What is the integer nearest to <math>\displaystyle S</math>? |
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== Solution == | == Solution == | ||
| − | {{ | + | The [[prime factorization]] of <math>100000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) - 1 = 48</math> proper divisors (the subtracted 1 to ignore <math>1000000</math> itself). The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers. |
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| + | Writing out the first few terms, we see that the answer is equal to <math>\log 1 + \log 2 + \log 4 + \log 5 \ldots = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots</math>. Each powers of 2 from 0 to 5 in this equation appear <math>7</math> times (excluding 6, which only appears 6 times due to the exclusion of <math>100000</math>). Therefore, it appears <math>(0 + 1 + 2 + 3 + 4 + 5) \cdot 7 + 6 \cdot 6 = 15 \times 7 + 36 = 141</math>. The same goes for <math>5</math>. | ||
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| + | The answer is thus <math>\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1986|num-b=7|num-a=9}} | |
| − | + | [[Category:Intermediate Algebra Problems]] | |
Revision as of 20:47, 23 March 2007
Problem
Let 
be the sum of the base 
logarithms of all the proper divisors (all divisors of a number excluding itself) of 
. What is the integer nearest to ?
Solution
The prime factorization of 
, so there are 
proper divisors (the subtracted 1 to ignore 
itself). The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.
Writing out the first few terms, we see that the answer is equal to 
. Each powers of 2 from 0 to 5 in this equation appear 
times (excluding 6, which only appears 6 times due to the exclusion of 
). Therefore, it appears 
. The same goes for 
.
The answer is thus 
.
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
