Difference between revisions of "1986 AIME Problems/Problem 8"
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The [[prime factorization]] of <math>1000000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) = 49</math> divisors, of which <math>48</math> are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers. | The [[prime factorization]] of <math>1000000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) = 49</math> divisors, of which <math>48</math> are proper. 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 <cmath>\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^05^1)\cdots (2^65^6).</cmath> Each power of <math>2</math> appears <math>7</math> times; and the same goes for <math>5</math>. So the overall power of <math>2</math> and <math>5</math> is <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math>. However, since the question asks for proper divisors, we exclude <math>2^65^6</math>, so each power is actually <math>141</math> times. The answer is thus <math>S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}</math>. | + | Writing out the first few terms, we see that the answer is equal to <cmath>\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots(2^05^1)(2^15^1)\cdots (2^55^6)(2^65^6).</cmath> Each power of <math>2</math> appears <math>7</math> times; and the same goes for <math>5</math>. So the overall power of <math>2</math> and <math>5</math> is <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math>. However, since the question asks for proper divisors, we exclude <math>2^65^6</math>, so each power is actually <math>141</math> times. The answer is thus <math>S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}</math>. |
=== Solution 2 === | === Solution 2 === | ||
Revision as of 13:09, 25 January 2014
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
Solution 1
The prime factorization of 
, so there are 
divisors, of which 
are proper. 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 ![\[\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots(2^05^1)(2^15^1)\cdots (2^55^6)(2^65^6).\]](http://latex.artofproblemsolving.com/a/3/1/a3179ba092f2d1363dac1159794c8cddfe1466db.png)
Each power of 
appears 
times; and the same goes for 
. So the overall power of and is 
. However, since the question asks for proper divisors, we exclude 
, so each power is actually 
times. The answer is thus 
.
Solution 2
Since the prime factorization of 
is 
, the number of factors in is 
. You can pair them up into groups of two so each group multiplies to . Note that 
. Thus, the sum of the logs of the divisors is half the number of divisors of 
(since they are asking only for proper divisors), and the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
