Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1986 AIME Problems/Problem 8"

m (→‎See also)
m (→‎Solution: not principle)
Line 3: Line 3:
  
 
== 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.  
+
The [[prime factorization]] of <math>1000000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) = 49</math> divisors (note the question asks for  proper divisors, so we ignore <math>1000000</math> and get <math>\displaystyle 49-1=48\displaystyle</math>). 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 <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 power of 2 from <math>0</math> to <math>5</math> in this equation appear <math>7</math> times (excluding 6, which only appears <math>6</math> 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>.
+
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 1000000 = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots (2^65^6)</math>. Each power of <math>2</math> appears <math>7</math> times; and the same goes for <math>5</math>. So they appear <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math> times. However, since the question asks for proper divisors, we exclude <math>2^65^6</math>, so each number appears <math>\displaystyle 141</math> times. The answer is thus <math>\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141</math>.
−
 
 
−
The answer is thus <math>\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 20:58, 11 September 2007

Problem

Let $\displaystyle S$ be the sum of the base $\displaystyle 10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $\displaystyle 1000000$. What is the integer nearest to $\displaystyle S$?

Solution

The prime factorization of $1000000 = 2^65^6$, so there are $(6 + 1)(6 + 1) = 49$ divisors (note the question asks for proper divisors, so we ignore $1000000$ and get $\displaystyle 49-1=48\displaystyle$). 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 $\log 1 + \log 2 + \log 4 + \log 5 + \ldots + \log 1000000 = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots (2^65^6)$. Each power of $2$ appears $7$ times; and the same goes for $5$. So they appear $7(1+2+3+4+5+6) = 7 \cdot 21 = 147$ times. However, since the question asks for proper divisors, we exclude $2^65^6$, so each number appears $\displaystyle 141$ times. The answer is thus $\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_8&oldid=16072"