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: box)
(→‎Solution 2)
 
(25 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Let <math>S</math> be the sum of the base <math>10</math> [[logarithm]]s of all the [[proper divisor]]s (all [[divisor]]s of a number excluding itself) of <math>1000000</math>. What is the integer nearest to <math>S</math>?
  
−
== Solution ==
+
__TOC__
 +
 
 +
== Solution 1 ==
 +
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 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\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>.
 +
 
 +
=== Simplification ===
 +
 
 +
The formula for the product of the divisors of <math>n</math> is <math>n^{(d(n))/2}</math>, where <math>d(n)</math> is the number of divisor of <math>n</math>.
 +
We know that <math>\log_{10} a + \log_{10} b + \log_{10} c + \log_{10} d...</math> and so on equals <math>\log_{10} (abcd...)</math> by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of the product of the proper divisors of <math>10^6</math>. The product of the divisors, by the earlier formula, is <math>{(10^6)}^{49/2} = 10^{49*3}</math>, and since we need the product of only the proper divisors, which means the divisors NOT including the number, <math>10^6</math>, itself, we divide <math>10^{(49*3)}</math> by <math>10^6</math> to get <math>10^{(49*3-6)} = 10^{(141)}</math>. The base-10 logarithm of this value, in base 10, is clearly <math>\boxed{141}</math>
 +
 
 +
== Solution 2 ==
 +
 
 +
Since the prime factorization of <math>10^6</math> is <math>2^6 \cdot 5^6</math>, the number of factors in <math>10^6</math> is <math>7 \cdot 7=49</math>. You can pair them up into groups of two so each group multiplies to <math>10^6</math>. Note that <math>\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6</math>. Thus, the sum of the logs of the divisors is half the number of divisors of <math>10^6 \cdot 6 -6</math> (since they are asking only for proper divisors), and the answer is <math>(49/2)\cdot 6-6=\boxed{141}</math>.
 +
 
 +
 
 +
 
 +
 
 +
== Solution 3 ==
 +
 
 +
Note that we can just pair terms up such that the product is <math>10^{6}.</math> Now, however, note that <math>10^{3}</math> is not included. Therefore we first exclude. We have <math>\displaystyle\frac{49-1}{2} = 24</math> pairs that all multiply to <math>10^{6}.</math> Now we include <math>10^{3}</math> so our current product is <math>24 \cdot 6 - 3.</math> However we dont want to include <math>10^6</math> since we are considering proper factors only so the final answer is <math>144 + 3 - 6 = \boxed{141}.</math>
  
 
== See also ==
 
== See also ==
−
* [[1986 AIME Problems]]
+
{{AIME box|year=1986|num-b=7|num-a=9}}
  
−
{{AIME box|year=1986|num-b=7|num-a=9}}
+
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:52, 16 September 2020

Problem

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

Solution 1

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

Simplification

The formula for the product of the divisors of $n$ is $n^{(d(n))/2}$, where $d(n)$ is the number of divisor of $n$. We know that $\log_{10} a + \log_{10} b + \log_{10} c + \log_{10} d...$ and so on equals $\log_{10} (abcd...)$ by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of the product of the proper divisors of $10^6$. The product of the divisors, by the earlier formula, is ${(10^6)}^{49/2} = 10^{49*3}$, and since we need the product of only the proper divisors, which means the divisors NOT including the number, $10^6$, itself, we divide $10^{(49*3)}$ by $10^6$ to get $10^{(49*3-6)} = 10^{(141)}$. The base-10 logarithm of this value, in base 10, is clearly $\boxed{141}$

Solution 2

Since the prime factorization of $10^6$ is $2^6 \cdot 5^6$, the number of factors in $10^6$ is $7 \cdot 7=49$. You can pair them up into groups of two so each group multiplies to $10^6$. Note that $\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6$. Thus, the sum of the logs of the divisors is half the number of divisors of $10^6 \cdot 6 -6$ (since they are asking only for proper divisors), and the answer is $(49/2)\cdot 6-6=\boxed{141}$.



Solution 3

Note that we can just pair terms up such that the product is $10^{6}.$ Now, however, note that $10^{3}$ is not included. Therefore we first exclude. We have $\displaystyle\frac{49-1}{2} = 24$ pairs that all multiply to $10^{6}.$ Now we include $10^{3}$ so our current product is $24 \cdot 6 - 3.$ However we dont want to include $10^6$ since we are considering proper factors only so the final answer is $144 + 3 - 6 = \boxed{141}.$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_8&oldid=133733"