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

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 $144 + 3 - 6 = \boxed{141}.$

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 == Problem ==
-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>?
+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__
-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>49-1=48</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 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>141</math> times. The answer is thus <math>S = \log 2^{141}5^{141} = \log 10^{141} = 141</math>.
+== 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+log10^6-logn=6</math>. Thus, the sum of the logs of the divisors is half the number of divisors of <math>10^6 \cdot6 -6</math> (since they are asking only for proper divisors), and the answer is <math>(49/2)\cdot6-6=141</math>.
+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 ==
 {{AIME box|year=1986|num-b=7|num-a=9}}
-* [[AIME Problems and Solutions]]
-* [[American Invitational Mathematics Examination]]
-* [[Mathematics competition resources]]
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

1986 AIME (Problems • Answer Key • Resources)
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