Difference between revisions of "1986 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
− | 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> | + | 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> | + | 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>. |
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+ | == Solution 2 == | ||
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+ | 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>. | ||
== See also == | == See also == |
Revision as of 12:36, 30 December 2007
Contents
[hide]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 divisors (note the question asks for proper divisors, so we ignore and get ). 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 power of appears times; and the same goes for . So they appear times. However, since the question asks for proper divisors, we exclude , so each number appears 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 |