1986 AIME Problems/Problem 8: Difference between revisions

Revision as of 13:49, 6 May 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 $100000 = 2^65^6$ , so there are $(6 + 1)(6 + 1) - 1 = 48$ proper divisors (the subtracted 1 to ignore $1000000$ itself). 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 $\log 1 + \log 2 + \log 4 + \log 5 \ldots = \log 1 \cdot 2 \cdot 4 \cdot 5 \cdots = \log (2^05^0)(2^15^0)(2^05^1)(2^25^0) \ldots$ . Each power of 2 from $0$ to $5$ in this equation appear $7$ times (excluding 6, which only appears $6$ times due to the exclusion of $100000$ ). Therefore, it appears $(0 + 1 + 2 + 3 + 4 + 5) \cdot 7 + 6 \cdot 6 = 15 \times 7 + 36 = 141$ . The same goes for $5$ .

The answer is thus $\displaystyle S = \log 2^{141}5^{141} = \log 10^{141} = 141$ .

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 == 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]]

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1986 AIME Problems/Problem 8: Difference between revisions

Revision as of 13:49, 6 May 2007

Problem

Solution

See also