Difference between revisions of "1986 AIME Problems/Problem 8"
m (→Solution 1) |
(→Solution 2) |
||
(11 intermediate revisions by 3 users not shown) | |||
Line 3: | Line 3: | ||
__TOC__ | __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. | 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 + \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>. | + | 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 == |
Latest revision as of 20:52, 16 September 2020
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 1
The prime factorization of , so there are divisors, of which 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 Each power of appears times; and the same goes for . So the overall power of and is . However, since the question asks for proper divisors, we exclude , so each power is actually times. The answer is thus .
Simplification
The formula for the product of the divisors of is , where is the number of divisor of . We know that and so on equals 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 . The product of the divisors, by the earlier formula, is , and since we need the product of only the proper divisors, which means the divisors NOT including the number, , itself, we divide by to get . The base-10 logarithm of this value, in base 10, is clearly
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 .
Solution 3
Note that we can just pair terms up such that the product is Now, however, note that is not included. Therefore we first exclude. We have pairs that all multiply to Now we include so our current product is However we dont want to include since we are considering proper factors only so the final 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.