Difference between revisions of "2019 AIME II Problems/Problem 9"
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2. <math>v_n(2)=4</math>. Then <math>\frac{20}{5}=4 = 4\times 1 = 2\times 2</math>. | 2. <math>v_n(2)=4</math>. Then <math>\frac{20}{5}=4 = 4\times 1 = 2\times 2</math>. | ||
The <math>4\times 1</math> case gives one solution, <math>2^4 \times 5^3 = 2000</math>. | The <math>4\times 1</math> case gives one solution, <math>2^4 \times 5^3 = 2000</math>. | ||
− | The <math>2\times 2</math> case gives <math>2^4\times 5 \times (3+7+11+13+17+19+23)</math>. | + | The <math>2\times 2</math> case gives <math>2^4\times 5 \times (3+7+11+13+17+19+23)</math>.Using any prime greater than <math>23</math> will make <math>n</math> greater than <math>2019</math>. |
The answer is <math>\frac{1}{20}(2000+80(3+7+..+23)) = 472</math>. | The answer is <math>\frac{1}{20}(2000+80(3+7+..+23)) = 472</math>. |
Revision as of 00:04, 23 July 2020
Problem 9
Call a positive integer -pretty if has exactly positive divisors and is divisible by . For example, is -pretty. Let be the sum of positive integers less than that are -pretty. Find .
Solution
Every 20-pretty integer can be written in form , where , , , and , where is the number of divisors of . Thus, we have , using the fact that the divisor function is multiplicative. As must be a divisor of 20, there are not many cases to check.
If , then . But this leads to no solutions, as gives .
If , then or . The first case gives where is a prime other than 2 or 5. Thus we have . The sum of all such is . In the second case and , and there is one solution .
If , then , but this gives . No other values for work.
Then we have .
-scrabbler94
Solution 2
For to have exactly positive divisors, can only take on certain prime factorization forms: namely, . No number that is a multiple of can be expressed in the first form, and the only integer divisible by that has the second form is , which is greater than .
For the third form, the only -pretty numbers are and , and only is small enough.
For the fourth form, any number of the form where is a prime other than or will satisfy the -pretty requirement. Since , . Therefore, can take on or .
Thus, .
Solution 3
The divisors of are . must be because . This means that can be exactly or .
1. . Then . The smallest is which is . Hence there are no solution in this case.
2. . Then . The case gives one solution, . The case gives .Using any prime greater than will make greater than .
The answer is .
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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