Difference between revisions of "2019 AIME I Problems/Problem 9"
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In order to obtain a sum of 7, we must have: | In order to obtain a sum of 7, we must have: | ||
| − | * either a number with 5 divisors and a number with 2 divisors, or | + | * either a number with 5 divisors (''a prime fourth power'') and a number with 2 divisors (''a prime''), or |
| − | * a number with 4 divisors and a number with 3 divisors. (No number greater than 1 can have fewer than 2 divisors.) | + | * a number with 4 divisors (''a semiprime'') and a number with 3 divisors (''a prime square''). (No number greater than 1 can have fewer than 2 divisors.) |
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square, like <math>3^2</math> with 3 divisors, or a fourth power, like <math>2^4</math>, with 5 divisors. We then find the smallest such values by hand. | Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square, like <math>3^2</math> with 3 divisors, or a fourth power, like <math>2^4</math>, with 5 divisors. We then find the smallest such values by hand. | ||
* <math>2^2</math> has two possibilities: 3 and 4, or 4 and 5. Neither works. | * <math>2^2</math> has two possibilities: 3 and 4, or 4 and 5. Neither works. | ||
Revision as of 21:55, 14 March 2019
Contents
Problem 9
Let 
denote the number of positive integer divisors of 
. Find the sum of the six least positive integers that are solutions to 
.
Solution
Essentially, you realize that to get 7 you need an odd amount of divisors + and even amount of divisors. This means that one of our n needs to be a square. Furthermore it must either be a prime squared to get 3 divisors or a prime to the fourth to get 5 divisors. Any more factors in a square would be to large. Thus n/n+1 is in the form p^2 or p^4. The rest of the solution is bashing left to the reader.
~~ paliwalar.21
Solution 2
In order to obtain a sum of 7, we must have:
- either a number with 5 divisors (a prime fourth power) and a number with 2 divisors (a prime), or
- a number with 4 divisors (a semiprime) and a number with 3 divisors (a prime square). (No number greater than 1 can have fewer than 2 divisors.)
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square, like 
with 3 divisors, or a fourth power, like 
, with 5 divisors. We then find the smallest such values by hand.
has two possibilities: 3 and 4, or 4 and 5. Neither works.
has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.
has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.
has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
has two possibilities: 48 and 49, or 49 and 50. Neither works.
has two possibilities: 80 and 81, or 81 and 82. Neither works.
has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.
has two possibilities: 3 and 4, or 4 and 5. Neither works.
has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
has two possibilities: 48 and 49, or 49 and 50. Neither works.
has two possibilities: 80 and 81, or 81 and 82. Neither works.
has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.Having computed the working possibilities, we sum them: 
. ~Kepy.
See Also
| 2019 AIME I (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
