Difference between revisions of "2019 AIME I Problems/Problem 9"
(→Solution 2) |
(→Solution 2) |
||
Line 18: | Line 18: | ||
* <math>3^4</math> has two possibilities: 80 and 81, or 81 and 82. Neither works. | * <math>3^4</math> has two possibilities: 80 and 81, or 81 and 82. Neither works. | ||
* <math>11^2</math> has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works. | * <math>11^2</math> has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works. | ||
− | Having computed the working possibilities, we sum them: <math>8+9+16+25+121 = \boxed{ | + | * <math>13^2</math> has two possibilities: 168 and 169, or 169 and 170. Neither works. |
+ | * <math>17^2</math> has two possibilities: 288 and 289, or 289 and 290. Neither works. | ||
+ | * <math>19^2</math> has two possibilities: 360 and 361, or 361 and 362. Only (361,362) works. | ||
+ | Having computed the working possibilities, we sum them: <math>8+9+16+25+121+361 = \boxed{540}</math>. ~Kepy. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=8|num-a=10}} | {{AIME box|year=2019|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:58, 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 fourth power of a prime) and a number with 2 divisors (a prime), or
- a number with 4 divisors (a semiprime) and a number with 3 divisors (a square of a prime). (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: 168 and 169, or 169 and 170. Neither works.
- has two possibilities: 288 and 289, or 289 and 290. Neither works.
- has two possibilities: 360 and 361, or 361 and 362. Only (361,362) 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.