Difference between revisions of "2019 AIME I Problems/Problem 9"
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Note: Bashing would work for this problem, but it would be very tedious. | Note: Bashing would work for this problem, but it would be very tedious. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let the ordered pair <math>(a,b)</math> represent the number of divisors of <math>n</math> and <math>n+1</math> respectively. | ||
+ | We see that to obtain a sum of <math>7</math>, we can have <math>(2,5), (3,4), (4,3),</math> and <math>(5,2)</math>. | ||
+ | |||
+ | |||
+ | Case 1: When we have <math>(2,5)</math> | ||
+ | For <math>n</math> to have 2 divisors, it must be a prime number. | ||
+ | For <math>n+1</math> to have 5 divisors, it must be in the form <math>a^4</math>. | ||
+ | If <math>n+1</math> is in the form <math>a^4</math>, then <math>n = a^4-1 = (a^2+1)(a-1)(a+1)</math>. This means that <math>n</math>, or <math>a^4-1</math> has factors other than 1 and itself; n is not prime. | ||
+ | No cases work in this case | ||
+ | |||
+ | |||
+ | Case 2: When we have <math>(4,3)</math> | ||
+ | For <math>n</math> to have 4 divisors, it must be in the form <math>a^3</math> or <math>ab</math>, where <math>a</math> and <math>b</math> are distinct prime numbers . | ||
+ | For <math>n+1</math> to have 3 divisors, it must be a square number. | ||
+ | Let <math>n+1 = A^2</math> (A is a prime number). When <math>n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3</math>. | ||
+ | We see that the only case when it works is when <math>a=2, A=3</math>, so <math>n=8</math> works. | ||
+ | |||
+ | |||
+ | Case 3: When we have <math>(5,2)</math> | ||
+ | For <math>n</math> to have 5 divisors, it must be in the form <math>a^4</math>, where <math>a</math> is a prime number. | ||
+ | For <math>n+1</math> to have 2 divisors, it must be a prime number. | ||
+ | Notice that <math>a</math> and <math>a^4</math> have the same parity (even/odd). Since every prime greater than 2 are odd, <math>n = a^4</math> must be even. Since <math>a^4</math> is even, <math>a</math> must be even as well, and the only prime number that is even is 2. When <math>a=2, n=16</math>. | ||
+ | |||
+ | |||
+ | Case 4: When we have <math>(3,4)</math> | ||
+ | For <math>n</math> to have 3 divisors, it must be a square number. | ||
+ | For <math>n+1</math> to have 4 divisors, it must be in the form <math>a^3</math> or <math>ab</math>, where <math>a</math> and <math>b</math> are distinct prime numbers. | ||
+ | Similar to Case 2, let <math>n = A^2</math> (A is a prime number). | ||
+ | |||
+ | * When <math>n+1 = a^3, A^2+1 = a^3</math>. | ||
+ | There are no cases that satisfy this equation. | ||
+ | |||
+ | * When <math>n+1=ab, A^2+1 = ab</math>. | ||
+ | We test squares of primes to find the answer | ||
+ | * <math>A=2</math>, <math>4+1=5</math>. Doesn't work. | ||
+ | * <math>A=3</math>, <math>9+1=10=2*5</math>. It works. <math>n=9</math> | ||
+ | * <math>A=5</math>, <math>25+1=26=2*13</math>. It works. <math>n=25</math> | ||
+ | * <math>A=7</math>, <math>49+1=50=2*5^2</math>. Doesn't work. | ||
+ | * <math>A=11</math>, <math>121+1=122=2*61</math>. It works. <math>n=121</math> | ||
+ | * <math>A=13</math>, <math>169+1=170=2*5*17</math>. Doesn't work | ||
+ | * <math>A=17</math>, <math>289+1=290=2*5*29</math>. Doesn't work | ||
+ | * <math>A=19</math>, <math>361+1=362=2*181</math>. It works. <math>n=361</math> | ||
+ | |||
+ | Now we add up the values of <math>n</math> to get the answer: <math>8+16+9+25+121+361 = \boxed{540}</math>. | ||
+ | ~toastybaker | ||
==See Also== | ==See Also== |
Revision as of 19:56, 7 November 2020
Contents
Problem 9
Let denote the number of positive integer divisors of (including and ). Find the sum of the six least positive integers that are solutions to .
Solution
In order to obtain a sum of , we must have:
- either a number with divisors (a fourth power of a prime) and a number with divisors (a prime), or
- a number with divisors (a semiprime or a cube of a prime) and a number with divisors (a square of a prime). (No integer greater than can have fewer than 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 divisors, or a fourth power like with divisors. We then find the smallest such values by hand.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . and both work.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Only works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Neither works.
- has two possibilities: and or and . Only works.
Having computed the working possibilities, we take the sum of the corresponding values of : . ~Kepy.
Possible improvement: since all primes are odd, their fourth powers are odd as well, which cannot be adjacent to any primes because both of the adjacent numbers will be even. Thus, we only need to check for the fourth power case. - mathleticguyyy
Note: Bashing would work for this problem, but it would be very tedious.
Solution 2
Let the ordered pair represent the number of divisors of and respectively. We see that to obtain a sum of , we can have and .
Case 1: When we have
For to have 2 divisors, it must be a prime number.
For to have 5 divisors, it must be in the form .
If is in the form , then . This means that , or has factors other than 1 and itself; n is not prime.
No cases work in this case
Case 2: When we have
For to have 4 divisors, it must be in the form or , where and are distinct prime numbers .
For to have 3 divisors, it must be a square number.
Let (A is a prime number). When .
We see that the only case when it works is when , so works.
Case 3: When we have
For to have 5 divisors, it must be in the form , where is a prime number.
For to have 2 divisors, it must be a prime number.
Notice that and have the same parity (even/odd). Since every prime greater than 2 are odd, must be even. Since is even, must be even as well, and the only prime number that is even is 2. When .
Case 4: When we have
For to have 3 divisors, it must be a square number.
For to have 4 divisors, it must be in the form or , where and are distinct prime numbers.
Similar to Case 2, let (A is a prime number).
- When .
There are no cases that satisfy this equation.
- When .
We test squares of primes to find the answer
- , . Doesn't work.
- , . It works.
- , . It works.
- , . Doesn't work.
- , . It works.
- , . Doesn't work
- , . Doesn't work
- , . It works.
Now we add up the values of to get the answer: . ~toastybaker
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 |
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