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 <math>7</math>, we must have: |
− | * either a number with 5 divisors (''a fourth power of a prime'') and a number with 2 divisors (''a prime''), or | + | * either a number with <math>5</math> divisors (''a fourth power of a prime'') and a number with <math>2</math> divisors (''a prime''), or |
− | * a number with 4 divisors (''a semiprime or a cube of a prime'') and a number with 3 divisors (''a square of a prime''). (No integer greater than 1 can have fewer than 2 divisors.) | + | * a number with <math>4</math> divisors (''a semiprime or a cube of a prime'') and a number with <math>3</math> divisors (''a square of a prime''). (No integer greater than <math>1</math> can have fewer than <math>2</math> 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 <math>3</math> divisors, or a fourth power like <math>2^4</math> with <math>5</math> divisors. We then find the smallest such values by hand. |
− | * <math>2^2</math> has two possibilities: 3 and 4 | + | * <math>2^2</math> has two possibilities: <math>3</math> and <math>4</math> or <math>4</math> and <math>5</math>. Neither works. |
− | * <math>3^2</math> has two possibilities: 8 and 9 | + | * <math>3^2</math> has two possibilities: <math>8</math> and <math>9</math> or <math>9</math> and <math>10</math>. <math>(8,9)</math> and <math>(9,10)</math> both work. |
− | * <math>2^4</math> has two possibilities: 15 and 16 | + | * <math>2^4</math> has two possibilities: <math>15</math> and <math>16</math> or <math>16</math> and <math>17</math>. Only <math>(16,17)</math> works. |
− | * <math>5^2</math> has two possibilities: 24 and 25 | + | * <math>5^2</math> has two possibilities: <math>24</math> and <math>25</math> or <math>25</math> and <math>26</math>. Only <math>(25,26)</math> works. |
− | * <math>7^2</math> has two possibilities: 48 and 49 | + | * <math>7^2</math> has two possibilities: <math>48</math> and <math>49</math> or <math>49</math> and <math>50</math>. Neither works. |
− | * <math>3^4</math> has two possibilities: 80 and 81 | + | * <math>3^4</math> has two possibilities: <math>80</math> and <math>81</math> or <math>81</math> and <math>82</math>. Neither works. |
− | * <math>11^2</math> has two possibilities: 120 and 121 | + | * <math>11^2</math> has two possibilities: <math>120</math> and <math>121</math> or <math>121</math> and <math>122</math>. Only <math>(121,122)</math> works. |
− | * <math>13^2</math> has two possibilities: 168 and 169 | + | * <math>13^2</math> has two possibilities: <math>168</math> and <math>169</math> or <math>169</math> and <math>170</math>. Neither works. |
− | * <math>17^2</math> has two possibilities: 288 and 289 | + | * <math>17^2</math> has two possibilities: <math>288</math> and <math>289</math> or <math>289</math> and <math>290</math>. Neither works. |
− | * <math>19^2</math> has two possibilities: 360 and 361 | + | * <math>19^2</math> has two possibilities: <math>360</math> and <math>361</math> or <math>361</math> and <math>362</math>. Only <math>(361,362)</math> works. |
Having computed the working possibilities, we take the sum of the corresponding values of <math>n</math>: <math>8+9+16+25+121+361 = \boxed{540}</math>. ~Kepy. | Having computed the working possibilities, we take the sum of the corresponding values of <math>n</math>: <math>8+9+16+25+121+361 = \boxed{540}</math>. ~Kepy. | ||
− | Possible improvement: since all primes <math>>2</math> 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 16 for the fourth power case. - mathleticguyyy | + | Possible improvement: since all primes <math>>2</math> 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 <math>16</math> for the fourth power case. - mathleticguyyy |
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. |
Revision as of 12:03, 26 December 2019
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
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.
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.