Difference between revisions of "2019 AIME I Problems/Problem 9"

(Solution)
(Solution 2)
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==Solution 2==
 
==Solution 2==
 
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 22:55, 14 March 2019

Problem 9

Let $\tau (n)$ denote the number of positive integer divisors of $n$. Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$.

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 $3^2$ with 3 divisors, or a fourth power, like $2^4$, with 5 divisors. We then find the smallest such values by hand.

  • $2^2$ has two possibilities: 3 and 4, or 4 and 5. Neither works.
  • $3^2$ has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.
  • $2^4$ has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.
  • $5^2$ has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
  • $7^2$ has two possibilities: 48 and 49, or 49 and 50. Neither works.
  • $3^4$ has two possibilities: 80 and 81, or 81 and 82. Neither works.
  • $11^2$ has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.

Having computed the working possibilities, we sum them: $8+9+16+25+121 = \boxed{179}$. ~Kepy.

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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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