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

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==Problem 9==
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==Problem==
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Let <math>\tau (n)</math> denote the number of positive integer divisors of <math>n</math>. Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>.
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Let <math>\tau(n)</math> denote the number of positive integer divisors of <math>n</math>. Find the sum of the six least positive integers <math>n</math> that are solutions to <math>\tau (n) + \tau (n+1) = 7</math>.
  
 
==Solution==
 
==Solution==
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In order to obtain a sum of 7, we must have:
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In order to obtain a sum of <math>7</math>, we must have:
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* either a number with 5 divisors (''a fourth power of a prime'') and a number with 2 divisors (''a prime''), or
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* 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
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* a number with 4 divisors (''a semiprime'') and a number with 3 divisors (''a square of a prime''). (No integer greater than 1 can have fewer than 2 divisors.)
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* 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.)
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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.
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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.
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* <math>2^2</math> has two possibilities: 3 and 4, or 4 and 5. Neither works.
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* <math>2^2</math> has two possibilities: <math>3</math> and <math>4</math> or <math>4</math> and <math>5</math>. Neither works.
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* <math>3^2</math> has two possibilities: 8 and 9, or 9 and 10. (8,9) and (9,10) both work.
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* <math>3^2</math> has two possibilities: <math>8</math> and <math>9</math> or <math>9</math> and <math>10</math>. <math>\boxed{(8,9)}</math> and <math>\boxed{(9,10)}</math> both work.
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* <math>2^4</math> has two possibilities: 15 and 16, or 16 and 17. Only (16,17) works.
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* <math>2^4</math> has two possibilities: <math>15</math> and <math>16</math> or <math>16</math> and <math>17</math>. Only <math>\boxed{(16,17)}</math> works.
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* <math>5^2</math> has two possibilities: 24 and 25, or 25 and 26. Only (25,26) works.
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* <math>5^2</math> has two possibilities: <math>24</math> and <math>25</math> or <math>25</math> and <math>26</math>. Only <math>\boxed{(25,26)}</math> works.
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* <math>7^2</math> has two possibilities: 48 and 49, or 49 and 50. Neither works.
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* <math>7^2</math> has two possibilities: <math>48</math> and <math>49</math> or <math>49</math> and <math>50</math>. Neither works.
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* <math>3^4</math> has two possibilities: 80 and 81, or 81 and 82. Neither works.
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* <math>3^4</math> has two possibilities: <math>80</math> and <math>81</math> or <math>81</math> and <math>82</math>. Neither works.
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* <math>11^2</math> has two possibilities: 120 and 121, or 121 and 122. Only (121,122) works.
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* <math>11^2</math> has two possibilities: <math>120</math> and <math>121</math> or <math>121</math> and <math>122</math>. Only <math>\boxed{(121,122)}</math> works.
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* <math>13^2</math> has two possibilities: 168 and 169, or 169 and 170. Neither works.
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* <math>13^2</math> has two possibilities: <math>168</math> and <math>169</math> or <math>169</math> and <math>170</math>. Neither works.
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* <math>17^2</math> has two possibilities: 288 and 289, or 289 and 290. Neither works.
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* <math>17^2</math> has two possibilities: <math>288</math> and <math>289</math> or <math>289</math> and <math>290</math>. Neither works.
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* <math>19^2</math> has two possibilities: 360 and 361, or 361 and 362. Only (361,362) works.
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* <math>19^2</math> has two possibilities: <math>360</math> and <math>361</math> or <math>361</math> and <math>362</math>. Only <math>\boxed{(361,362)}</math> works.
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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.
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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{\textbf{540}}</math>. ~Kepy.
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 +
 
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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
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==Solution 2==
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Let the ordered pair <math>(a,b)</math> represent the number of divisors of <math>n</math> and <math>n+1</math> respectively.
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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>.
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Case 1: When we have <math>(2,5)</math>
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For <math>n</math> to have 2 divisors, it must be a prime number.
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For <math>n+1</math> to have 5 divisors, it must be in the form <math>a^4</math>.
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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; <math>n</math> is not prime.
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No cases work in this case
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 +
 
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Case 2: When we have <math>(4,3)</math>
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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 .
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For <math>n+1</math> to have 3 divisors, it must be a square number.
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Let <math>n+1 = A^2</math> (<math>A</math> is a prime number). When <math>n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3</math>.
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We see that the only case when it works is when <math>a=2, A=3</math>, so <math>n=8</math> works.
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Case 3: When we have <math>(5,2)</math>
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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.
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For <math>n+1</math> to have 2 divisors, it must be a prime number.
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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>.
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Case 4: When we have <math>(3,4)</math>
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For <math>n</math> to have 3 divisors, it must be a square number.
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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.
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Similar to Case 2, let <math>n = A^2</math> (<math>A</math> is a prime number).
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* When <math>n+1 = a^3, A^2+1 = a^3</math>.
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There are no cases that satisfy this equation.
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* When <math>n+1=ab, A^2+1 = ab</math>.
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We test squares of primes to find values of n that work.
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* <math>A=2</math>,  <math>4+1=5</math>.  Doesn't work.
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* <math>A=3</math>,  <math>9+1=10=2*5</math>.  It works.  <math>n=9</math>
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* <math>A=5</math>,  <math>25+1=26=2*13</math>.  It works. <math>n=25</math>
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* <math>A=7</math>,  <math>49+1=50=2*5^2</math>.  Doesn't work.
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* <math>A=11</math>,  <math>121+1=122=2*61</math>.  It works. <math>n=121</math>
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* <math>A=13</math>,  <math>169+1=170=2*5*17</math>.  Doesn't work
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* <math>A=17</math>,  <math>289+1=290=2*5*29</math>.  Doesn't work
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* <math>A=19</math>,  <math>361+1=362=2*181</math>.  It works. <math>n=361</math>
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Now we add up the values of <math>n</math> to get the answer: <math>8+16+9+25+121+361 = \boxed{540}</math>.
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~toastybaker
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==Solution 3 (Official MAA)==
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Let <math>p,\,q,</math> and <math>r</math> represent primes. Because <math>\tau(n)=1</math> only for <math>n=1,</math> there is no <math>n</math> for which <math>\{\tau(n),\tau(n+1)\}=\{1,6\}</math>. If <math>\{\tau(n),\tau(n+1)\}=\{2,5\},</math> then <math>\{n,n+1\}=\{p,q^4\},</math> so <math>|p-q^4|=1.</math> Checking <math>q=2</math> and <math>p=17</math> yields the solution <math>n=16.</math> If <math>q>2,</math> then <math>q</math> is odd, and <math>p=q^4\pm 1</math> is even, so <math>p</math> cannot be prime.
  
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If <math>\{\tau(n),\tau(n+1)\}=\{3,4\},</math> then <math>\{n,n+1\}=\{p^2,q^3\}</math> or <math>\{p^2,qr\}.</math> Consider <math>|p^2-q^3|=1.</math> If <math>p^2-1=(p-1)(p+1)=q^3,</math> Then <math>q=2.</math> This yields the solution <math>p=3</math> and <math>q=2,</math> so <math>n=8.</math> If <math>q^3-1=(q-1)(q^2+q+1)=p^2,</math> then <math>q-1=1,</math> which does not give a solution. Consider <math>|p^2-qr|=1.</math> If <math>p^2-1=(p-1)(p+1)=qr,</math> then if <math>p>2,</math> the left side is divisible by 8, so there are no solutions. Finding the smallest four primes such that <math>p^2+1=qr</math> gives <math>3^2+1=10,\,5^2+1=26,\,11^2+1=122,</math> and <math>19^2+1=362.</math> The six least values of <math>n</math> are <math>8,9,16,25,121,</math> and <math>361,</math> whose sum is <math>540.</math>
  
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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
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==Video Solution ==
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https://www.youtube.com/watch?v=2ouOexOnG1A
  
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Note: Bashing would work for this problem, but it would be very tedious.
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~ North America Math COntest Go Go Go
  
 
==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}}
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[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 08:21, 4 October 2022

Problem

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

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 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.)

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 $10$. $\boxed{(8,9)}$ and $\boxed{(9,10)}$ both work.
  • $2^4$ has two possibilities: $15$ and $16$ or $16$ and $17$. Only $\boxed{(16,17)}$ works.
  • $5^2$ has two possibilities: $24$ and $25$ or $25$ and $26$. Only $\boxed{(25,26)}$ 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 $\boxed{(121,122)}$ works.
  • $13^2$ has two possibilities: $168$ and $169$ or $169$ and $170$. Neither works.
  • $17^2$ has two possibilities: $288$ and $289$ or $289$ and $290$. Neither works.
  • $19^2$ has two possibilities: $360$ and $361$ or $361$ and $362$. Only $\boxed{(361,362)}$ works.

Having computed the working possibilities, we take the sum of the corresponding values of $n$: $8+9+16+25+121+361 = \boxed{\textbf{540}}$. ~Kepy.


Possible improvement: since all primes $>2$ 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

Solution 2

Let the ordered pair $(a,b)$ represent the number of divisors of $n$ and $n+1$ respectively. We see that to obtain a sum of $7$, we can have $(2,5), (3,4), (4,3),$ and $(5,2)$.


Case 1: When we have $(2,5)$ For $n$ to have 2 divisors, it must be a prime number. For $n+1$ to have 5 divisors, it must be in the form $a^4$. If $n+1$ is in the form $a^4$, then $n = a^4-1 = (a^2+1)(a-1)(a+1)$. This means that $n$, or $a^4-1$ has factors other than 1 and itself; $n$ is not prime. No cases work in this case


Case 2: When we have $(4,3)$ For $n$ to have 4 divisors, it must be in the form $a^3$ or $ab$, where $a$ and $b$ are distinct prime numbers . For $n+1$ to have 3 divisors, it must be a square number. Let $n+1 = A^2$ ($A$ is a prime number). When $n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3$. We see that the only case when it works is when $a=2, A=3$, so $n=8$ works.


Case 3: When we have $(5,2)$ For $n$ to have 5 divisors, it must be in the form $a^4$, where $a$ is a prime number. For $n+1$ to have 2 divisors, it must be a prime number. Notice that $a$ and $a^4$ have the same parity (even/odd). Since every prime greater than 2 are odd, $n = a^4$ must be even. Since $a^4$ is even, $a$ must be even as well, and the only prime number that is even is 2. When $a=2, n=16$.


Case 4: When we have $(3,4)$ For $n$ to have 3 divisors, it must be a square number. For $n+1$ to have 4 divisors, it must be in the form $a^3$ or $ab$, where $a$ and $b$ are distinct prime numbers. Similar to Case 2, let $n = A^2$ ($A$ is a prime number).

  • When $n+1 = a^3, A^2+1 = a^3$.

There are no cases that satisfy this equation.

  • When $n+1=ab, A^2+1 = ab$.

We test squares of primes to find values of n that work.

  • $A=2$, $4+1=5$. Doesn't work.
  • $A=3$, $9+1=10=2*5$. It works. $n=9$
  • $A=5$, $25+1=26=2*13$. It works. $n=25$
  • $A=7$, $49+1=50=2*5^2$. Doesn't work.
  • $A=11$, $121+1=122=2*61$. It works. $n=121$
  • $A=13$, $169+1=170=2*5*17$. Doesn't work
  • $A=17$, $289+1=290=2*5*29$. Doesn't work
  • $A=19$, $361+1=362=2*181$. It works. $n=361$

Now we add up the values of $n$ to get the answer: $8+16+9+25+121+361 = \boxed{540}$. ~toastybaker

Solution 3 (Official MAA)

Let $p,\,q,$ and $r$ represent primes. Because $\tau(n)=1$ only for $n=1,$ there is no $n$ for which $\{\tau(n),\tau(n+1)\}=\{1,6\}$. If $\{\tau(n),\tau(n+1)\}=\{2,5\},$ then $\{n,n+1\}=\{p,q^4\},$ so $|p-q^4|=1.$ Checking $q=2$ and $p=17$ yields the solution $n=16.$ If $q>2,$ then $q$ is odd, and $p=q^4\pm 1$ is even, so $p$ cannot be prime.

If $\{\tau(n),\tau(n+1)\}=\{3,4\},$ then $\{n,n+1\}=\{p^2,q^3\}$ or $\{p^2,qr\}.$ Consider $|p^2-q^3|=1.$ If $p^2-1=(p-1)(p+1)=q^3,$ Then $q=2.$ This yields the solution $p=3$ and $q=2,$ so $n=8.$ If $q^3-1=(q-1)(q^2+q+1)=p^2,$ then $q-1=1,$ which does not give a solution. Consider $|p^2-qr|=1.$ If $p^2-1=(p-1)(p+1)=qr,$ then if $p>2,$ the left side is divisible by 8, so there are no solutions. Finding the smallest four primes such that $p^2+1=qr$ gives $3^2+1=10,\,5^2+1=26,\,11^2+1=122,$ and $19^2+1=362.$ The six least values of $n$ are $8,9,16,25,121,$ and $361,$ whose sum is $540.$

Video Solution

https://www.youtube.com/watch?v=2ouOexOnG1A

~ North America Math COntest Go Go Go

See Also

2019 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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