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Difference between revisions of "2005 AMC 10A Problems/Problem 24"

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m (→‎Solution 2)
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Looking at pairs of [[divisor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:
 
Looking at pairs of [[divisor]]s of <math>48</math>, we have several possibilities to solve for <math>p_{1}</math> and <math>p_{2}</math>:
  
−
(Note: you can skip several cases below by observing that <math>p_1+p_2</math> and <math>p_1-p_{2}</math> must be even, and p_1+p_2  \neq  p_1-p_2 (\mod 4) <math>.
+
(Note: you can skip several cases below by observing that <math>p_1+p_2</math> and <math>p_1-p_{2}</math> must be even, and <math>p_1+p_2  \neq  p_1-p_2 (\mod 4) </math>.
  
−
</math> (p_{2}+p_{1}) = 48 <math>
+
<math> (p_{2}+p_{1}) = 48 </math>
  
−
</math> (p_{2}-p_{1}) = 1 <math> (impossible)
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<math> (p_{2}-p_{1}) = 1 </math> (impossible)
  
−
</math> p_{1} = \frac{47}{2} <math>
+
<math> p_{1} = \frac{47}{2} </math>
  
−
</math> p_{2} = \frac{49}{2} <math>
+
<math> p_{2} = \frac{49}{2} </math>
  
  
−
</math> (p_{2}+p_{1}) = 24 <math>
+
<math> (p_{2}+p_{1}) = 24 </math>
  
−
</math> (p_{2}-p_{1}) = 2 <math>
+
<math> (p_{2}-p_{1}) = 2 </math>
  
−
</math> p_{1} = 11 <math>
+
<math> p_{1} = 11 </math>
  
−
</math> p_{2} = 13 <math>
+
<math> p_{2} = 13 </math>
  
 
(Valid!)
 
(Valid!)
  
−
</math> (p_{2}+p_{1}) = 16 <math>
+
<math> (p_{2}+p_{1}) = 16 </math>
  
−
</math> (p_{2}-p_{1}) = 3 <math>  (impossible)
+
<math> (p_{2}-p_{1}) = 3 </math>  (impossible)
  
−
</math> p_{1} = \frac{13}{2} <math>
+
<math> p_{1} = \frac{13}{2} </math>
  
−
</math> p_{2} = \frac{19}{2} <math>
+
<math> p_{2} = \frac{19}{2} </math>
  
  
−
</math> (p_{2}+p_{1}) = 12 <math>  
+
<math> (p_{2}+p_{1}) = 12 </math>  
  
−
</math> (p_{2}-p_{1}) = 4 <math>  (impossible)
+
<math> (p_{2}-p_{1}) = 4 </math>  (impossible)
  
−
</math> p_{1} = 4 <math>
+
<math> p_{1} = 4 </math>
  
−
</math> p_{2} = 8 <math>
+
<math> p_{2} = 8 </math>
  
  
−
</math> (p_{2}+p_{1}) = 8 <math>
+
<math> (p_{2}+p_{1}) = 8 </math>
  
−
</math> (p_{2}-p_{1}) = 6 <math>
+
<math> (p_{2}-p_{1}) = 6 </math>
  
−
</math> p_{1} = 1 <math>  (not prime)
+
<math> p_{1} = 1 </math>  (not prime)
  
−
</math> p_{2} = 7 <math>
+
<math> p_{2} = 7 </math>
  
  
−
The only solution </math> (p_{1} , p_{2}) <math> where both numbers are primes is </math>(11,13)<math>.  
+
The only solution <math> (p_{1} , p_{2}) </math> where both numbers are primes is <math>(11,13)</math>.  
  
−
Therefore the number of [[positive integer]]s </math>n<math> that satisfy both statements is </math>\boxed{\textbf{(B) }1}.$
+
Therefore the number of [[positive integer]]s <math>n</math> that satisfy both statements is <math>\boxed{\textbf{(B) }1}.</math>
  
 
==Solution 3==
 
==Solution 3==

Revision as of 00:15, 9 July 2024

Problem

For each positive integer $n > 1$, let $P(n)$ denote the greatest prime factor of $n$. For how many positive integers $n$ is it true that both $P(n) = \sqrt{n}$ and $P(n+48) = \sqrt{n+48}$?

$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution 1

If $P(n) = \sqrt{n}$, then $n = p_{1}^{2}$, where $p_{1}$ is a prime number.

If $P(n+48) = \sqrt{n+48}$, then $n + 48$ is a square, but we know that n is $p_{1}^{2}$.


This means we just have to check for squares of primes, add $48$ and look whether the root is a prime number. We can easily see that the difference between two consecutive square after $576$ is greater than or equal to $49$, Hence we have to consider only the prime numbers till $23$.

Squaring prime numbers below $23$ including $23$ we get the following list.

$4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529$

But adding $48$ to a number ending with $9$ will result in a number ending with $7$, but we know that a perfect square does not end in $7$, so we can eliminate those cases to get the new list.

$4 , 25 , 121 , 361$

Adding $48$, we get $121$ as the only possible solution. Hence the answer is $\boxed{\textbf{(B) }1}$.

edited by mobius247

Note: Solution 1

Since all primes greater than $2$ are odd, we know that the difference between the squares of any two consecutive primes greater than $2$ is at least $(p+2)^2-p^2=4p+4$, where p is the smaller of the consecutive primes. For $p>11$, $4p+4>48$. This means that the difference between the squares of any two consecutive primes both greater than $11$ is greater than $48$, so $n$ and $n+48$ can't both be the squares of primes if $n=p^2$ and $p>11$. So, we only need to check $n=2^2, 3^2, 5^2, 7^2,$ and $11^2$.

~apsid

Video Solution

https://youtu.be/IsqrsMkR-mA

~rudolf1279

Solution 2

If $P(n) = \sqrt{n}$, then $n = p_{1}^{2}$, where $p_{1}$ is a prime number.

If $P(n+48) = \sqrt{n+48}$, then $n+48 = p_{2}^{2}$, where $p_{2}$ is a different prime number.

So:

$p_{2}^{2} = n+48$

$p_{1}^{2} = n$

$p_{2}^{2} - p_{1}^{2} = 48$

$(p_{2}+p_{1})(p_{2}-p_{1})=48$

Since $p_{1} > 0$ : $(p_{2}+p_{1}) > (p_{2}-p_{1})$.

Looking at pairs of divisors of $48$, we have several possibilities to solve for $p_{1}$ and $p_{2}$:

(Note: you can skip several cases below by observing that $p_1+p_2$ and $p_1-p_{2}$ must be even, and $p_1+p_2 \neq p_1-p_2 (\mod 4)$.

$(p_{2}+p_{1}) = 48$

$(p_{2}-p_{1}) = 1$ (impossible)

$p_{1} = \frac{47}{2}$

$p_{2} = \frac{49}{2}$


$(p_{2}+p_{1}) = 24$

$(p_{2}-p_{1}) = 2$

$p_{1} = 11$

$p_{2} = 13$

(Valid!)

$(p_{2}+p_{1}) = 16$

$(p_{2}-p_{1}) = 3$ (impossible)

$p_{1} = \frac{13}{2}$

$p_{2} = \frac{19}{2}$


$(p_{2}+p_{1}) = 12$

$(p_{2}-p_{1}) = 4$ (impossible)

$p_{1} = 4$

$p_{2} = 8$


$(p_{2}+p_{1}) = 8$

$(p_{2}-p_{1}) = 6$

$p_{1} = 1$ (not prime)

$p_{2} = 7$


The only solution $(p_{1} , p_{2})$ where both numbers are primes is $(11,13)$.

Therefore the number of positive integers $n$ that satisfy both statements is $\boxed{\textbf{(B) }1}.$

Solution 3

For the statement to be true, we must have both $n$ and $n + 48$ be squares of primes. Support we have the number $x^2$, where $x$ is a positive integer. Then the next perfect square, $(x+1)^2$, is $(x+1)^2 - x^2 = 2x+1$ greater than $x^2$. The next perfect square after that will be $(x+2)^2 = 4x + 4$ greater than $x^2$. In general, the prime $(x+n)^2$ will be $nx + n^2$ greater than $x^2$. However, we must have that $nx + n^2 = 48$. $n$ can take on any value between $1$ and $6$ (if $n$ is equal to $7$, we have $14x + 49$, where $x$ would have to be negative for the difference to be $48$). However, we can eliminate all the cases where $n$ is odd, because we would then have a number of the form $even + odd$, which is odd because $x$ can take only integral values. As such, we consider $n = 2$, $n = 4$, and $n = 6$. If $n = 2$, then $4x + 4 = 48 \implies x = 11$. Then our squares are $11^2$ and $13^2$, both of which are squares of primes. If $n = 4$, then $8x + 16 = 48 \implies x = 4$. However, $4$ isn't prime, so we discard this case. Finally, if $n = 6$, then $12x + 36 = 48 \implies x = 1$. Again, $1$ isn't prime, so we discard this case as well. Thus, we only have $\boxed{\textbf{(B)}~1}$ valid case.

~ cxsmi

Video Solution 2

https://youtu.be/-8t5xaths_Q

~savannahsolver

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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