Difference between revisions of "2005 AMC 10A Problems/Problem 24"
(A note about a quicker solution using similar methods to Solution 1) |
Dairyqueenxd (talk | contribs) (→Solution 2) |
||
(5 intermediate revisions by 2 users not shown) | |||
Line 2: | Line 2: | ||
For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>? | For each positive integer <math> n > 1 </math>, let <math>P(n)</math> denote the greatest prime factor of <math>n</math>. For how many positive integers <math>n</math> is it true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>? | ||
− | <math> \ | + | <math> \textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5 </math> |
==Solution 1== | ==Solution 1== | ||
Line 10: | Line 10: | ||
− | This means we just have to check for squares of primes, add 48 and look whether the root is a prime number. | + | This means we just have to check for squares of primes, add <math>48</math> 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, | + | We can easily see that the difference between two consecutive square after <math>576</math> is greater than or equal to <math>49</math>, |
− | Hence we have to consider only the prime numbers till 23. | + | Hence we have to consider only the prime numbers till <math>23</math>. |
− | + | Squaring prime numbers below <math>23</math> including <math>23</math> we get the following list. | |
− | Squaring prime numbers below 23 including 23 we get the following list. | ||
<math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math> | <math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math> | ||
− | + | But adding <math>48</math> to a number ending with <math>9</math> will result in a number ending with <math>7</math>, but we know that a perfect square does not end in <math>7</math>, so we can eliminate those cases to get the new list. | |
− | 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. | ||
<math>4 , 25 , 121 , 361</math> | <math>4 , 25 , 121 , 361</math> | ||
− | + | Adding <math>48</math>, we get <math>121</math> as the only possible solution. | |
− | Adding 48, we get 121 as the only possible solution. | + | Hence the answer is <math>\boxed{\textbf{(B) }1}</math>. |
− | Hence the answer is (B) | ||
− | |||
− | |||
edited by mobius247 | edited by mobius247 | ||
==Note: Solution 1== | ==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 <math>(p+2)^2-p^2=4p+4</math>, where p is the smaller of the consecutive primes. For <math>p>11</math>, <math>4p+4>48</math>. This means that the difference between the squares of any two consecutive primes both greater than 11 is greater than 48, so <math>n</math> and <math>n+48</math> can't both be the squares of primes if <math>n=p^2</math> and p>11. So, we only need to check <math>n=2^2, 3^2, 5^2, 7^2, 11^2</math>. | + | Since all primes greater than <math>2</math> are odd, we know that the difference between the squares of any two consecutive primes greater than <math>2</math> is at least <math>(p+2)^2-p^2=4p+4</math>, where p is the smaller of the consecutive primes. For <math>p>11</math>, <math>4p+4>48</math>. This means that the difference between the squares of any two consecutive primes both greater than <math>11</math> is greater than <math>48</math>, so <math>n</math> and <math>n+48</math> can't both be the squares of primes if <math>n=p^2</math> and <math>p>11</math>. So, we only need to check <math>n=2^2, 3^2, 5^2, 7^2,</math> and <math>11^2</math>. |
+ | |||
~apsid | ~apsid | ||
Line 59: | Line 55: | ||
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>: | ||
− | |||
<math> (p_{2}+p_{1}) = 48 </math> | <math> (p_{2}+p_{1}) = 48 </math> | ||
Line 108: | Line 103: | ||
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> | + | Therefore the number of [[positive integer]]s <math>n</math> that satisfy both statements is <math>\boxed{\textbf{(B) }1}.</math> |
==See Also== | ==See Also== |
Revision as of 13:36, 14 December 2021
Problem
For each positive integer , let denote the greatest prime factor of . For how many positive integers is it true that both and ?
Solution 1
If , then , where is a prime number.
If , then is a square, but we know that n is .
This means we just have to check for squares of primes, add and look whether the root is a prime number.
We can easily see that the difference between two consecutive square after is greater than or equal to ,
Hence we have to consider only the prime numbers till .
Squaring prime numbers below including we get the following list.
But adding to a number ending with will result in a number ending with , but we know that a perfect square does not end in , so we can eliminate those cases to get the new list.
Adding , we get as the only possible solution. Hence the answer is .
edited by mobius247
Note: Solution 1
Since all primes greater than are odd, we know that the difference between the squares of any two consecutive primes greater than is at least , where p is the smaller of the consecutive primes. For , . This means that the difference between the squares of any two consecutive primes both greater than is greater than , so and can't both be the squares of primes if and . So, we only need to check and .
~apsid
Video Solution
CHECK OUT Video Solution:https://youtu.be/IsqrsMkR-mA
~rudolf1279
Solution 2
If , then , where is a prime number.
If , then , where is a different prime number.
So:
Since : .
Looking at pairs of divisors of , we have several possibilities to solve for and :
The only solution where both numbers are primes is .
Therefore the number of positive integers that satisfy both statements is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.