Difference between revisions of "2005 AMC 10A Problems/Problem 24"
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− | 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 23 including 23 we get the following list. | + | Squaring prime numbers below <math>23</math> including <math>23</math> 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 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. | + | 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. |
<math>4 , 25 , 121 , 361</math> | <math>4 , 25 , 121 , 361</math> | ||
− | Adding 48, we get 121 as the only possible solution. | + | Adding <math>48</math>, we get <math>121</math> as the only possible solution. |
− | Hence the answer is (B). | + | Hence the answer is <math>\boxed{\textbf{(B) }121}</math>. |
− | |||
edited by mobius247 | edited by mobius247 |
Revision as of 13:31, 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 2 are odd, we know that the difference between the squares of any two consecutive primes greater than 2 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 11 is greater than 48, so
and
can't both be the squares of primes if
and p>11. So, we only need to check
.
~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.