2005 AMC 10A Problems/Problem 24
Contents
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
Solution 3
For the statement to be true, we must have both and be squares of primes. Support we have the number , where is a positive integer. Then the next perfect square, , is greater than . The next perfect square after that will be greater than . In general, the prime will be greater than . However, we must have that . can take on any value between and (if is equal to , we have , where would have to be negative for the difference to be ). However, we can eliminate all the cases where is odd, because we would then have a number of the form , which is odd because can take only integral values. As such, we consider , , and . If , then . Then our squares are and , both of which are squares of primes. If , then . However, isn't prime, so we discard this case. Finally, if , then . Again, isn't prime, so we discard this case as well. Thus, we only have valid case.
~ cxsmi
Video Solution 2
~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 | ||
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