Difference between revisions of "2005 AMC 10A Problems/Problem 24"
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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>: | ||
− | |||
<math> (p_{2}+p_{1}) = 48 </math> | <math> (p_{2}+p_{1}) = 48 </math> | ||
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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 |
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