Difference between revisions of "2005 AMC 10A Problems/Problem 24"
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==Video Solution== | ==Video Solution== | ||
− | + | https://youtu.be/IsqrsMkR-mA | |
~<i>rudolf1279</i> | ~<i>rudolf1279</i> | ||
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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 <math>p_1+p_2 \neq p_1-p_2 \pmod 4 </math>.) | ||
<math> (p_{2}+p_{1}) = 48 </math> | <math> (p_{2}+p_{1}) = 48 </math> | ||
− | <math> (p_{2}-p_{1}) = 1 </math> | + | <math> (p_{2}-p_{1}) = 1 </math> (impossible) |
<math> p_{1} = \frac{47}{2} </math> | <math> p_{1} = \frac{47}{2} </math> | ||
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<math> p_{1} = 11 </math> | <math> p_{1} = 11 </math> | ||
− | <math> p_{2} = 13 </math> | + | <math> p_{2} = 13 </math> (Valid!) |
<math> (p_{2}+p_{1}) = 16 </math> | <math> (p_{2}+p_{1}) = 16 </math> | ||
− | <math> (p_{2}-p_{1}) = 3 </math> | + | <math> (p_{2}-p_{1}) = 3 </math> (impossible) |
<math> p_{1} = \frac{13}{2} </math> | <math> p_{1} = \frac{13}{2} </math> | ||
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− | <math> (p_{2}+p_{1}) = 12 </math> | + | <math> (p_{2}+p_{1}) = 12 </math> |
− | <math> (p_{2}-p_{1}) = 4 </math> | + | <math> (p_{2}-p_{1}) = 4 </math> (impossible) |
<math> p_{1} = 4 </math> | <math> p_{1} = 4 </math> | ||
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<math> (p_{2}-p_{1}) = 6 </math> | <math> (p_{2}-p_{1}) = 6 </math> | ||
− | <math> p_{1} = 1 </math> | + | <math> p_{1} = 1 </math> (not prime) |
<math> p_{2} = 7 </math> | <math> p_{2} = 7 </math> | ||
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For the statement to be true, we must have both <math>n</math> and <math>n + 48</math> be squares of primes. Support we have the number <math>x^2</math>, where <math>x</math> is a positive integer. Then the next perfect square, <math>(x+1)^2</math>, is <math>(x+1)^2 - x^2 = 2x+1</math> greater than <math>x^2</math>. The next perfect square after that will be <math>(x+2)^2 = 4x + 4</math> greater than <math>x^2</math>. In general, the prime <math>(x+n)^2</math> will be <math>nx + n^2</math> greater than <math>x^2</math>. However, we must have that <math>nx + n^2 = 48</math>. <math>n</math> can take on any value between <math>1</math> and <math>6</math> (if <math>n</math> is equal to <math>7</math>, we have <math>14x + 49</math>, where <math>x</math> would have to be negative for the difference to be <math>48</math>). However, we can eliminate all the cases where <math>n</math> is odd, because we would then have a number of the form <math>even + odd</math>, which is odd because <math>x</math> can take only integral values. As such, we consider <math>n = 2</math>, <math>n = 4</math>, and <math>n = 6</math>. If <math>n = 2</math>, then <math>4x + 4 = 48 \implies x = 11</math>. Then our squares are <math>11^2</math> and <math>13^2</math>, both of which are squares of primes. If <math>n = 4</math>, then <math>8x + 16 = 48 \implies x = 4</math>. However, <math>4</math> isn't prime, so we discard this case. Finally, if <math>n = 6</math>, then <math>12x + 36 = 48 \implies x = 1</math>. Again, <math>1</math> isn't prime, so we discard this case as well. Thus, we only have <math>\boxed{\textbf{(B)}~1}</math> valid case. | For the statement to be true, we must have both <math>n</math> and <math>n + 48</math> be squares of primes. Support we have the number <math>x^2</math>, where <math>x</math> is a positive integer. Then the next perfect square, <math>(x+1)^2</math>, is <math>(x+1)^2 - x^2 = 2x+1</math> greater than <math>x^2</math>. The next perfect square after that will be <math>(x+2)^2 = 4x + 4</math> greater than <math>x^2</math>. In general, the prime <math>(x+n)^2</math> will be <math>nx + n^2</math> greater than <math>x^2</math>. However, we must have that <math>nx + n^2 = 48</math>. <math>n</math> can take on any value between <math>1</math> and <math>6</math> (if <math>n</math> is equal to <math>7</math>, we have <math>14x + 49</math>, where <math>x</math> would have to be negative for the difference to be <math>48</math>). However, we can eliminate all the cases where <math>n</math> is odd, because we would then have a number of the form <math>even + odd</math>, which is odd because <math>x</math> can take only integral values. As such, we consider <math>n = 2</math>, <math>n = 4</math>, and <math>n = 6</math>. If <math>n = 2</math>, then <math>4x + 4 = 48 \implies x = 11</math>. Then our squares are <math>11^2</math> and <math>13^2</math>, both of which are squares of primes. If <math>n = 4</math>, then <math>8x + 16 = 48 \implies x = 4</math>. However, <math>4</math> isn't prime, so we discard this case. Finally, if <math>n = 6</math>, then <math>12x + 36 = 48 \implies x = 1</math>. Again, <math>1</math> isn't prime, so we discard this case as well. Thus, we only have <math>\boxed{\textbf{(B)}~1}</math> valid case. | ||
− | ~ cxsmi | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] |
==Video Solution 2== | ==Video Solution 2== |
Latest revision as of 01:37, 9 July 2024
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
~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
:
(Note: you can skip several cases below by observing that and
must be even, and
.)
(impossible)
(Valid!)
(impossible)
(impossible)
(not prime)
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.