Difference between revisions of "2015 AIME I Problems/Problem 3"
Pi is 3.141 (talk | contribs) (→Solution 5) |
Firebolt360 (talk | contribs) |
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | ==Solution 2== | + | ==Solution 2 (Similar to 1)== |
+ | |||
+ | Observe that this is the same as <math>16p+1=n^3</math> for some integer <math>n</math>. | ||
+ | So: | ||
+ | <cmath>\begin{align*} | ||
+ | 16p &= n^3-1\\ | ||
+ | 16p &= n^3-1^3\\ | ||
+ | 16p &= (n-1)(n^2+n+1)\\ | ||
+ | \end{align*}</cmath> | ||
+ | Observe that either <math>p=n-1</math> or <math>p=n^2+n+1</math> because <math>p</math> and <math>16</math> share no factors (<math>p</math> can't be <math>2</math>). | ||
+ | Let <math>p=n-1</math>. | ||
+ | Then: | ||
+ | <cmath>\begin{align*} | ||
+ | p &= n-1\\ | ||
+ | 16 &= n^2+n+1\\ | ||
+ | n^2+n &= 15\\ | ||
+ | n(n+1) &= 15\\ | ||
+ | \end{align*}</cmath> | ||
+ | Which is obviously impossible for integer n. So <math>p=n^2+n+1</math> and | ||
+ | <cmath>\begin{align*} | ||
+ | 16 &= n-1\\ | ||
+ | n &= 17\\ | ||
+ | p &= 17^2+17+1\\ | ||
+ | p = 289+17+1 &= \boxed{307}\\ | ||
+ | \end{align*}</cmath> - firebolt360 | ||
+ | |||
+ | ==Solution 3== | ||
Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>. Therefore, <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>. From this, we get <math>8p=a(4a^2+6a+3)</math>. We know <math>p</math> is a prime number and it is not an even number. Since <math>4a^2+6a+3</math> is an odd number, we know that <math>a=8</math>. | Since <math>16p+1</math> is odd, let <math>16p+1 = (2a+1)^3</math>. Therefore, <math>16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1</math>. From this, we get <math>8p=a(4a^2+6a+3)</math>. We know <math>p</math> is a prime number and it is not an even number. Since <math>4a^2+6a+3</math> is an odd number, we know that <math>a=8</math>. | ||
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Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>. | Therefore, <math>p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}</math>. | ||
− | ==Solution | + | ==Solution 4== |
Let <math>16p+1=a^3</math>. Realize that <math>a</math> congruent to <math>1\mod 4</math>, so let <math>a=4n+1</math>. Expansion, then division by 4, gets <math>16n^3+12n^2+3n=4p</math>. Clearly <math>n=4m</math> for some <math>m</math>. Substitution and another division by 4 gets <math>256m^3+48m^4+3m=p</math>. Since <math>p</math> is prime and there is a factor of <math>m</math> in the LHS, <math>m=1</math>. Therefore, <math>p=\boxed{307}</math>. | Let <math>16p+1=a^3</math>. Realize that <math>a</math> congruent to <math>1\mod 4</math>, so let <math>a=4n+1</math>. Expansion, then division by 4, gets <math>16n^3+12n^2+3n=4p</math>. Clearly <math>n=4m</math> for some <math>m</math>. Substitution and another division by 4 gets <math>256m^3+48m^4+3m=p</math>. Since <math>p</math> is prime and there is a factor of <math>m</math> in the LHS, <math>m=1</math>. Therefore, <math>p=\boxed{307}</math>. | ||
− | ==Solution | + | ==Solution 5== |
Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>. | Notice that <math>16p+1</math> must be in the form <math>(a+1)^3 = a^3 + 3a^2 + 3a + 1</math>. Thus <math>16p = a^3 + 3a^2 + 3a</math>, or <math>16p = a\cdot (a^2 + 3a + 3)</math>. Since <math>p</math> must be prime, we either have <math>p = a</math> or <math>a = 16</math>. Upon further inspection and/or using the quadratic formula, we can deduce <math>p \neq a</math>. Thus we have <math>a = 16</math>, and <math>p = 16^2 + 3\cdot 16 + 3 = \boxed{307}</math>. | ||
− | ==Solution | + | ==Solution 6== |
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. | Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. | ||
Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 | Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 |
Revision as of 13:48, 30 May 2020
Contents
Problem
There is a prime number such that
is the cube of a positive integer. Find
.
Solution 1
Let the positive integer mentioned be , so that
. Note that
must be odd, because
is odd.
Rearrange this expression and factor the left side (this factoring can be done using or synthetic divison once it is realized that
is a root):
Because is odd,
is even and
is odd. If
is odd,
must be some multiple of
. However, for
to be any multiple of
other than
would mean
is not a prime. Therefore,
and
.
Then our other factor, , is the prime
:
Solution 2 (Similar to 1)
Observe that this is the same as for some integer
.
So:
Observe that either
or
because
and
share no factors (
can't be
).
Let
.
Then:
Which is obviously impossible for integer n. So
and
- firebolt360
Solution 3
Since is odd, let
. Therefore,
. From this, we get
. We know
is a prime number and it is not an even number. Since
is an odd number, we know that
.
Therefore, .
Solution 4
Let . Realize that
congruent to
, so let
. Expansion, then division by 4, gets
. Clearly
for some
. Substitution and another division by 4 gets
. Since
is prime and there is a factor of
in the LHS,
. Therefore,
.
Solution 5
Notice that must be in the form
. Thus
, or
. Since
must be prime, we either have
or
. Upon further inspection and/or using the quadratic formula, we can deduce
. Thus we have
, and
.
Solution 6
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here
Hence, is our only answer
pi_is_3.141
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.