Difference between revisions of "2002 AMC 12B Problems/Problem 11"
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\qquad\mathrm{(E)}\ \mathrm{prime}</math> | \qquad\mathrm{(E)}\ \mathrm{prime}</math> | ||
− | == Solution == | + | == Solution 1 == |
− | Since <math>A-B</math> and <math>A+B</math> must have the same [[parity]], and since there is only one even prime number, it follows that <math>A-B</math> and <math>A+B</math> are both odd. Thus one of <math>A, B</math> is odd and the other even. Since <math>A+B > A > A-B > 2</math>, it follows that <math>A</math> (as a prime greater than <math>2</math>) is odd. Thus <math>B = 2</math>, and <math>A-2, A, A+2</math> are consecutive odd primes. At least one of <math>A-2, A, A+2</math> is divisible by <math>3</math>, from which it follows that <math>A-2 = 3</math> and <math>A = 5</math>. The sum of these numbers is thus <math>17</math>, a prime, so the answer is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>. | + | |
+ | Since <math>A-B</math> and <math>A+B</math> must have the same [[parity]], and since there is only one even prime number, it follows that <math>A-B</math> and <math>A+B</math> are both odd. Thus one of <math>A, B</math> is odd and the other even. Since <math>A+B > A > A-B > 2</math>, it follows that <math>A</math> (as a prime greater than <math>2</math>) is odd. Thus <math>B = 2</math>, and <math>A-2, A, A+2</math> are consecutive odd primes. At least one of <math>A-2, A, A+2</math> is divisible by <math>3</math>, from which it follows that <math>A-2 = 3</math> and <math>A = 5</math>. The sum of these numbers is thus <math>17</math>, a prime, so the answer is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | In order for both <math>A - B</math> and <math>A + B</math> to be prime, one of <math>A, B</math> must be 2, or else both <math>A - B</math>, <math>A + B</math> would be even numbers. | ||
+ | |||
+ | If <math>A = 2</math>, then <math>A < B</math> and <math>A - B < 0</math>, which is not possible. Thus <math>B = 2</math>. | ||
+ | |||
+ | Since <math>A</math> is prime and <math>A > A - B > 2</math>, we can infer that <math>A > 3</math> and thus <math>A</math> can be expressed as <math>6n \pm 1</math> for some natural number <math>n</math>. | ||
+ | |||
+ | However in either case, one of <math>A - B</math> and <math>A + B</math> can be expressed as <math>6n \pm 3 = 3(2n \pm 1)</math> which is a multiple of 3. Therefore the only possibility that works is when <math>A - B = 3</math> and <cmath>A + B + (A - B) + (A + B) = 5 + 2 + 3 + 7 = 17</cmath> | ||
+ | |||
+ | Which is a prime number. <math>\boxed{(E)}</math> | ||
+ | |||
+ | ~ Nafer | ||
+ | |||
+ | == Solution 3 (intuitive)== | ||
+ | |||
+ | Trying out some primes for <math>A</math> and <math>B</math> such that <math>A-B</math> and <math>A+B</math> are prime, <math>A=5</math> and <math>B=2</math> can be found almost immediately. Summing the four primes, the result is <math>5+2+3+7=17</math>, which is <math>\boxed{\mathrm{(E)}\ \text{prime}}</math>. | ||
== See also == | == See also == | ||
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | Note: | ||
+ | Simple trail and error gives us the primes 5 and 2 which fits the description the question asks for; 5, 2, 3, 7 are all primes. |
Latest revision as of 11:57, 10 August 2022
- The following problem is from both the 2002 AMC 12B #11 and 2002 AMC 10B #15, so both problems redirect to this page.
Problem
The positive integers and
are all prime numbers. The sum of these four primes is
Solution 1
Since and
must have the same parity, and since there is only one even prime number, it follows that
and
are both odd. Thus one of
is odd and the other even. Since
, it follows that
(as a prime greater than
) is odd. Thus
, and
are consecutive odd primes. At least one of
is divisible by
, from which it follows that
and
. The sum of these numbers is thus
, a prime, so the answer is
.
Solution 2
In order for both and
to be prime, one of
must be 2, or else both
,
would be even numbers.
If , then
and
, which is not possible. Thus
.
Since is prime and
, we can infer that
and thus
can be expressed as
for some natural number
.
However in either case, one of and
can be expressed as
which is a multiple of 3. Therefore the only possibility that works is when
and
Which is a prime number.
~ Nafer
Solution 3 (intuitive)
Trying out some primes for and
such that
and
are prime,
and
can be found almost immediately. Summing the four primes, the result is
, which is
.
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Note: Simple trail and error gives us the primes 5 and 2 which fits the description the question asks for; 5, 2, 3, 7 are all primes.