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 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. | 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. |
Revision as of 15:51, 30 April 2021
- The following problem is from both the 2002 AMC 12B #11 and 2002 AMC 10B #15, so both problems redirect to this page.
Contents
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
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.