2002 AMC 12B Problems/Problem 11
- 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
Solution1 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
,
and
can be expressed as
for some natural number
. Moreover for all
,
,
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 |
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