Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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+ | {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}} | ||
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== Problem == | == Problem == | ||
Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? | Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? | ||
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<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } </math> | <math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } </math> | ||
− | == Solution 1 == | + | ==Solution== |
+ | === Solution 1 === | ||
Let the primes be <math>p</math> and <math>q</math>. | Let the primes be <math>p</math> and <math>q</math>. | ||
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But <math>(p-1) \cdot (q-1)=120=2^3\cdot 3 \cdot 5</math> could be <math>2 \cdot 60,4 \cdot 30,6 \cdot 20</math> or <math>10 \cdot 12.</math> Of these, three have <math>p</math> and <math>q</math> prime, but only the last has them both small enough. Therefore the answer is <math> C </math>. | But <math>(p-1) \cdot (q-1)=120=2^3\cdot 3 \cdot 5</math> could be <math>2 \cdot 60,4 \cdot 30,6 \cdot 20</math> or <math>10 \cdot 12.</math> Of these, three have <math>p</math> and <math>q</math> prime, but only the last has them both small enough. Therefore the answer is <math> C </math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
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All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is <math>(13)(17)-(13+17) = 221 - 30 = 191</math>. Thus, we can eliminate E. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is <math>(5)(7)-(5+7) = 23</math>. Therefore, A cannot be an answer. So, the answer must be <math> C </math>. | All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is <math>(13)(17)-(13+17) = 221 - 30 = 191</math>. Thus, we can eliminate E. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is <math>(5)(7)-(5+7) = 23</math>. Therefore, A cannot be an answer. So, the answer must be <math> C </math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2000|num-b=5|num-a=7}} | {{AMC12 box|year=2000|num-b=5|num-a=7}} | ||
+ | {{AMC10 box|year=2000|num-b=10|num-a=12}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 22:43, 26 November 2011
- The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.
Contents
[hide]Problem
Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
Solution
Solution 1
Let the primes be and .
The problem asks us for possible values of where
Using Simon's Favorite Factoring Trick:
Possible values of and are:
The possible values for (formed by multipling two distinct values for and ) are:
So the possible values of are:
The only answer choice on this list is
Note: once we apply the factoring trick we see that, since and are even, should be a multiple of .
These means that only and are possible.
We can't have with and below . Indeed, would have to be or .
But could be or Of these, three have and prime, but only the last has them both small enough. Therefore the answer is .
Solution 2
All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is . Thus, we can eliminate E. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is . Therefore, A cannot be an answer. So, the answer must be .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
2000 AMC 10 (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 10 Problems and Solutions |