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 23:43, 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.
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 | ||
