Difference between revisions of "2000 AMC 12 Problems/Problem 6"

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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==
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==Solution 1==
  
 
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>\mathrm{(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>\mathrm{(C)}</math>.
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==Solution 2==
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Let the two primes be <math>p</math> and <math>q</math>. We wish to obtain the value of <math>pq-(p+q)</math>, or <math>pq-p-q</math>. Using Simon's Favorite Factoring Trick, we can rewrite this expression as <math>(1-p)(1-q) -1</math> or <math>(p-1)(q-1) -1</math>. Noticing that <math>(13-1)(11-1) - 1 = 120-1 = 119</math>, we see that the answer is <math>\mathrm{(C)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:15, 24 December 2015

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 $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$\mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$

Solution 1

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 $(13)(17)-(13+17) = 221 - 30 = 191$. 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 $(5)(7)-(5+7) = 23$. Therefore, A cannot be an answer. So, the answer must be $\mathrm{(C)}$.


Solution 2

Let the two primes be $p$ and $q$. We wish to obtain the value of $pq-(p+q)$, or $pq-p-q$. Using Simon's Favorite Factoring Trick, we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$. Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$, we see that the answer is $\mathrm{(C)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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