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

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== Problem ==
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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}}
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?
 
  
<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }  </math>
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==Problem==
  
== Solution ==
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Two different prime numbers 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?
Let the primes be <math>p</math> and <math>q</math>.
 
  
The problem asks us for possible values of <math>K</math> where <math>K=pq-p-q</math>
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<math>\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231</math>
  
Using [[Simon's Favorite Factoring Trick]]:
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==Solution 1==
  
<math>K+1=pq-p-q+1</math>
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Any two 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 A, 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. So, the answer must be <math>\boxed{\textbf{(C) }119}</math>.
  
<math>K+1=(p-1)(q-1)</math>
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==Solution 2==
  
Possible values of <math>(p-1)</math> and <math>(q-1)</math> are:
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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>\boxed{\textbf{(C) }119}</math>.
  
<math>4,6,10,12,16</math>
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==Solution 3==
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The answer must be in the form <math>pq - p - q</math> = <math>(p - 1)(q - 1) - 1</math>. Since <math>p - 1</math> and <math>q - 1</math> are both even, <math>(p - 1)(q - 1) - 1</math> is <math>3 \pmod 4</math>, and the only answer that is <math>3 \pmod 4</math> is <math>\boxed{\textbf{(C) }119}</math>.
  
The possible values for <math>K+1</math> (formed by multipling two distinct values for <math>(p-1)</math> and <math>(q-1)</math>) are:
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==Videos:==
 
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https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s
<math>24,40,48,60,64,72,96,120,160,192</math>
 
 
 
So the possible values of <math>K</math> are:
 
 
 
<math>23,39,47,59,63,71,95,119,159,191</math>
 
 
 
The only answer choice on this list is <math> 119 \Rightarrow C </math>
 
  
 
== See also ==
 
== See also ==
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{{AMC10 box|year=2000|num-b=10|num-a=12}}
 
{{AMC12 box|year=2000|num-b=5|num-a=7}}
 
{{AMC12 box|year=2000|num-b=5|num-a=7}}
 
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{{MAA Notice}}
[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]

Revision as of 18:18, 31 December 2021

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?

$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$

Solution 1

Any two 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 A, 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. So, the answer must be $\boxed{\textbf{(C) }119}$.

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 $\boxed{\textbf{(C) }119}$.

Solution 3

The answer must be in the form $pq - p - q$ = $(p - 1)(q - 1) - 1$. Since $p - 1$ and $q - 1$ are both even, $(p - 1)(q - 1) - 1$ is $3 \pmod 4$, and the only answer that is $3 \pmod 4$ is $\boxed{\textbf{(C) }119}$.

Videos:

https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s

See also

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
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

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