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

Latest revision as of 01:47, 13 August 2024

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

Video by RMM Club

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

Video Solution by Daily Dose of Math

https://youtu.be/xmiLZVy0JVM?si=DJitB2xBZOuyMkWu

~Thesmartgreekmathdude

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-== Problem ==
+{{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>
+==Problem==
-== Solution 1 ==
+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>
+<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]]:
+==Solution 1==
-<math>K+1=pq-p-q+1</math>
+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>
+==Solution 2==
-Possible values of <math>(p-1)</math> and <math>(q-1)</math> are:
+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>
+==Video by RMM Club==
+https://www.youtube.com/watch?v=ddE5GO1RNLw&t=1s
-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:
+==Video Solution by Daily Dose of Math==
-<math>24,40,48,60,64,72,96,120,160,192</math>
+https://youtu.be/xmiLZVy0JVM?si=DJitB2xBZOuyMkWu
-So the possible values of <math>K</math> are:
+~Thesmartgreekmathdude
-<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>
-Note: once we apply the factoring trick we see that, since <math>p-1</math> and <math>q-1</math> are even, <math>K+1</math> should be a multiple of <math>4</math>.
-These means that only <math> 119 \Rightarrow C </math> and <math>231 \Rightarrow E</math> are possible.
-We can't have <math>(p-1) \cdot (q-1)=232=2^3\cdot 29</math> with <math>p</math> and <math>q</math> below <math>18</math>. Indeed, <math>(p-1) \cdot (q-1)</math> would have to be <math>2 \cdot 116</math> or <math>4 \cdot 58</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 ==
-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 ==
+{{AMC10 box|year=2000|num-b=10|num-a=12}}
 {{AMC12 box|year=2000|num-b=5|num-a=7}}
+{{MAA Notice}}
-[[Category:Introductory Algebra Problems]]
+[[Category:Introductory Number Theory Problems]]

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