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

Revision as of 17:55, 6 January 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

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 $C$ .

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

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

@@ Line 7: / Line 7: @@
 ==Solution==
-=== Solution 1 ===
-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>
-Using [[Simon's Favorite Factoring Trick]]:
-<math>K+1=pq-p-q+1</math>
-<math>K+1=(p-1)(q-1)</math>
-Possible values of <math>(p-1)</math> and <math>(q-1)</math> are:
-<math>4,6,10,12,16</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:
-<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>
-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>.

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

Revision as of 17:55, 6 January 2015

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Solution

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