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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>
  
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== Solution ==
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== 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>.
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== Solution 2 ==
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Since 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 A, B, and D. We can eliminate E because the highest two prime we can pick are 13 and 17. their product is 221 and their sum is 30, so the highest number we can make is 191. Therefore, the answer is <math> C </math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:29, 19 October 2010

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

Let the primes be $p$ and $q$.

The problem asks us for possible values of $K$ where $K=pq-p-q$

Using Simon's Favorite Factoring Trick:

$K+1=pq-p-q+1$

$K+1=(p-1)(q-1)$

Possible values of $(p-1)$ and $(q-1)$ are:

$4,6,10,12,16$

The possible values for $K+1$ (formed by multipling two distinct values for $(p-1)$ and $(q-1)$) are:

$24,40,48,60,64,72,96,120,160,192$

So the possible values of $K$ are:

$23,39,47,59,63,71,95,119,159,191$

The only answer choice on this list is $119 \Rightarrow C$

Note: once we apply the factoring trick we see that, since $p-1$ and $q-1$ are even, $K+1$ should be a multiple of $4$.

These means that only $119 \Rightarrow C$ and $231 \Rightarrow E$ are possible.

We can't have $(p-1) \cdot (q-1)=232=2^3\cdot 29$ with $p$ and $q$ below $18$. Indeed, $(p-1) \cdot (q-1)$ would have to be $2 \cdot 116$ or $4 \cdot 58$.

But $(p-1) \cdot (q-1)=120=2^3\cdot 3 \cdot 5$ could be $2 \cdot 60,4 \cdot 30,6 \cdot 20$ or $10 \cdot 12.$ Of these, three have $p$ and $q$ prime, but only the last has them both small enough. Therefore the answer is $C$.

Solution 2

Since 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 A, B, and D. We can eliminate E because the highest two prime we can pick are 13 and 17. their product is 221 and their sum is 30, so the highest number we can make is 191. Therefore, the answer is $C$.

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
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