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

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== Solution 2 ==
 
== Solution 2 ==
  
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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. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is 221 - 30, or 191. Thus, we can eliminate E. Therefore, the answer is <math> C </math>.
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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 ==
 
== See also ==

Revision as of 23:48, 21 February 2011

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

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 $C$.

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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