Difference between revisions of "2000 AMC 12 Problems/Problem 6"
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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>. | 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> | + | 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>. |
== See also == | == See also == |
Revision as of 20:29, 31 January 2009
Problem
Two different prime numbers between and
are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
Solution
Let the primes be and
.
The problem asks us for possible values of where
Using Simon's Favorite Factoring Trick:
Possible values of and
are:
The possible values for (formed by multipling two distinct values for
and
) are:
So the possible values of are:
The only answer choice on this list is
Note: once we apply the factoring trick we see that, since and
are even,
should be a multiple of
.
These means that only and
are possible.
We can't have with
and
below
. Indeed,
would have to be
or
.
But could be
or
Of these, three have
and
prime, but only the last has them both small enough. Therefore the answer is
.
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 |