Difference between revisions of "2000 AMC 10 Problems/Problem 11"
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Two prime numbers between <math>4</math> and <math>18</math> are both odd. | Two prime numbers between <math>4</math> and <math>18</math> are both odd. | ||
| − | odd | + | <math>\text{odd}\cdot \text{odd}=\text{odd}</math> |
| − | odd-odd-odd=odd | + | <math>\text{odd}-\text{odd}-\text{odd}=\text{odd}</math> |
Thus, we can discard the even choices. | Thus, we can discard the even choices. | ||
| − | <math>ab-a-b=(a-1)(b-1)-1</math> | + | <math>ab-a-b=(a-1)(b-1)-1</math> |
| − | <math> | + | Both <math>a-1</math> and <math>b-1</math> are even, so one more than <math>ab-a-b</math> is a multiple of four. |
| − | + | <math>119</math> is the only possible choice. | |
| − | <math> | + | <math>11,13</math> satisfy this, <math>11\cdot 13-11-13=143-24=119</math>. |
| − | <math> | + | <math>\boxed{\text{C}}</math> |
| − | |||
| − | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=10|num-a=12}} | {{AMC10 box|year=2000|num-b=10|num-a=12}} | ||
Revision as of 21:44, 9 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
Two prime numbers between and are both odd.
Thus, we can discard the even choices.
Both 
and 
are even, so one more than 
is a multiple of four.
is the only possible choice.
satisfy this, 
.
See Also
| 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 | ||
