Difference between revisions of "2000 AMC 10 Problems/Problem 11"
(New page: Two prime numbers between <math>4</math> and <math>18</math> are both odd. odd*odd=odd. odd-odd-odd=odd. Thus, we can discard the even choices. <math>ab-a-b=(a-1)(b-1)-1</math>. <mat...) |
5849206328x (talk | contribs) m |
||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
+ | ==Solution== | ||
+ | |||
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. | ||
Line 17: | Line 21: | ||
<math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>. | <math>11,13</math> satisfy this, <math>11*13-11-13=143-24=119</math>. | ||
− | C | + | C |
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2000|num-b=10|num-a=12}} |
Revision as of 18:40, 8 January 2009
Problem
Solution
Two prime numbers between and are both odd.
odd*odd=odd.
odd-odd-odd=odd.
Thus, we can discard the even choices.
.
.
Both of these factors are even, so the number +1 must be a multiple of .
is the only possiblity.
satisfy this, .
C
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 |