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...) |
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| + | ==Problem== | ||
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| + | ==Solution== | ||
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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. | ||
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<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 |
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2000|num-b=10|num-a=12}} | ||
Revision as of 19:40, 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 | ||
