Difference between revisions of "2005 AMC 10B Problems/Problem 22"
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− | + | How many integers <math>n</math> satisfy both of the inequalities <math>4n + 3 < 25</math> and <math>-7n + 5 < 24</math>? | |
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<math>\text{(A) 8 } \text{(B) 12 } \text{(C) 16 } \text{(D) 17 } \text{(E) 21 }</math> | <math>\text{(A) 8 } \text{(B) 12 } \text{(C) 16 } \text{(D) 17 } \text{(E) 21 }</math> |
Revision as of 17:47, 19 October 2020
Problem
How many integers satisfy both of the inequalities and ?
Solution
Since , the condition is equivalent to having an integer value for . This reduces, when , to having an integer value for . This fraction is an integer unless is an odd prime. There are 8 odd primes less than or equal to 24, so there are numbers less than or equal to 24 that satisfy the condition.
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
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