Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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== Solution 2 == | == Solution 2 == | ||
− | From <math>1</math>-<math>12</math>, the multiples of <math>3</math> or <math>4</math> but not <math>12</math> are <math>3, 4, 6, 8, </math>and <math>9</math>, a total of five numbers. Since <math>\frac{5}{12}</math> of positive integers are multiples of <math>3</math> or <math>4</math> but not <math>12</math> from <math>1</math>-<math>12</math>, the answer is approximately <math>\frac{5}{12} \cdot 2005</math> = <math>\boxed{\ | + | From <math>1</math>-<math>12</math>, the multiples of <math>3</math> or <math>4</math> but not <math>12</math> are <math>3, 4, 6, 8, </math>and <math>9</math>, a total of five numbers. Since <math>\frac{5}{12}</math> of positive integers are multiples of <math>3</math> or <math>4</math> but not <math>12</math> from <math>1</math>-<math>12</math>, the answer is approximately <math>\frac{5}{12} \cdot 2005</math> = <math>\boxed{\textbf{(C) }835}</math> |
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:44, 15 December 2021
Contents
[hide]Problem
How many numbers between and are integer multiples of or but not ?
Solution 1
To find the multiples of or but not , you need to find the number of multiples of and , and then subtract twice the number of multiples of , because you overcount and do not want to include them. The multiples of are The multiples of are . The multiples of are So, the answer is
Solution 2
From -, the multiples of or but not are and , a total of five numbers. Since of positive integers are multiples of or but not from -, the answer is approximately =
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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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