Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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== Problem == | == Problem == | ||
| − | == Solution == | + | How many numbers between <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>? |
| + | |||
| + | <math>\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169 </math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | |||
| + | To find the multiples of <math>3</math> or <math>4</math> but not <math>12</math>, you need to find the number of multiples of <math>3</math> and <math>4</math>, and then subtract twice the number of multiples of <math>12</math>, because you overcount and do not want to include them. The multiples of <math>3</math> are <math>\frac{2005}{3} = 668\text{ }R1.</math> The multiples of <math>4</math> are <math>\frac{2005}{4} = 501 \text{ }R1</math>. The multiples of <math>12</math> are <math>\frac{2005}{12} = 167\text{ }R1.</math> So, the answer is <math>668+501-167-167 = \boxed{\textbf{(C) } 835}</math> | ||
| + | |||
| + | == 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{\textbf{(C) }835}</math> | ||
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== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=12|num-a=14}} | |
| + | {{MAA Notice}} | ||
Latest revision as of 13:44, 15 December 2021
Contents
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
