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{\textbf{(C) }835}</math> | 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> | ||
| + | |||
| + | \textbf{Solution 3:} | ||
| + | |||
| + | For every multiple of 12, there are a total of 7 multiples of 3 and 4 combined. However, we do not want to include the 2 multiples of 12, so we subtract 2 from 7, leaving us with 5. | ||
| + | |||
| + | We divide 2005 by 12 to determine how many such blocks exist: | ||
| + | <cmath>\left\lfloor \frac{2005}{12} \right\rfloor = 167</cmath> | ||
| + | |||
| + | Multiplying that by 5, we get: | ||
| + | <cmath>167 \times 5 = 835</cmath> | ||
| + | |||
| + | So our answer is:<cmath>\boxed{\textbf{(C) } 835}</cmath> | ||
| + | |||
| + | -LittleWavelet | ||
== 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}} | ||
Revision as of 18:19, 6 August 2025
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 
= 
\textbf{Solution 3:}
For every multiple of 12, there are a total of 7 multiples of 3 and 4 combined. However, we do not want to include the 2 multiples of 12, so we subtract 2 from 7, leaving us with 5.
We divide 2005 by 12 to determine how many such blocks exist:
![\[\left\lfloor \frac{2005}{12} \right\rfloor = 167\]](http://latex.artofproblemsolving.com/4/4/6/4462c6552a0fd51def60049af75378523c0c981d.png)
Multiplying that by 5, we get:
![\[167 \times 5 = 835\]](http://latex.artofproblemsolving.com/a/e/3/ae30528617ff5d566b333b481c8a17e894661d49.png)
So our answer is:![\[\boxed{\textbf{(C) } 835}\]](http://latex.artofproblemsolving.com/e/7/f/e7ffb4df50c1218013662f8fddf96d29cf36cddd.png)
-LittleWavelet
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
