Difference between revisions of "2005 AMC 10B Problems/Problem 13"
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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>? | 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>\ | + | <math>\textbf{(A) } 501 \qquad \textbf{(B) } 668 \qquad \textbf{(C) } 835 \qquad \textbf{(D) } 1002 \qquad \textbf{(E) } 1169 </math> |
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== Solution 1 == | == 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{\ | + | 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 == | == Solution 2 == | ||
− | From 1-12, the multiples of 3 or 4 but not 12 are 3, 4, 6, 8, and 9, a total of five numbers. Since <math>\frac{5}{12}</math> of positive integers are multiples of 3 or 4 but not 12, 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 14:44, 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 |
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