Difference between revisions of "2012 AMC 8 Problems/Problem 10"

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<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12 </math>
 
<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12 </math>
  
==Solution==
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==Solution 1==
 
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of <math> 2012 </math>, since all of the valid 4-digit number will always be greater than <math> 1000 </math>. The best way to solve this problem is by using casework.
 
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of <math> 2012 </math>, since all of the valid 4-digit number will always be greater than <math> 1000 </math>. The best way to solve this problem is by using casework.
  
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Summing the amounts of numbers, we find that there are <math> \boxed{\textbf{(D)}\ 9} </math> such numbers.
 
Summing the amounts of numbers, we find that there are <math> \boxed{\textbf{(D)}\ 9} </math> such numbers.
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==Solution 2==
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Notice that the first digit cannot be <math> 0 </math>, as the number is greater than <math> 1000 </math>. Therefore, there are three digits that can be in the thousands.
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The rest three digits of the number have no restrictions, and therefore there are <math> 3! \implies 6 </math> for each leading digit.
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Since the two <math> 2 </math>'s are indistinguishable, there are <math> \frac {3*6}{2} </math> such numbers <math> \implies \boxed{\textbf{(D)}\ 9} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=9|num-a=11}}
 
{{AMC8 box|year=2012|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:13, 21 November 2013

Problem

How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$

Solution 1

For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$, since all of the valid 4-digit number will always be greater than $1000$. The best way to solve this problem is by using casework.

There can be only two leading digits, namely $1$ or $2$.

When the leading digit is $1$, you can make $\frac{3!}{2!1!} \implies 3$ such numbers.

When the leading digit is $2$, you can make $3! \implies 6$ such numbers.

Summing the amounts of numbers, we find that there are $\boxed{\textbf{(D)}\ 9}$ such numbers.

Solution 2

Notice that the first digit cannot be $0$, as the number is greater than $1000$. Therefore, there are three digits that can be in the thousands.

The rest three digits of the number have no restrictions, and therefore there are $3! \implies 6$ for each leading digit.

Since the two $2$'s are indistinguishable, there are $\frac {3*6}{2}$ such numbers $\implies \boxed{\textbf{(D)}\ 9}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AJHSME/AMC 8 Problems and Solutions

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