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

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==Problem 10==
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==Problem==
 
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
 
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
  
 
<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\cdot6}{2} </math> such numbers <math> \implies \boxed{\textbf{(D)}\ 9} </math>.
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==Solution 3(a simpler version of Solution 2)==
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We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than <math>1000</math>. This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are <math>3</math> choices for the thousands place, <math>3</math> for the hundreds, <math>2</math> for the tens, and <math>1</math> for the ones. Adding this up we get <math>9</math> which is choice <math>\boxed{\textbf{(D)} }</math>.
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—-jason.ca
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==Video Solution==
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https://youtu.be/OBKIeTgw4Zg ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=9|num-a=11}}
 
{{AMC8 box|year=2012|num-b=9|num-a=11}}
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{{MAA Notice}}

Revision as of 17:27, 2 January 2023

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 $\frac {3\cdot6}{2}$ such numbers $\implies \boxed{\textbf{(D)}\ 9}$.

Solution 3(a simpler version of Solution 2)

We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than $1000$. This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are $3$ choices for the thousands place, $3$ for the hundreds, $2$ for the tens, and $1$ for the ones. Adding this up we get $9$ which is choice $\boxed{\textbf{(D)} }$.

—-jason.ca

Video Solution

https://youtu.be/OBKIeTgw4Zg ~savannahsolver

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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