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> | ||
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+ | ==Solution== | ||
+ | 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. | ||
+ | |||
+ | There can be only two leading digits, namely <math> 1 </math> or <math> 2 </math>. | ||
+ | |||
+ | When the leading digit is <math> 1 </math>, you can make <math> \frac{3!}{2!1!} \implies 3 </math> such numbers. | ||
+ | |||
+ | When the leading digit is <math> 2 </math>, you can make <math> 3! \implies 6 </math> such numbers. | ||
+ | |||
+ | Summing the amounts of numbers, we find that there are <math> \boxed{\textbf{(D)}\ 9} </math> such numbers. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=9|num-a=11}} | {{AMC8 box|year=2012|num-b=9|num-a=11}} |
Revision as of 10:37, 24 November 2012
Problem 10
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
Solution
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 , since all of the valid 4-digit number will always be greater than . The best way to solve this problem is by using casework.
There can be only two leading digits, namely or .
When the leading digit is , you can make such numbers.
When the leading digit is , you can make such numbers.
Summing the amounts of numbers, we find that there are such numbers.
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
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 |