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== | + | ==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== | ||
+ | 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. | ||
+ | |||
+ | 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
Contents
Problem
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
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 , 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.
Solution 2
Notice that the first digit cannot be , as the number is greater than . 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 for each leading digit.
Since the two 's are indistinguishable, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.