Difference between revisions of "2007 AMC 12A Problems/Problem 16"

(this is a combinatorics problem LOL)
(Video solutions go at the bottom.)
 
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== Problems ==
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== Problem ==
 
How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two?  
 
How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two?  
  
 
<math>\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256</math>
 
<math>\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256</math>
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== Solution 1==
 
== Solution 1==
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Thus the total is <math>96 + 16 = 112 \Longrightarrow \mathrm{(C)}</math>. (by scrabbler94)
 
Thus the total is <math>96 + 16 = 112 \Longrightarrow \mathrm{(C)}</math>. (by scrabbler94)
  
== Solution 3==
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== Video Solution ==
This solution takes advantage of the choices available, and as such, will get you nowhere on the AIME and most other contests.
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https://youtu.be/0W3VmFp55cM?t=2012
Observe that there's a common mistake where people forget that 0 could be part of the number. 4 valid permutations of <math>840,630,420,</math> and <math>210</math> result in <math>16</math> total that they miss. Looking at the answers, only two differ by <math>16</math>, namely <math>A</math> and <math>C</math>. So it's safe to bet that the answer is <math>\mathrm{(C)}</math>
 
 
 
That's what I did. Then, I had time at the end to go back and verify.
 
-Rowechen.
 
 
 
  
B and D do too...
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~ pi_is_3.14
  
xD lol
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== See Also ==
  
== See also ==
 
[[2005 AMC 12A Problems/Problem 11|similar problem]]
 
 
{{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}
  
[[Category: Introductory Combinatorics Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:10, 16 February 2021

Problem

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$


Solution 1

We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.

Common difference Sequences possible Number of sequences
1 $012, \ldots, 789$ 8
2 $024, \ldots, 579$ 6
3 $036, \ldots, 369$ 4
4 $048, \ldots, 159$ 2

This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.

However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.

Solution 2

Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives $3! = 6$ possible 3-digit numbers; otherwise, $4$ possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.

Case 1: 0 is not in the number. Then there are $\binom{5}{2} + \binom{4}{2} = 16$ ways to choose two nonzero digits of the same parity, and each choice generates $3! = 6$ 3-digit numbers, giving $16 \times 6 = 96$ numbers.

Case 2: 0 is in the number. Then there are $4$ ways to choose the largest digit (2, 4, 6, or 8), and each choice generates $4$ 3-digit numbers, giving $4 \times 4 = 16$ numbers.

Thus the total is $96 + 16 = 112 \Longrightarrow \mathrm{(C)}$. (by scrabbler94)

Video Solution

https://youtu.be/0W3VmFp55cM?t=2012

~ pi_is_3.14

See Also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions

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