2007 AMC 12A Problems/Problem 16

Problems

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)

Solution 3

This solution takes advantage of the choices available, and as such, will get you nowhere on the AIME and most other contests. Observe that there's a common mistake where people forget that 0 could be part of the number. 4 valid permutations of $840,630,420,$ and $210$ result in $16$ total that they miss. Looking at the answers, only two differ by $16$ , namely $A$ and $C$ . So it's safe to bet that the answer is $\mathrm{(C)}$

That's what I did. Then, I had time at the end to go back and verify. -Rowechen.

B and D do too...

xD lol

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2007 AMC 12A Problems/Problem 16

Contents

Problems

Solution 1

Solution 2

Solution 3

See also