How many three-digit numbers are composed of three distinct digits suc

Question

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

(A) 96 
(B) 104 
(C) 112 
(D) 120 
(E) 256

siyeezy · Accepted Answer

Pick 3 single digit numbers such that the gap between them are the same, which can be 1, 2, 3, or 4. That way, the middle will be the average of the other two. Remember to subtract the numbers that begin with 0 since we want three digit numbers.

0,1,2; 1,2,3 ... 7,8,9 = 8 groups
0,2,4; ... 5,7,9 = 6 groups
0,3,6; ... 3,6,9 = 4 groups
0,4,8; 1,5,9 = 2 groups

So 20 different groupings of numbers, each with 3! ways to arrange them or 120. Subtract the 8 numbers that began with 0 since those are two digit numbers, and you get 112

Ans C

ScottTargetTestPrep · Answer

If the “average” digit is 1, then the other two digits are 0 and 2. For example, 210.

If the “average” digit is 2, then the other two digits are 1 and 3 OR 0 and 4. For example, 321 and 420.

If the “average” digit is 3, then the other two digits are 2 and 4 OR 1 and 5 OR 0 and 6. For example, 432, 531 and 630.

If the “average” digit is 4, then the other two digits are 3 and 5 OR 2 and 6 OR 1 and 7 OR 0 and 8. For example, 543, 642, 741 and 840.

If the “average” digit is 5, then the other two digits are 4 and 6 OR 3 and 7 OR 2 and 8 OR 1 and 9. For example, 654, 753, 852 and 951.

If the “average” digit is 6, then the other two digits are 5 and 7 OR 4 and 8 OR 3 and 9. For example, 765, 864 and 963.

If the “average” digit is 7, then the other two digits are 6 and 8 OR 5 and 9. For example, 876 and 975.

If the “average” digit is 8, then the other two digits are 7 and 9. For example, 987.

For 20 numbers listed above as examples, all but 4 have nonzero distinct digits. So each of them has 3! = 6 ways to permute its digits (for example, 321 could be 123, 132, 213, 231 and 312, in addition to itself). For the 4 numbers that have zero as one of the digits, each of them only has 4 ways to permute its digits (for example, 210 could be 102, 120 and 201, in addition to itself). Therefore, the total number of numbers is 16 x 6 + 4 x 4 = 96 + 16 = 112.

Answer: C

GRIT999 · Answer

Bunuel  please help

rocky620 · Answer

ScottTargetTestPrep 
Sir is there any mathematical approach to solve these type of questions or making cases is the only approach?

ScottTargetTestPrep · Answer

Response: 
We have to make cases at some point in the solution because of the restrictions implied by the question (for instance, the hundreds digit of a three digit integer cannot be equal to zero). However, we can reduce the number of cases to just three if we reason as follows:

If we have a three digit integer with distinct digits and one digit is the average of the others, the digits other than the average digit must either be both even or both odd. As long as the other two digits are both even or both odd and distinct, the average will also be distinct from the other two digits and will be between them, so the two digits together with the average will form three digit integers (assuming the hundreds digit is not zero).

Case 1. Zero is one of the digits

In this case, the digit other than the average digit must be even, so that digit is 2, 4, 6, or 8. Since there are four possibilities for the non-average digit and since these digits can be arranged in four ways (as explained in the solution above), there are 4 x 4 such integers.

Case 2. Two non-zero even digits

We can choose two digits from among 2, 4, 6, and 8 in 4C2 = 4!/(2!*2!) = 6 ways. Together with the average digit, the three digits can be arranged in 3! = 6 ways. Thus, there are 6 x 6 such integers.

Case 3: Two odd digits

We can choose any two of 1, 3, 5, 7, and 9. There are 5C2 = 5!/(2!*3!) = 10 ways to make such a choice. Since each choice of two digits together with the average can be arranged in 3! = 6 ways, there are 10 x 6 such integers.

In total, there are 4 x 4 + 6 x 6 + 10 x 6 = 16 x 6 + 4 x 4 such integers, which is the same answer as we obtained before.

Pooki · Answer

Answer C:
the number we need to find out is ABC
The question ask if any digit is average of 2 others
so the two other number must be both even or both odd
if both odd, so we have 1,3,5,7,9 :
2C5*3!=60
If both even, so we have 0,2,4,6,8
3C5*3!=60 but this way we may end up with 0BC which is not a 3 digits number, so we have to deduct these cases= 021.012.042.024.063.036.084.048
so answer is 60+60-8=112

chetan2u · Answer

Let the three numbers be a, b and c, where  \frac{a+b}{2}=c . 
This means a+b must be even, and  when will a+b be even.

1) When both are even 
So choose 2 even digits out of possible 5=5C2=10.
As the digits are distinct, each set of a, b and c can be arranged in 3! ways. 
Total ways = 10*3!=60 
However, it also consists of numbers that become 2-digits when hundreds digit is 0. Let us subtract these. 
When hundreds digit is 0, the remaining even digit can be found in 4C1 ways and this digit can take any of the two places, so 4*2 or 8 ways. 
Total ways = 60-8=52.

2) When both are odd.  
So choose 2 odd digits out of possible 5=5C2=10.
As the digits are distinct, each set of a, b and c can be arranged in 3! ways. 
Total ways = 10*3!=60

Total combined ways = 52+60=112

bumpbot · Answer

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How many three-digit numbers are composed of three distinct digits suc

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