How many three-digit positive integers are there such that a

Question

How many three-digit positive integers are there such that at least one digit is equal to the average of the other two digits?

A) 116
B) 120
C) 121
D) 124
E) 125

MarkSullivan · Answer

Fun one! I'll start with a caveat, though: I'd advise most of my students to bail quickly here (and would even do so myself if I were short on time!)

A quick glance shows a few things that make this an excellent bail candidate: 1) it's a counting question, and the answer choices indicate the count will be too high to literally list the possibilities out (so you'll either need to find and apply a formula or come up with a way to organize possibilities into more manageable chunks), and 2) the choices are very close together, so estimation is likely out.

Behind time? Guess quickly and come back (or don't) at the end.
On time, but struggle with combinatorics? Guess quickly and come back (or don't) at the end.
On time and love/do well with combinatorics? Give it a try, but be ready to bail if a sure-fire path doesn't emerge after a minute or so.

Here was my thought process to solve:

We have three digit numbers, at least one of which is equal to the average of the other two. Immediately, that "at least one of which" language struck me as odd... if you're averaging two of them doesn't that leave exactly one other one? Language that seems odd is often important to the problem, so I thought more about that and that lead to the examples of the form xxx (111, 222, ..., 999). That's 9 three-digit numbers with this property right there.

What about those that are not all the same digit repeated? In this case, we only have digits to work with so they must be evenly-spaced: 123, 357, 258, and such. Of course, each of these will have 3! = 6 versions to count (123, 132, 213, 231, 312, 321 are the six corresponding to 123). But wait, what about 0's? The set {0, 2, 4} is evenly-spaced and can form some three-digit numbers with the desired property (like 240), but not six of them because the 0 cannot go in the hundreds place (024 = 24 is a two-digit number). So we'll need to separate those out, then deal with the other possibilities systematically.

It's only at this point that I felt I had a workable PLAN: deal with the examples with 0's, organize the others according to the size of the gap (gap of one: 123, gap of two: 135, gap of three: 147, etc.), and don't forget the nine of the form xxx. (phew!) If I were at this point 1-1.5 minutes into the problem, I'd go for it. If it had already been more than 1.5 minutes, I'd mark it to return to if I end up with time at the end.

111, 222, ..., 999:  9 three-digit numbers

{0, 1, 2}, {0, 2, 4}, {0, 3, 6}, {0, 4, 8} are the four sets of numbers with zero, each of which will have 2! = 4 three-digit numbers we can form (for example, {0, 1, 2} can form 102, 120, 201, and 210):  16 three-digit numbers

Each other possible set will have 3! = 6 variations because we don't need to worry about what goes in the hundreds digit:

gap of 1: 123, 234, ..., 789 (7 sets)
gap of 2: 135, 246, ..., 579 (5 sets)
gap of 3: 147, 258, 369 (3 sets)
gap of 4: 159 (1 set

And bigger gaps aren't possible, so that 7 + 5 + 3 + 1 = 16 sets, or 16 * 6 =  96 three-digit numbers

In total, we have 96 + 16 + 9 =  121 three-digit numbers  with this property.

HarshavardhanR · Answer

This is a tricky question under test-timing. Solving this was a lot of fun!

---

Say, the average of the 3 numbers is "a".

Case 1:   All 3 digits are equal to the average of the 3 numbers

(a,a,a) -> 9 possible numbers (111, 222, etc.....upto 999)

Case 2:   Exactly 2 digits are equal to the average of the 3 numbers

(a,a,b) -> Avg = a =  \frac{(a + a + b) }{ 3}

=> b = a.

In other words, if 2 digits are equal to the average of the 3 numbers, then, by default, the third digit is also equal to the average.

i.e., there is no (a,a,b). It boils down to (a,a,a), which we have already counted.

Case 3: Exactly 1 digit is equal to the average of the 3 numbers

(b,c,a) -> where a =  \frac{(b+c) }{ 2}

Here, observe that b and c must be different numbers. if b and c are both the same, we will again end up with (b,b,b). Those cases are already counted.

So, let's count the possible (b,c,a)'s.

a =  \frac{(b+c) }{ 2}

Insight: for "a" to be a proper digit (integer), (b+c) must be an even number.

Case 3 ->Sub-Case 1: b and c are two different odd numbers

(1,3,5,7,9)

- 5C2 ways of selecting two different odd numbers for b and c
- Once b and c are chosen, a =  \frac{b+c}{2}  is fixed. Only 1 possibility
- Once we have a,b, and c, there are 3! different numbers that can be created.

Total = 5C2 x 1 x 3! = 60 possible numbers

Case 3 ->Sub-Case 2: b and c are two different non-0 even numbers

We need to worry about 0 ending up as the left-most digit, which is not allowed, so we can count the cases which include 0, separately. Here, let's focus on the non-0 cases.

(2,4,6,8)

- 4C2 ways of selecting two different even numbers for b and c
- Once b and c are chosen, a =  \frac{b+c}{2}  is fixed. Only 1 possibility
- Once we have a,b, and c, there are 3! different numbers that can be created.

Total = 4C2 x 1 x 3! = 36 possible numbers

Case 3 ->Sub-Case 3: One of the 2 numbers (b or c) is 0.

Some considerations ->  
- b and c cannot both be 0. Then, the avg is also 0. 000 is not to be counted.
- 0 can never be the left-most digit (because then the number would not be 3-digit).
- The avg a cannot be 0, because it is derived from the other digits (at least one of them non-0, none of them negative)

So, the only real possibility is ->

(b,0,a =  \frac{(b+0)}{2}  =  \frac{b}{2} )

Given "a" is a digit, b has to be a non-0 even number.

b -> (2,4,6,8)

- 4C1 = 4 ways of selecting b
- Once b is chosen, the other numbers are locked in. The other numbers are b/2 and 0. 
- Now, that we have (b,0,b/2), the number of possible numbers is  2  x  2  x  1  (remember to exclude 0 from the left-most digit)

Total = 4C1 x 1 x 2 x 2 = 16 possible numbers.

Final answer:  9 + 60 + 36 + 16 =   121   3-digit positive integers.

---
Harsha

kevincan · Answer

What makes this question complicated is the fact that a 3-digit positive integer cannot have 0 as a hundreds digit. Let’s forget that restriction and deal with it later.

We can have the digit three times : 000,111,...999 : 10 options

We can also have digits that, when arranged numerically, would form an arithmetic sequence with a positive common difference d, which can be 1,2,3 or 4

012, 123, ....,789: 8 options 
024, 135, ...,579: 6 options 
036, ...369: 4 options 
048,159 : 2 options

20 content options * 3! = 120 distinct orderings

130 in total , including the ones that begin with 0.

Reject 000 and 2 from each of the 012, 024, 036, 048 options .

That we reject 9 of the 130 because they start with 0 and are left with 121.

How many three-digit positive integers are there such that a

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