Ten people are seated around a table...?

Question

The average income of these ten people is $10,000. Each person's income is the average o the income of the people sitting immediately to his or her left and right. What is the possible range of incomes for each person?

Answer 1

Here's one way to prove that everyone must earn $10,000.

Let n be the income of someone at the table. Then the people to their left and right must have incomes of n + k and n - k.

The person next to the n - k person, already has an n next to them, so the other person will have to earn n - 2k. The person next to them will have an income of n - 3k, then n - 4k, etc.

Going the other direction the n + k person will be next to the n person and the n + 2k person. The n + 2k person will be next to the n + 3k who is next to the n + 4k person, etc.

The last person in the chain must therefore have an income of both n + 5k and n - 5k. This only works if k = 0.

In other words, there is no difference in income between each person and the people next to them. With 10 people earning n each, the average will be 10n / 10 = n. So in order for the average to be 10,000 they must all have incomes of $10,000.

The range is [10000 , 10000] which really means n = 10,000 for everyone.

Answer 2

The incomes for individuals would seem at first blush to be able to range from $0 to $100,000. This is because with 10 people, each averaging $10,000, the whole table makes $100,000. So if one person makes that and the rest $0, then the table averages $10,000 each.

However, that violates the second sentence of the problem. Call the person making $100,000 Person 1. Then Persons 2 and 3 each make $0. But Person 2's income was defined to be the average of Persons 1 and 3's incomes: ($100,000 + $0) / 2 = $50,000. Which is NOT $0. So this distribution cannot occur and the actual range must be narrower.

Consider then Persons 1 through 5 making $20,000 a year and Persons 6 through 10 making $0. Seems fine with Persons 1, 2, and 3: Person 2's income is now ($20,000 + $20,000) / 2 = $20,000 which matches. But we look at Persons 5, 6, and 7 and the same problem as above arises: Person 6 = ($20,000 + $0) / 2 = $10,000 which does not match his $0 income. Adjust either way, to four people making $25,000 each, the others $0, or eight people making $12,500 each, the others $0, and we still get the same difficulty at the edge of the similar incomes.

This leads us inexorably to each person making exactly $10,000. But this might not be correct: we could have passed over a point that would be an inflection point on a graph. We don't think so because the values we are calculating are linear, so there shouldn't be any surprise inflection points that we passed over. But let's get more detailed.

For instance, consider mixing the high and low earners, rather than keeping them together in groups, like we did above. Then the problem we had at the "edges" just spreads around the table. For instance, if the five people making $20,000 each have a person making $0 between them (so the earning go: $20,000 - $0 - $20,000 - $0 - etc.), then the problem we had at the edge (Persons 5, 6, and 7) simply becomes the same problem but everywhere: Persons 1, 2, and 3: $20,000 - $0 - $20,000 and Person 2's income has to be ($20,000 + $20,000) / 2 = $20,000 but it is actually $0. If it were $20,000, then the average for the table would be $20,000 and not $10,000. Let's extend that approach and see what the algebra yields.

(This last part is fairly obnoxious but is the actual (fairly) rigorous part.)

We have considered sets of same incomes for simplicity, but consider how a wide variety of incomes would affect things (even perhaps everyone different). Let's set Person 1's income at ($10,000 + a). "a" can be any amount permissible, positive of negative. The actual constraints don't really matter, but as we established earlier, the extreme range permitted by the first sentence was $0 to $100,000 for any individual so the variable in each person's income could be –$10,000 to +$90,000 (because I follow the form of "$10,000 + variable). So we'll set five persons' incomes and learn the other five from the averages (following sentence 2's formula):

Person 1: $10,000 + a
Person 3: $10,000 + b
Person 5: $10,000 + c
Person 7: $10,000 + d
Person 9: $10,000 + e

and therefore:

Person 2: {($10,000 + a) + ($10,000 + b)} / 2 = ($20,000 + a + b) / 2 = $10,000 + (a + b) / 2
Person 4: {($10,000 + b) + ($10,000 + c)} / 2 = ($20,000 + b + c) / 2 = $10,000 + (b + c) / 2
Person 6: {($10,000 + c) + ($10,000 + d)} / 2 = ($20,000 + c + d) / 2 = $10,000 + (c + d) / 2
Person 8: {($10,000 + d) + ($10,000 + e)} / 2 = ($20,000 + d + e) / 2 = $10,000 + (d + e) / 2
Person 10: {($10,000 + e) + ($10,000 + a)} / 2 = ($20,000 + e + a) / 2 = $10,000 + (e + a) / 2

So, as you can see, adding them all up we get: $100,000 + a + b + c + d + e + a/2 + b/2 + b/2 + c/2 + c/2 + d/2 + d/2 + e/2 + e/2 + a/2. Combining, we get: $100,000 + 2a + 2b + 2c + 2d + 2e. So the incomes of all the persons added together equals $100,000 plus twice each of the variables added to the five incomes we defined (remember, the variables could be positive or negative values). Earlier we established that the sum of the incomes at the table is $100,000 so we find that: 2a + 2b + 2c + 2d + 2e = $0 = a + b + c + d + e.

$0 = a + b + c + d + e. With the variables constrained to be no lower than –$10,000 (because no one can earn less than $0) and no higher than +$90,000 (because of no one being able to earn less than $0, no one else can earn more than $100,000 or the table would exceed 10 people times $10,000 per person and the average would exceed $10,000; since we defined each income as $10,000 plus some combination of variables, the actual constraint is $100,000 - $10,000 = $90,000).

So, to further constrain things, consider "a" through "d" being equal to –$10,000 each. Since the sum of all five equals $0, the fifth variable ("e") cannot be higher than –(4 * (–$10,000)) = $40,000. So the final constraints before considering the effects of these extremes on the individual incomes are: –$10,000 <= a, b, c, d, e <= $40,000.

Now, what happens if we let, say, "c" be $40,000 (so Person 5's income is $50,000) and the rest be –$10,000? Then Persons 1, 2, and 3 and 7, 8, 9, and 10 make $0, Persons 4 and 6 make $25,000 and Person 5 makes $50,000. Ouch. If Person 5's income is the average of Persons 4 and 6, it can't very well be double both! So this extreme cannot actually work.

And if we substitute lower and lower values for, say, "c", we find ourselves gradually approaching $0 for it and still having slight errors in satisfying sentence 2 in the problem. In fact, it is only at c = $0 that we see the errors disappear. Since we chose "c" to simplify looking around the table in the answer and so for no reason that matters to the problem itself, we can expect the same result regardless of the variable chosen. So we again inexorably come to each person's income being exactly $10,000.

Answer 3

I concur with the first answer, since they all have to earn the average of the two people on either side of them and they all average out to $10,000, they must all earn $10,000.

Answer 4

Possible range of income for each person is 0 to 10000.

Answer 5

10000

Answer 6

$10,000 the left, right, all the people have to get the same amount.

Answer 7

They all must earn exactly $10,000.