Finding the Order of 5 Numbers

Question

Mr. James was finding the averages of his monthly profits for his restaurant. He noticed as he calculated the new average of his profits every time he entered one month's profits, the average always returned an integer.

His monthly profits for the 5 months he had his restaurant were 71, 76, 80, 82, 91 (dollars). What was the particular order that Mr. James added the profits in to output an integer average each time?

Work so far:

$\alpha_1, \alpha_2,\alpha_3,\alpha_4,\alpha_5$ represents the sequence of the profits.

$\frac{\alpha_1 + \alpha_2}{2} \in Z$

$\frac{\alpha_1 + \alpha_2 + \alpha_3}{3} \in Z$

$\frac{\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4}{4} \in Z$

$\frac{\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5}{5} \in Z$ (given)

$\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 +\alpha_5 = 400 $

This seems like a problem where one has to use a large amount of deductive reasoning. However, I really don't know where to start and how to prove one term of the sequence must be whatever value.

Answer 1

Start from the beginning. The first two months need to both be odd or both be even. As there are only two odd numbers, try that first. If the first two are $71,91$, the sum so far is $162$ and we need the third month to be a multiple of $3$, but there isn't one available, so the first two month are even. Among the evens there are two that are $1 \pmod 3$ and one that is $2 \pmod 3$ so you only have two choices for the sum of the first two. Only one works. Keep going.

Note that nothing can tell you the order of the first two months.

Answer 2

The first two are divisble by two, so they are both even or both odd. If both odd, they are 71 and 91, the sum of whose digits adds up to 18, so divisible by 3, so if they were it, there should be another number divisble by 3 for the next month ... But there isn't. So, the first two numbers were even. The fifth number was even too, for else the first 4 are not divisble by 4. Which means numbers 3 and 4 were the odd ones. The fifth number is either 80 or 76, or else the first four are not divisble by 4 given that the sum of all 5 is divisble by 4. So, for the first three numbers we're looking for two even numbers (but not both 80 and 76) and an odd number where the sum is divisible by 3 ... It can't be 80 and 82, since their digits add up to 18, so the odd one would have to be divisble by 3 by itself, and neither one isnt't. So, the two even numbers at the start are 76 and 82 ... Sum of digits 23. If we add 71, sum is 31, which is not divisible by 3, the the third number has to be 91, which gives sum of 33 so indeed divisible by 3. So, there you go:

First two: 76 and 82

Third: 91

Fourth: 71

Fifth: 80

The reasoning shows that this is the only possible solution.