I couldn’t do the bonus cause I can’t read it on that pdf but Keacher, you can’t be serious about the main problem. You know from inference that the solution must be a perfect integer. This is the average equals 35 problem right?

You don’t even have to do the calculations for the answer. For consecutive integers to have an average of 35 then they have to go at least from 1 to 69. 35 is right in the middle, knock it off and the average is the same.

Specifically, it’s the “average equals 35 and 7/16” problem — and therein lies the rub. Yes, the average of a sequence of number from 1 to n = (n+1)/2. The challenge, as I understand it, is solving for k and n such that (n*(n+1)/2-k)/(n-1) = 567/16

…where k = the number to eliminate and n = the largest number in the sequence

Oh, you can’t see that 7/16 on the pdf either. Don’t I feel like an ass. Still saying that 39.8125 is 40 in a math teaser is like saying that Pi is exactly 3.

So the way your doing it involves 1 equation and 2 unknowns, I’m curious as to you think n is.

Using your equation we can justify that (n-1) is some multiple of 16. The only values of n that would therefore give averages even close to 35 is 65 and 81 but in neither of them do the numbers work out. I’m stumped.

Indeed. However, the problem is sufficiently constrained. To wit:

From the problem, we know that k must be less than or (arguably) equal to n, and k must be greater than or equal to 1. We can further deduce that n must be 69 or 70. For example, the greatest possible result with n = 70 occurs when k = 1, at which point the average of the remaining integers is 36; when k = 70, the smallest result occurs: 35. In the same manner, we find the upper and lower limits for n = 69 are 35.5 and 34.5; for n = 68, 35 and 34; and for n = 71, 36.5 and 35.5. Therefore, the only values of n which can produce a result where 567/16 (35.4375) is an average are 69 and 70.

Solving the equation (n*(n+1)/2-k)/(n-1)=567/16 for k when n = 69 gives k as 5.25, which, as 5.25 is not an integer, is not a solution. Solving when n = 70 gives the aforementioned 39.8125 result, which is also not an integer, although it is quite a bit closer to 40 than 5.25 is to 5.

I doubt the problem is flawed. Therefore, there must be a problem with my technique. Unfortunately, I don’t see where it might be.

I’m thinking it might be with the wording an how we’re interpretting it. On the warm-up it doesn’t ask how old the mother and son is it asks Where are the mom and dad. And indeed, doing the math without regard to solving with only integers yields that the mother would be conceiving the child at the time in question. Which means the answer is that the mom and dad are getting busy. Maybe we’re missing something with this problem.

In case anybody wants to play along but can’t read the PDF file, the problem states: “A list of consecutive integers, starting with 1, is written on the blackboard. One of the numbers is erased. If the average of the remaining numbers is 35 7/16, then what number was erased?” (Bailey 2005)

Maybe they’re as stumped as we are

I’ve been worked on it earlier then saw your post and pretty much have come to all the same conclusions. I even got fed up and tried to write a program to just brute force it and I got nowhere.

Maybe the blackboard is significant? Maybe there’s some significance in the expression of the average as a mixed number?

Just a thought…Maybe when it says you erase one of the numbers, it only means you erase one digit. Like, one of the numbers is 35. Say, when “one of the numbers is erased,” the 5 in 35 is erased. I don’t know how to factor that into equations, but it seems it would allow for a fairly challenging solution.

Perhaps. My initial reasoning is that eliminating something like the 5 from 35 would result in a net loss of 32 (35-3=32), which would be almost the same as eliminating the number 32. The key word there is “almost;” by eliminating one of the digits instead of the entire number, the divisor for the mean calculation becomes n instead of n-1. Could be significant…

Well…when you go through it like that, with the same equation you used before, except dividing by n, you get to 39.9375…which is closer to 40, but still no cigar. Must be missing something else about the problem. (Oh…it works if 405 is somehow an integer between 1 and 80 😉 )

Last week, I emailed Professor Bailey about the problem. And the answer is… there was a typo. The average should have been 35 6/17, not 35 7/16. With that correction, the solution becomes trivial.

I think I speak for all of us when I say, “Bitches!”

Just checking in to your site. Anyways, after 6 beers, its 60 milliseconds for myself. Hope all is going well for you.

– Eric