Nah, I just fixated on the wrong bit. I should have known that wasn't it as it probably would have fallen other a trick question rather than a pure maths problem.

Interesting.

I remember studying the derivation of the least significant digit of Grahams number. I'll see if I can spare this some time tomorrow, if it's not solved by then.

Sorry but as I was going to sleep I thought of this. The answer is: 6.

Powers of 2 go: 2, 4, 8, 16, 32, 64, 128, 256... etc
The least significant digit follows a pattern 2, 4, 8, 6, 2, 4, 8, 6...

The exponent of Moser's number (base 2) is clearly a multiple of 4. Hence the last digit is 6.

I'm really going to bed now. I'll try and whip up a better proof if the above is not enough tomorrow. Unless my answer is completely wrong.

The answer is correct. It isn't hard to prove it rigorously, however.

Your post does have a small mistake: You use the word "patter" when I assume you mean "pattern".

OK, fixed.

Question is up for for whoever wants to have a go. I'll post something later if no one comes up by then.

Ok, I made up this puzzle and it should be non-ambiguous, but please don’t stone me if I made a mistake. I had noone to test it on, yet.

---------------------------------------------------------------

There are three friends of different age, counted in full years.

None of the friends is older than 99 years, but all of them together are older than 99 years.

The digit sum of the age of each friend is the same for all three, and it equals the digit sum of their comined age.

The youngest friend is half as old as the middle friend.

The 3rd root of the product of all their ages equals half the oldest friend’s age.

How old are the three friends?

y young, m middle, o old
99 < y + m + o
99 > o > m > y
digitsum(y) = digitsum(m) = digitsum(o) = digitsum(y + m + o)
m = 2 * y
(1 / 2 * o) ^ 3 = y * m * o
=>
1 / 8 * o ^ 2 = y * m = 2 * y ^ 2
=>
o = 4 * y

From digitsum(y) = digitsum(m) and 2 * y = m
=>
m = 10 * am + bm
y = 10 * ay + by
=>
by = 2 * n > bm
=>
am > ay
=> am = 2 * ay + 1
With:
o = 4 * y and o < 99 and by > 4:
15 <= y <= 19

digitsum(y) = digitsum(m)
=>
15: 6, 2 *15 = 30: 3
16: 7, 2* 16 = 32: 5
17: 8, 2 * 17 = 34: 7
18: 9, 2 * 18 = 36: 9
19: 10, 2 * 19 = 38: 11

y = 18
m = 36
o = 72

y + m + o = 126 => also digitsum 9

That is correct. Great job!

Your turn to ask one then, or pass your turn.

Okay, from the regional level of the mathematical olympiad in Germany:
A school wants to go on a trip with 482 pupils. 30 teachers shall guard them. There are busses with 20 and 22 guest seats each and cars with 5 seats available (driven by a teacher). Every seat should be used. If one (and not more) teacher should be in each transport vehicle, what are the combinations that fulfill the conditions?

If I understood correctly, we have a choice of

_1 teacher + 4 students
_1 teacher + 19 students
_1 teacher + 21 students

We need exactly 30 vehicles.

So...

The average students per vehicle is 482/30 is roughly 16.

Given that a bus takes roughly 20, while a car takes 4, it would mean that we're looking for a rough 1:3 car to bus ratio (3 buses + 1 car give roughly 16 average as well)

So out of 30, lets say we take 8 cars with 32 students.

That leaves 22 buses for 450 students.
The average 20 per bus would means we can fit only 440.

So we need more of the 21 seaters
21*16 + 19*6 gives us what we need.

So
_16 of 22-seater buses
_6 of 20-seater buses
_8 cars

I wonder if there is a faster way of doing this.

a = 482
b = 30
c + d + e = 30
5 * c + 20 * d + 22 * e = 512
5 * (30 - d - e) + 20 * d + 22 * e = 512
512 = 150 - 5 * d - 5 * e + 20 * d + 22 * e
362 = 15 * d + 17 * e
Here the 15 means that e has to be 10 * n + (1 or 6), so 362 - 17 * e ends with 5 or 0 and might have a change to be divisible by 15. Testing this leaves these solutions:
e = 1, d = 23, c = 6: 22 + 23 * 20 + 6 * 5
e = 16, d = 6, c = 8: 16 * 22 + 6 * 20 + 8 * 5

Ah, so I did get one of them.

Something quick then, it's fairly easy but the result is pretty interesting.

A̶ ̶c̶a̶m̶e̶l̶ ̶m̶e̶r̶c̶h̶a̶n̶t̶... A football (that's soccer for you Americans) team plays better at home than it does away. One year, in order to qualify for the finals they have to beat a team from across the country twice (at least) in three matches. They are given a choice of playing either
Home then Away then Home
or
Away then Home then Away
Naturally they select the first option in order to play twice at home.

Next year however the rules change. They have to win two games (at least) in a row in order to qualify. They are given the same choices. Which one should they choose?

Since there is a 50% chance of getting the right answer by guessing (and since the way the problem is posed seems to suggest what the correct answer is), I'll need some sort of explanation or calculation besides the answer.