last day (15 days later) » 

Stack Exchange
Stack Exchange
12:43
0
A: Teacher, teacher on the wall, Who's the dumbest of them all?

Ivo BeckersBecause the two pupils are consecutive one of those is divisable by 2. Lets call that pupil $x$. This means that $x/2$ also can't be a divisor because if $x/2$ is a divisor and $2$ is a divisor then $x$ is also a divisor, contradiction. This either means that the consecutive number are $x$ and $x...

StephenTG
StephenTG
12 is not divisible by 8 but it is divisible by 4, so your first argument isn't necessarily true
Ivo Beckers
Ivo Beckers
I don't understand what you're saying. All i say is that if $a$ and $b$ are divisors of a number then $a \cdot b$ is also a divisor of that number
StephenTG
StephenTG
If we use 12 as that number (just as an example), and have a = 2, b = 4, then a and b are divisors of 12, but a*b is not
Ivo Beckers
Ivo Beckers
But a*b does not equal 12. In my example I say the following. Let's say that x=12. This means that the number is not divisable by 12. This means that number also not divisable by 6. 6 and 12 are not consecutive so this can't be the case.
StephenTG
StephenTG
What I'm saying is that just because a number is not divisible by 12, does not mean that it is not divisible by 6. 18 is not divisible by 12, but it is divisible by 6.
Ivo Beckers
Ivo Beckers
12:43
Aah. now I understand. You're right. thanks for making me see that. I guess my answer is incorrect. I'l just leave it here for others who might make the same mistake
StephenTG
StephenTG
I think this does give us somewhere to go from, though, since I think the only case where a and b divide a number x, but ab doesn't, is if ab > x/2.
I think this means the numbers that are wrong might have to be the two numbers just after half the number written on the board
Ivo Beckers
Ivo Beckers
I think the correct thing is if a and b are divisors of a number then a*b is also a divisor of that number if and only if a is no divisor of b and b is no divisor of a.
StephenTG
StephenTG
what about 4,6, and 12?
Ivo Beckers
Ivo Beckers
yeah youre right (again). uhm.. the thing is they can't have the same prime factors I gues. 4 = 2*2 and 6 = 2*3. both have a 2 in it. Something like that I think
StephenTG
StephenTG
That probably works
Ivo Beckers
Ivo Beckers
12:52
I think that we can say that x will be between half n and n
StephenTG
StephenTG
x being one of the incorrect students, yes?
Ivo Beckers
Ivo Beckers
yes
StephenTG
StephenTG
if x < n/2, then 2x divides but x does not, which is definitely impossible
Ivo Beckers
Ivo Beckers
exactly
StephenTG
StephenTG
One case that works is if n = 4, since 1,2,3 divide 6, but 4,5 don't.
f''
f''
13:01
Every even number that isn't a power of 2 is divisible by an odd number greater than 2
Ivo Beckers
Ivo Beckers
maybe it's always the case that the last 2 students are incorrect. Let me see if I can find a counter example
no that's not true. with n=4. and the number is 10 is also possible but then it's not divisable by 3 and 4
CodeNewbie
CodeNewbie
There is a counter example. You could have 7,8 as incorrect factors, and still be able to construct a number till n=12.
Ivo Beckers
Ivo Beckers
yeah
f''
f''
if the even number is a and a isn't a power of 2, then a=b*c where b is a power of 2 and c>2 is odd
but then if b and c divide n, a has to divide n
*the number, not n
I think that means the even number which is wrong has to be a power of 2
both wrong numbers have to be prime or prime powers?
CodeNewbie
CodeNewbie
Yes, indeed. I too just stumbled upon that realization.
I'm just not math-equipped enough to write an elegant answer to prove it.
Ivo Beckers
Ivo Beckers
13:14
I think this might be true yes. But I am too no mathemathician
f''
f''
so the even one has to be the largest power of 2 up to n+1, and has to have a prime or prime power next to it
9 is the only possible prime power because en.wikipedia.org/wiki/Catalan%27s_conjecture so other than 8/9 the odd number has to be prime
Ivo Beckers
Ivo Beckers
yeah. So these are the Mersenne primes en.wikipedia.org/wiki/Mersenne_prime and the primes in the form 2^n + 1
f''
f''
2^n+1 are the Fermat primes en.wikipedia.org/wiki/Fermat_number
Ivo Beckers
Ivo Beckers
guess all that's left is write an elegant answer
f''
f''
So either: 2^x-1 is prime, n+1 is at least 2^x and at most 2^(x+1)-3
or: 2^x+1 is prime, n+1 is at least 2^x+1 and at most 2^(x+1)-1
or the wrong ones are 8/9 and n+1 is 9 to 15
Ivo Beckers
Ivo Beckers
13:35
maybe also nice to know that if x and x+1 are the 2 non-divisors then the number on the bord is at least divisable by the LCM(1,2,..,x-1)

last day (15 days later) »