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# Annoyingly simple maths problem watch

1. An old maths lecturer of mine (who is awesome btw) posted this problem on his facebook, saying that he'd being asked for some help on it and couldn't see where to go.

I've tried umpteen different ideas and got nowhere, i've tried numerous methods of induction and ended up proving nothing.

Any ideas?
Show is divisible by for all
2. Interesting. I can get 1000 as a factor fairly easily.
3. Hint: Consider the factorisation of

Edit:

Made a slip. I can still only get 1000 as a factor.

Edit again: Yep, definitely 2000 as a factor.

Further hint:

Spoiler:
Show

You can pair the numbers in two different ways, each producing a factor.
These factors are coprime.

4. (Original post by ghostwalker)
Hint: Consider the factorisation of

Edit:

Made a slip. I can still only get 1000 as a factor.

Edit again: Yep, definitely 2000 as a factor.

Further hint:

Spoiler:
Show

You can pair the numbers in two different ways, each producing a factor.
These factors are coprime.

Okay so i've said that for some a and how can I say that a=b though?
5. (Original post by kexy)
Okay so i've said that for some a and how can I say that a=b though?

You can't and it probably doesn't.

for some a,b

Hence 16 is a factor.

Now try pairing the terms differently to get another factor.
6. the title of this thread is misleading... i don't think the majority people of tsr would find this problem simple at all
7. (Original post by 3006)
the title of this thread is misleading... i don't think the majority people of tsr would find this problem simple at all
Unlucky.
(Original post by ghostwalker)
...
Thanks for your help, I got it! Rep tomorrow, remind me.
8. (Original post by kexy)
An old maths lecturer of mine (who is awesome btw) posted this problem on his facebook, saying that he'd being asked for some help on it and couldn't see where to go.

I've tried umpteen different ideas and got nowhere, i've tried numerous methods of induction and ended up proving nothing.

Any ideas?
Show is divisible by for all
Alternative route:
Mod 125, 121 = -4, 1900 = 2000 - 100 = -100 = 25.
So mod 125 we have 121^n -25^n + 25^n - 121^n = 0.
Mod 16, 121 = 9, 25 = 9, 1900 = 2000 - 100 = -(96+4) = -4.
So mod 16 we have 9^n - 9^n + (-4)^n -(-4)^n = 0.

So 121^n -25^n +1900^n - (-4)^n is divisible by 16 and 125 and therefore by 16 x 125 = 2000.
9. (Original post by DFranklin)
...

So 121^n -25^n +1900^n - (-4)^n is divisible by 16 and 125 and therefore by 16 x 125 = 2000 ...
I'm unsure about this last bit, it's just been pointed out to me.
2 and 6 are bot factors of 30 but 2*6=12 isn't.
10. Chinese remainder, they're both coprime. Ignore meeeee.
11. Here's the solution:

Notice that 2000 = 16 x 125 and that these factors are relatively prime.

121 is congruent to 9 modulo 16 and so is 25.
1900 is congruent to (-4) modulo 16 and so is (-4).
Therefore 9^n-9^n+(-4)^n-(-4)^n is congruent to 0 modulo 16.

Consequently, your expression is divisible by 16.

121 is congruent to (-4) modulo 125 and so is (-4).
25 is congruent to 25 modulo 125 and so is 1900.

etc, etc.
12. (Original post by kexy)
An old maths lecturer of mine (who is awesome btw) posted this problem on his facebook, saying that he'd being asked for some help on it and couldn't see where to go.

I've tried umpteen different ideas and got nowhere, i've tried numerous methods of induction and ended up proving nothing.

Any ideas?
Show is divisible by for all
I would say that the title of this thread " Annoyingly simple maths problem" is annoyingly stupid, since the problem is not simple.
13. Grime was awesome. Is he going to do research under Simon Singh? From what I've heard this "enigma" thing is just temporary.
14. Induction?
15. (Original post by Simplicity)
Induction?
I'd like to see you try.
16. (Original post by generalebriety)
I'd like to see you try.
Okay.

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