A new cool riddle/problem to solve

[MasterShin]

There was another math contest, and bring you a new problem from there.
Now, I did not solve this one... Lets see you:

Prove that for any normal (1,2,3 not 1.5) n at n^2 + 5n + 16, the number will never devide 169 exactly.

Wish you lots of luck!

[webprodesign]

Master Shin's back with his riddles!!

Lets see if LuxFX gets this one first, but I'll try to figure it out.

[calpomatt]

for any normal (1,2,3 not 1.5) n at n^2 + 5n + 16, the number will never devide 169 exactly.

169 is only divisible by +/-1, +/-13, and +/-169 (to get whole numbers)

If you can only use normal numbers, you can only use 1, 13, and 169.

For x=1
n2+5n+16=1 (n is normal)
n does not exist

For x=13
n2+5n+16=13 (n is normal)
n does not exist

For x=169
n2+5n+16=169 (n is normal)
n is some number between 10 and 11....like 10.1 or something.....anyways, not a 'normal' number

therefore...there is no n which will divide 169 exactly.

DOMINATED!!

ya....it's probably not even right

- Matt

[TrIzKuT]

isn't the actual problem something like this:

(n x n) + (5 x n) + 16 = ?

??

[MasterShin]

calpomatt - I didn't get your explenation.

lets say 169000 - it can't be into 'n*n + 5*n + 16' with a normal n... But what about 169169 and so on?

[calpomatt]

Basically I was looking at the answer to prove it wrong and worked my way backwards.

I started with the idea that n*n + 5*n + 16 = x (where x is a number that divides 169 evenly)

The only natural numbers (x) that would divide 169 evenly are 1, 13, and 169. (prime factorization or something)

I then solved for (n) after placing (x) into the equation. When I found that there could be no (n) to satisfy the equation, it proved the the theorem was correct.

I have no clue if this is right though....so don't sweat it if it makes no sense.

- Matt

for my answer n2 -> 'n*n'
my keyboard is busted so I can't type that little /\ looking thing