A new cool riddle/problem to solve

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!

Master Shin's back with his riddles!!

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

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 :rolleyes:

- Matt

isn't the actual problem something like this:

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

??

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?

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