Show that \(2450^n-1370^n-1150^n-250^n\) is divisible by 1980 for all \(n\epsilon N\)
Are you trying to denote the set membership of n in N? If so use \in, also what is N?
n belongs to the set of Natural numbers
You should use \mathbb{N}
$$n\in \mathbb{N}$$
Do you know some modular arithmetic?
yeah.. I just don't know how to choose the correct figures ..
Since \(1980=4\cdot 9\cdot 5\cdot 11\), one way to do this is to try and show that the statement is divisible by 4, 9, 5 and 11.
oops its 1980 my bad
Reduce 2450 modulo 1980, then if you can show the expression is zero modulo 4,9,5 and 11 your done.
can't i directly show that the statement is divisible by 5 and 396 and hence prove that it is also divisible by 1980?
That works too.
yes, sense there both coprime
Thanks! So, if i come across such problems, all i need to do is find the factors first and then prove that the expression is divisible by a prime factor among them and the product of the rest all factors.. (as long as the product of rest all factors is not a multiple of the prime number itself)
Yes, but for a proof of this and other techniques I would read up on some elementary number theory, if you plan on using it in the future or somthing.
elementary number theory like?
Modular arithmetic
yeah, i know modular arithmetic.. but i didn't know how to apply that in solving this.. Now i do!
If gcd(a,b)=1, and a divides n, and b divides n. Then ab divides n, this is essentially what were using here, to simplify the problem into cases where the expression is divisible by smaller integers.
Ah, yes!! Thanks!! OS is always helpful! :)
I might be wrong, but I dont think your statement is divisible by 9. Maybe one of those minus signs should be a plus? Then it will work nicely.
If the problem was \[2450^n-1370^n+1150^n-250^n\]it would make more sense.
Ah yes, That was a typo.. Sorry! Its \(+1150^n\)
Join our real-time social learning platform and learn together with your friends!