Show that 43 to the power of n+83*92 to the power of 3n-1 is divisible by all of positive integers n?

Question

Could you please, instead of just giving the answer, take me step by step through the process?
Thanks! Your help is much appreciated ^^

Answer 1

Sorry, this statement is palpably false!
In fact, the only integer divisible by all positive integers
n is 0.
Proof:
1 is only divisible by 1.
Let k be any integer greater than 1.
Then we know
by the prime decomposition law,
k = p1^a1... pt^at,
where p1, ... , pt are distinct primes.
In particular, k is divisible by only a finite
number of primes.
But we know there are infinitely many primes,
so there are infinitely many primes not dividing k.
Now we have to prove that
43^(n+83) + 92^(3n-1) is not zero.
But that's clear, because both terms of
the sum are positive for all n.

Answer 2

enable's use mod 7 arithmetic(called calendar arithmetic). (think of of seven days as a result as touchdown on the comparable day of the week.) i assume you're finding at 40 3 ^n + eighty 3(ninety two ^(3n-a million) ) here. actual, extra is actual: 40 3^m + eighty 3*(ninety two^n) is divisible via 7 for ALL valuable integers m and n. look: 40 3 = a million(mod 7) So 40 3^m = a million(mod 7) for all m. next, ninety two = a million(mod 7) and eighty 3 = -a million(mod 7) So 40 3^m + eighty 3*(ninety two^n) = a million - a million = 0(mod 7).

Answer 3

{43^(n+83)} * {92^(3n-1)}

... will always work if n = 1.
... will still work if n = 2 (since 92 is even)
but 43^86 * 92^8 is not divisible by 3 since none of the original bases are divisible by 3 to start with.

The statement fails.
d:

Answer 4

This is time taking or I would do it for u..
sorryyyyyyyyyyyyyyyyyy/...............

Answer 5

No Thanks, I'm full.