Show that 43 to the power of n+83*92 to the power of 3n-1 is divisible by 7 for all 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

(P.S. Yes, I did ask a question similar to this before except I mistyped it and therefore needed to type it again)

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

Answer 1

I presume you mean:
F(n)= (43^n) + 83*(92^(3n-1))

Check if it is true for n=1
43 + 83*92^2
= 702555
= 7 * 100365

Check if it is true for n=2
43^2 + 83*92^5
= 547037666105
= 7 * 78148238015

It is true for n=1 and n=2

Assume it is true for n=m
F(m+1) - F(m)
= 43^(m+1) - 43^m + 83*(92^(3(m+1)-1) - 92^(3m-1))
= (43^m)(43-1) + 83*(92^(3m-1))*(92^3 - 1)
= (43^m)*42 + 83*(92^(3m-1))*(778687)
= 7*[(43^m)*6 + 83*(92^(3m-1))*(111241)]

If F(m) is divisible by 7 then F(m+1) will be divisible by 7
F(1) is divisible by 7
Therefore F(n) is divisible by 7 for all values of n by mathematical induction.

Answer 2

Let's use mod 7 arithmetic(also called calendar
arithmetic). (Think of 7 days hence as
landing on the same day of the week.)
I assume you are looking at 43 ^n + 83(92 ^(3n-1) ) here.
Actually, more is true:
43^m + 83*(92^n) is divisible by 7 for ALL positive
integers m and n.
Look: 43 = 1(mod 7)
So 43^m = 1(mod 7) for all m.
Next, 92 = 1(mod 7) and 83 = -1(mod 7)
So 43^m + 83*(92^n) = 1 - 1 = 0(mod 7).

Answer 3

if n is a natural number then
k=(n+83*92)^(3n-1) and k= (n+83*92)(3n-1) are also natural numbers for all n
43 is prime, its only divisors are 1 and 43
so 43^k also only has divisors 1 and 43
so 7 doesn't divide 43^k for any natural number k

||Repost||
I can't believe that people thumb me down for misreading a question which was miswritten.
In the case of 43^n + 83*92^(3n-1)
your question now makes sense.
Using the mod function with properties
(x+y)(mod7) = x(mod7) + y(mod7) and
(xy)(mod7) = x(mod7) * y(mod7)
gives
43(mod7) = 1(mod7)
so
[43^n] (mod7) = [1^n](mod7) = 1(mod7)
and
83(mod7) = -1(mod7)
and
92(mod7) = 1(mod7)
so
[92^(3n-1)] (mod7) = [1^(3n-1)](mod7) = 1(mod7)
therefore
[43^n + 83*92^(3n-1)](mod7) = [1 + (-1)*1] (mod7) = 0(mod7)
so 7 divides 43^n + 83*92^(3n-1) for all n

the end
.

Answer 4

Do you mean 43^{[(n+83)(92)]^3n-1}?

Answer 5

What level are you? Junior school, high school, undergrad?

Don't you even know how to use brackets to clarify what you mean?

There are several different interpretations of your question. Until you clarify what the expression is, I doubt you will get many sensible answers.
.