Determine all positive integers n for which 13n + 6 is divisible by 7. Prove tha

ID: 3724121 • Letter: D

Question

Determine all positive integers n for which 13n + 6 is divisible by 7. Prove that your answer is correct

The procedure of proof should follow the following format. The answer should like that.

Example

Prove:For all integers a,b,c,m where m > 0, if ab (mod m) and bc (mod m),then ac (mod m)

Proof. Let a,b,c,m be arbitrary integers, suppose that m > 0.

(1)Suppose that ab (mod m) and bc (mod m)

(2)From (1) and using a result from last class, we conclude that m(a-b) and m(b-c)

(3)From (2) and the definition of divides, a-b = mk1and b-c = mk2for some integers k1and k2

(4)From (3), we write (a-b) + (b-c) = mk1+ mk2= m(k1+ k2),which means that a - c = m(k1+ k2). By definition of divides, m(a-c)

(5)From (4) and a result from last class, m(a-c) implies that ac (mod m).

Proof: -

Given equation is 13n+6.

Rewriting the above equation

=> (14-1)n + (7-1)

=>14n+7 - n - 1

=> 7(2n+1) - (n+1)

Now the factor 7(2n+1) is always divisible by 7.

So, for the total "7(2n+1) - (n+1)" to be divisible by 7 , (n+1) should be divisible by 7

Let, n+1 = 7k, where k is some natual number, k (--- N.

So, n = 7k - 1.

For all k in set of natural numbers, this gives us all values of n, for which the equation

13n+6 is divisible by 7

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