Describe all n such that 3 devides n2^n+1

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In summary, the problem is to find all values of n such that n*2^n + 1 is divisible by 3. To solve this, it is helpful to first analyze the sequence of 2^n mod 3 and then determine the sequence of n*2^n + 1. The period of these two sequences will give the period for the description of n*2^n + 1. With this in mind, the original approach of assuming n*2^n + 1 = 0 (mod 3) and finding for which values of n this is possible is correct.
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rbetan
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Describe All n such that 3 divides n times 2 raised to the n, plus one.( n*2^n + 1)

I know that a number is divisible by 3 if the sum of its digits adds up to a number that itself is divisible by 3. But this is probably not helpful for this problem.

I also know that a number A is divisible by B if when A is divided by B the remainder is 0. In other words, A is congruent to 0 in mod B.

My original idea is to assume that n*2^n + 1= 0(mod 3). And then somehow try to see for which values of n this is possible. But i am stuck.

Maybe this is not the right approach.

Any help with this will be highly appreciated. Thanks
 
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  • #2
Break the problem down to smaller steps what is the sequence of 2^n mod 3? Then what is n time this plus 1? P.S. The period of the sequence for n mod 3 is different from the period for 2^n mod 3. Their product is the period for the description of n*2^n + 1. With this in mind your original approach is correct.
 

1. What is the significance of the expression n2^n+1 in this problem?

The expression n2^n+1 is important because it represents a number that is being tested for divisibility by 3. The goal of this problem is to find all possible values of n that make this expression divisible by 3.

2. How do you approach solving this problem?

To solve this problem, we can use the rules of divisibility to simplify the expression. We know that for a number to be divisible by 3, the sum of its digits must be divisible by 3. We can apply this rule to the expression n2^n+1 by looking at the sum of its digits: n+2^n+1. From there, we can see that if n is odd, the sum of its digits will be even, and therefore not divisible by 3. But if n is even, the sum of its digits will be odd, and therefore divisible by 3. This leads us to the conclusion that all even values of n will make the expression divisible by 3.

3. Are there any other ways to solve this problem?

Yes, there are other ways to approach this problem. Another method is to use modular arithmetic. We can rewrite the expression as 2^n mod 3, which will give us the possible remainders when 2^n is divided by 3. From there, we can see that when n is even, the remainder will be 1, which means the expression is divisible by 3. When n is odd, the remainder will be 2, which means the expression is not divisible by 3.

4. Is there a general solution for this problem?

Yes, there is a general solution for this problem. We can use algebraic manipulation to rewrite the expression as (2^n+1)/3. From there, we can see that this expression will be an integer when 2^n+1 is divisible by 3. By using the rules of exponents, we can simplify 2^n+1 to (2+2^n)/3. This means that any value of n that makes 2+2^n divisible by 3 will also make the original expression divisible by 3.

5. Can this problem be extended to other divisors?

Yes, this problem can be extended to any divisor. Instead of looking for values of n that make the expression divisible by 3, we can look for values that make it divisible by any other number. For example, if we want to find all values of n that make the expression divisible by 5, we can use the same approach and rewrite the expression as (2^n+1)/5. This means that any value of n that makes 2+2^n divisible by 5 will also make the original expression divisible by 5.

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