sequence and divisibility question

Question

A sequence begins 2, 5, 7, 12, 19, … Each term (after the first two terms) is equal to the sum of the two previous terms. How many of the following statements are true?

I. The 20th term is divisible by 2 II. The 40th term is divisible by 2 III. The 40th term is divisible by 3 IV. The 60th term is divisible by 3

How do we do this? Are there specific properties that we need to know?

Answer 1

Let's work modulo 2 and 3. 2 first.

The first term is 0 mod 2. The next is 1 mod 2. If we continue this pattern, we have:

0, 1, 1, 0, 1, 1, ...

This is easy to prove. Anyway, this makes the 20th term 1 mod 2 and the 40th term 0 mod 2, so I is false and II is true.

Now, modulo 3 we have:

2, 2, 1, 0, 1, 1, 2, 0, 2, 2, ...

This repeats with the sequence 22101120. So the 40th term is 0 mod 3, and the 60th term is also 0 mod 3. So statements III and IV are also true. The answer is 3.

Answer 2

A couple of hints for the first two:

1. Even + Odd = Odd
2. Odd + Odd = Even

For the last two, look for a pattern for the terms modulo $3$.

Answer 3

I assume the first term (not the zero-th term) is considered to be $2$.

Consider the sequence modulo $2$. It will be

$$0,1,1,0,1,1,0,1,1,\cdots$$ So the $3n+1$-st term is divisible by $2$. Thus statement I is false, but II is true.

Consider the sequence modulo $3$. It will be

$$2,2,1,0,1,1,2,0,2,2,1,0,1,1,2,0,\cdots$$ so terms $8k+4$ and $8k$ will be divisible by $3$. Thus statement III is true, and IV is also true.