# Modular Arithematic-see if my answer is correct

1. ### saadsarfraz

84
Q- What is the remainder when 1+2+2$$^{2}$$+...+2$$^{219}$$ is divided by 5.

Solution: 2$$^{0}$$=1 mod5
2$$^{1}$$=2 mod5
2$$^{2}$$=4 mod5
2$$^{3}$$=3 mod5
2$$^{4}$$=1 mod5

Now I take (1,2,4,3) to be a set numbers. Since the summation goes to 219, there are a total of 220/4 = 55 sets. So I add 1+2+4+3= 10 and 10*55 = 550 <- my answer.

2. Science news on Phys.org
3. ### saadsarfraz

84
Can anyone see if this is correct.

4. ### Gokul43201

11,141
Staff Emeritus
No, it has a couple of errors. First, how many residues did you compute (4 or 5)? Or how many congruence classes are there modulo 5?

Second, can the remainder exceed the divisor?

5. ### saadsarfraz

84
i computed 4 residues, the pattern 1,2,3,4 keeps on repeating. you are right about remainder exceeding the divisor. any ideas about how i should compute this?

6. ### Gokul43201

11,141
Staff Emeritus
Oops, I made a silly mistake. You are correct that the pattern 1,2,4,3 is repeating, since $2^{m+4} \equiv 2^m ~(mod 5)$.

So you have found out that:

$$\sum_{n=0}^{219}2^n \equiv 550~(mod 5)$$

What is the least residue of 550 (mod 5)? That is the required answer.

Now alternatively, you should also be capable of identifying the kind of series that is given to you and evaluating it directly (before looking at congruences).

7. ### saadsarfraz

84
so 550 mod 5 is 0, as that leaves no remainder right?

8. ### Gokul43201

11,141
Staff Emeritus
Correct.

Have you thought about the more direct approach, by summing the series?

9. ### NoMoreExams

626
Geometric series?

Doesn't that sum to $$2^{220} - 1$$?

Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?