Modular Arithematicsee if my answer is correctby saadsarfraz Tags: arithematicsee, correct, modular 

#1
Jan2509, 08:29 PM

P: 84

Q What is the remainder when 1+2+2[tex]^{2}[/tex]+...+2[tex]^{219}[/tex] is divided by 5.
Solution: 2[tex]^{0}[/tex]=1 mod5 2[tex]^{1}[/tex]=2 mod5 2[tex]^{2}[/tex]=4 mod5 2[tex]^{3}[/tex]=3 mod5 2[tex]^{4}[/tex]=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
Jan2509, 08:30 PM

P: 84

Can anyone see if this is correct.




#3
Jan2509, 10:58 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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? 



#4
Jan2609, 12:04 AM

P: 84

Modular Arithematicsee if my answer is correct
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?




#5
Jan2609, 06:50 AM

Emeritus
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PF Gold
P: 11,154

Oops, I made a silly mistake. You are correct that the pattern 1,2,4,3 is repeating, since [itex]2^{m+4} \equiv 2^m ~(mod 5)[/itex].
So you have found out that: [tex]\sum_{n=0}^{219}2^n \equiv 550~(mod 5) [/tex] 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). 



#6
Jan2609, 07:53 PM

P: 84

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




#7
Jan2609, 08:49 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

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



#8
Jan2609, 10:23 PM

P: 626

Geometric series?
Spoiler
Doesn't that sum to [tex] 2^{220}  1 [/tex]?



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