**Difficult Math Trivia Question - 9 September**

Can you find out the remainder when 3^300 is divided by 5?

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Can you find out the remainder when 3^300 is divided by 5?

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Well, trying to avoid modular arithmetic, we can think like this¨

ReplyDelete3^300=9^600=(10-1)^600

10 clearly divides 5 (in modular terms 10=0 mod(5) )

and 1^600=1 (even power of 1) , so the overall remainder is 5+1=6 mod(5)=1 . The remainder is 1.

Alternative way of getting the remainder:

3^300=3^(3*100) =3^(3*10*10) =3^(3*2*5*2*5)

3^3= 27= 2mod5 (27/5=5+ 2 remainder)

3^2=9=4 (mod5)

3^5=243 =3 (mod 5)

So we have: 2+4+4+3+3=16 =1 mod5 ,thus remainder =1.

typo correction: (-1)^600=1 (even power)

Delete3

Delete3 It is obvious that it is not feasible to calculate 3^300 as it will take too much of time. So we will use a trick to solve the question. We will calculate the remainder of each power till we find a pattern. 3^1 divided by 5 leaves the remainder 3. 3^2 divided by 5 leaves the remainder 4. 3^3 divided by 5 leaves the remainder 2. 3^4 divided by 5 leaves the remainder 1. 3^5 divided by 5 leaves the remainder 3. 3^6 divided by 5 leaves the remainder 4. As you can see that the pattern is now repeating itself and it will go on like this till 3^300 and beyond. Since every fourth remainder is same as the first, we will look for the power of 4 only. 300 is divisible by 4. Therefore at the power of 300, the first remainder will repeat itself and the remainder will be 3.

@admin: it's not "first reminder" you are taking.you should

Deletetake "zeroth reminder".i mean 3^0 % 5=1;will further be 3^4 % 5=1;3^8 % 5=1; and so on till 3^300 % 5=1;

i don't know why you are taking 3^1 % 5 which make no sense.so answer is obeviously 1.(% is reminder)

Admin, sorry but your solution is wrong. Since i don't intend to give a Number Theory lesson (my comment was explanatory enough), just check the result here...(3^300= 1 mod (5) means remainder 1.

ReplyDeletehttp://www.wolframalpha.com/input/?i=3%5E300+mod%285%29

Admin, your thinking is actually correct but you misinterpret the cyclicity of modulos. It is not "every forth term" BUT every forth power.

ReplyDeleteThat's why 3^4 = 1 mod5, 3^8=1 mod5, 3^12 =1 mod5....and 3^300=3^(4*75)= also 1 mod5, of course... you can check it out for small numbers yourself.

I meant to say above: "BUT every power which is a product of 4" (is 1 mod(5)

Delete