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Difficult Math Trivia Question

Difficult Math Trivia Question - 9 September

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

For Solution : Click Here

7 comments:

  1. Well, trying to avoid modular arithmetic, we can think like this¨
    3^300=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.

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

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    2. 3

      3 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.

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

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  2. 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.
    http://www.wolframalpha.com/input/?i=3%5E300+mod%285%29

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

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

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