SOLUTION: find the remainder when 3^3x33^33x333^333x3333^3333 is divided by 100

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Question 1111649: find the remainder when 3^3x33^33x333^333x3333^3333 is divided by 100
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
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All you need to know to solve the problem is "What are 2 last digits of each separate multiplier ?"


It is the same as to ask  "What is   mod 100 ?"


It is not so easy to answer the question.

It is not so easy even for MS Excel.

But if you use  MS Excel recursive function   = MOD( (33*MOD(a[n],100), 100), starting from   = 33,

you will get the answer easily:   mod 100   is periodic with the period  20.


In other words,   mod 100 =33,    

or, which is the same,  the last two digits of the number  sequence  , n = 1, 2, 3, . . .  repeat  with the period of  20.


If you do not believe me, make calculations on your own.


It is enough (for an advanced mind, which you, without any doubt, are, by bringing so advanced problem . . . ) to complete the solution.


I want to say that the rest is simple mechanical procedure . . . 

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The lesson of this site might be useful to you
    - Find the last three digits of these numbers

Similar problems were solved there, and you will find there the required ideas and techniques.


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