document.write( "Question 1111649: find the remainder when 3^3x33^33x333^333x3333^3333 is divided by 100 \n" ); document.write( "
Algebra.Com's Answer #726639 by ikleyn(52797)\"\" \"About 
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document.write( "All you need to know to solve the problem is \"What are 2 last digits of each separate multiplier ?\"\r\n" );
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document.write( "It is the same as to ask  \"What is  \"33%5En\" mod 100 ?\"\r\n" );
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document.write( "It is not so easy to answer the question.\r\n" );
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document.write( "It is not so easy even for MS Excel.\r\n" );
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document.write( "But if you use  MS Excel recursive function  \"a%5Bn%2B1%5D\" = MOD( (33*MOD(a[n],100), 100), starting from  \"a%5B1%5D\" = 33,\r\n" );
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document.write( "you will get the answer easily:  \"33%5En\" mod 100   is periodic with the period  20.\r\n" );
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document.write( "In other words,  \"33%5E21\" mod 100 =33,    \r\n" );
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document.write( "or, which is the same,  the last two digits of the number  sequence  \"33%5En\", n = 1, 2, 3, . . .  repeat  with the period of  20.\r\n" );
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document.write( "If you do not believe me, make calculations on your own.\r\n" );
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document.write( "It is enough (for an advanced mind, which you, without any doubt, are, by bringing so advanced problem . . . ) to complete the solution.\r\n" );
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document.write( "I want to say that the rest is simple mechanical procedure . . . \r\n" );
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\n" ); document.write( "The lesson of this site might be useful to you\r
\n" ); document.write( "\n" ); document.write( "    - Find the last three digits of these numbers \r
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\n" ); document.write( "\n" ); document.write( "Similar problems were solved there, and you will find there the required ideas and techniques.\r
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