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Divisibility by 7 (Posted on 2005-05-22) Difficulty: 2 of 5
(2222^5555 + 5555^2222) is or isn't divisible by 7 ?

Just pencil and paper.

See The Solution Submitted by pcbouhid    
Rating: 2.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Big numbers are NO PROBLEM for divisibility | Comment 7 of 8 |
(In reply to Big numbers are NO PROBLEM for divisibility by Rupesh Khandelwal)

RUPESH U WERE RIGHT BUT A BIT LENGTHY

hmmmm right upto

<TABLE width="100%" border=0> <TBODY> <TR> <TD class=content>Big numbers are NO PROBLEM for divisibility </TD> <TD class=content vAlign=top align=right><NOBR><IMG src="http://perplexus.info/images/perplexus/icons/up.gif" border=0> | Comment 4 of 6 | <IMG src="http://perplexus.info/images/perplexus/icons/down.gif" border=0> </NOBR></TD></TR></TBODY></TABLE>

Given number is divisible by 7.

This problem can be easily solved by focusing on the remainder when the given number is divided by 7.

Dividing (2222^5555 + 5555^2222) by 7 leaves following remainder

3^5555 + 4^2222 which is same as

THAN -----

 (3^2)[(3^3)^1851]  + (4^2)[(4^3)^740]

3^3 div by 7 gives remainder -1 , 4^3 div 7 gives rem +1

so we have 9 x (-1)^1851 + 16 x (1)^740 = 7 which wen divided by seven gives 0 as remainder 


  Posted by phi on 2006-03-09 04:43:36
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