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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)

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Solution Big numbers are NO PROBLEM for divisibility | Comment 4 of 8 |

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

(3^5)^1111 + (4^2)^1111

on dividing above number by 7 we get remainder as

5^1111 + 2^1111  ----------------------- Eqn 1

Since divisor is 7, remainder 5 is same as remainder -2. This will simplify calculations a lot. For simpleton souls I have also solved it taking Eqn 1. Those who are comfortable with above step, the riddle is almost solved, as remainder now becomes

= (-2)^1111 + 2^1111

= -(2^1111) + 2^1111

= 0

Thus the given number is divisible by 7.

*****************************

From Eqn 1, Remainder is : 5^1111 + 2^1111

5^1111 => (((5^3)*(5^3))^185)*5 => (36^185)*5 => 1*5 = 5

2^1111 => (8^370)*2 => 1*2 = 2 

Thus remiander becomes 5 + 2 = 7 = 0

 

 


  Posted by Rupesh Khandelwal on 2005-05-25 05:52:18
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