Thu March 11, 2010 By: Sashank Gadia

Explain ?

1 Comments

7 years ago

Dear student


The last 3 digits of any number is the remainder when that number is divided by 1000, so we just calculate powers of 17 up to 256 as efficiently as possible and reduce them modulo 1000 as we go along.



(From here on "=" will mean congruent to)

17^2 = 289 (mod 1000)

17^4 = (17^2)^2= 289^2 = 83521 = 521 (mod1000)

17^8 = (17^4)^2 = 521^2 = 271441 = 441 (mod 1000)

17^16 = (17^8)^2 = 441^2 = 481 (mod 1000)

17^32 = 481^2 = 361 (mod 1000)

17^64 = 361^2 = 321 (mod 1000)

17^128 = 321^2 = 41 (mod 1000)

17^256 = 41^2 = 681 (mod 1000)



So the last 3 digits of 17^256 are 681.


Regards


Team


Topperlearning.com

Kasim Razak 2 2 days ago
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Kasim Razak 3 2 days ago
Good. I like this video/chapter/test/question/greeting
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