CBSE Class 11-science Answered

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.

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Team

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