Division Algorithm question

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prove that when a perfect square is divided by 9, the remainder is never 2,3,5,6 or 8

Oct 1st, 2015

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All numbers will be of the form  9k + r

where r is the remainder when you divide by 9. The values of r must be in the range  0, 1, 2, 3, 4, 5, 6, 7, 8

(9k+ 0)(9k+0) = 81k^2 which has a remainder of zero when divided by 9

(9k+1)(9k+1) = 81k"2 + 18k + 1 so the remainder when divided by 9 is 1

(9k + 2)(9k+2) = 81k^2 + 36k + 4 so the remainder when divided by 9 is 4

(9k+3)(9k+3) = 81k^2 + 54k + 9 so the remainder is zero when divided by 9

(9k+4)(9k+4) = 81k^2 + 72k + 16 = 81k^2 + 72k + 9 + 7

so the remainder when divided by 9 s 7

(9k+5)(9K+5) = 81K^2 + 90K + 25 = 81K^2 + 90k + 18 + 7

so the remainder when divided by 9 is 7

(9k+6)(9k+6) = 81k^2 + 108k + 36 since 36 = 9*4 the remainder when be zero when divided by 9

(9k+7)(9k+7) = 81k^2 + 126k + 49 = 81k^2 + 126k+ 45 + 4

so the remainder when divided by 9 is 4

(9k+8)(9k+8) = 81k^2 + 144k+ 64 = 81k^2 + 144k + 63 + 1

so the remainder when divided by 9 is 1

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 1st, 2015

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