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If mm is a positive integer then the values of kk for which 6m+k6m + k cannot be a perfect square are

Solution

Correct Option: 3

Find which residues a perfect square can leave when divided by 66.

Check squares of 0,1,2,3,4,50, 1, 2, 3, 4, 5 modulo 66:

02=00^2 = 0

12=11^2 = 1

22=42^2 = 4

32=933^2 = 9 \equiv 3

42=1644^2 = 16 \equiv 4

52=2515^2 = 25 \equiv 1


Possible residues of a perfect square mod 66: {0,1,3,4}\{0, 1, 3, 4\}.

Residues 22 and 55 are impossible.


Since 6m+kk(mod6)6m + k \equiv k \pmod{6}, the values of kk for which 6m+k6m + k can never be a perfect square are exactly those kk giving impossible residues:

k=2k = 2 and k=5k = 5

Answer =2= 2 and 55

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