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If xx is the least positive integer satisfying 100≡x(mod6)100 \equiv x \pmod{6}, then (2x+1)(2x+1) is equal to :

Solution

✅ Correct Option: 3

100=6×16+4100 = 6 \times 16 + 4, so 100≡4(mod6)100 \equiv 4 \pmod{6}.

Least positive x=4x = 4.

2x+1=2(4)+1=92x + 1 = 2(4) + 1 = 9.

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