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(1) x divided by 2 leaves a remainder of 1. So x is odd and not divisible by 6 so remainder cannot be 0. x is a multiple of 3 and odd. Quick check shows 9 divided by 6 leaves a remainder of 3. 15 divided by 6 leaves a remainder of 3. Every odd multiple of 3 is 6 + the previous odd multiple. So any odd multiple of 3 will always leave the remainder 3. SO SUFFICIENT.
(2) x when divided by 12 leaves a remainder 3. so x-3 is divisible by 12. Means (x-3) is also divisible by 6. So x should leave the same remainder 3 when divided by 6. Mathematically:
x = 12a+3 = 6*(2a)+3= 6*b+3. So remainder when x is divided by 6 will always be 3. SO SUFFICIENT. |
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