I messed up with both the statements in order to prove them wrong.....Statements are always right one should not waste time to prove it wrong..
statements 1 and 2 both sufficient hence D
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wingsoffire
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prateek_guy2004
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Don't look for the incorrect things that you have done rather look for remedies....
https://www.beatthegmat.com/motivation-t90253.html
https://www.beatthegmat.com/motivation-t90253.html
How does statement 2 work out?
Plus in:
x = 15, therefore according to statement II x/12 = 1 R 3 and x/6 = 2 R 3
x = 18, therefore according to statement II x/12 = 1 R 6 and x/6 = 3 R 0
Therefore, as you change x, the remainder of x/6 does NOT remain consistent with the remainder of x/12. Isn't the answer A (Statement I alone is suffient).
Please explain why statement II is sufficient on its own.
Plus in:
x = 15, therefore according to statement II x/12 = 1 R 3 and x/6 = 2 R 3
x = 18, therefore according to statement II x/12 = 1 R 6 and x/6 = 3 R 0
Therefore, as you change x, the remainder of x/6 does NOT remain consistent with the remainder of x/12. Isn't the answer A (Statement I alone is suffient).
Please explain why statement II is sufficient on its own.
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ronnie1985
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Q. R(x/6) = ?
S1. R(x/2) = 1
R(x/3) = 0
x = 3*(2n+1) = 6n+3
R(x/6) = R((6n+3)/6) = 3
Sufficient
S2. R(x/12) = 3
x = 12n+3
R(x/6) = R((12n+3)/6) = 3
Sufficient
(D) is answer
S1. R(x/2) = 1
R(x/3) = 0
x = 3*(2n+1) = 6n+3
R(x/6) = R((6n+3)/6) = 3
Sufficient
S2. R(x/12) = 3
x = 12n+3
R(x/6) = R((12n+3)/6) = 3
Sufficient
(D) is answer
Follow your passion, Success as perceived by others shall follow you
D?aznmexicana wrote:What is the remainder when the positive integer x is divided by 6?
1) When x is divided by 2 the remainder is 1, and when x is divided by 3, the remainder is 0
2) When x is divided by 12 the remainder is 3
A LCM 6
B Factor of 6
Plug in numbers for each statement and you will find that both statement are sufficient independently.
1) x = 2m+1 = 3n
Try m = 3k, 3k+1 & 3k+2 => m = 3k+1
=> x = 6k+3 => remainder = 3
2) x = 12k+3 = 6(2k)+3 => remainder = 3
Choose D
Try m = 3k, 3k+1 & 3k+2 => m = 3k+1
=> x = 6k+3 => remainder = 3
2) x = 12k+3 = 6(2k)+3 => remainder = 3
Choose D
