If the two-digit number M contains the digits x and y, what is the rem

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

(1) M is not odd.

(2) x + y = 12

Clearly we have M=XY or YX
(1) M is not odd => no info if M is divisible by 3 => insufficient
(2) x + y = 12 => M can be (39) or (93), (48,84),(57,75),(66) => insufficient
(1) and (2) => (48,84,66) all are divisible by 6 =>C

Given that M is a two-digit number composed of x and y. We are to determine the remainder when M is divided by 6.

Statement 1: M is not odd
We can have many possibilities but all we need to do is to prove insufficiency with two possible numbers.
12 is a two-digit number which is not odd and divisible by 6, hence the remainder is 0. 14 is a two-digit number which is not odd, and not divisible by 6, therefore, it has a remainder of 2 when divided by 6. Hence statement 1 is not sufficient on its own.

Statement 2: x+y=12
Not sufficient. This because 93 is a two digit number in which the sum of the digits = 12; it leaves a remainder of 3 when it is divided by 6. Meanwhile 48 also satisfy statement 2 and leaves a remainder of 0 when it is divided by 6.

Combining both statements however is sufficient.
we now have only 48, 66, and 84 as the only possible values of M. Since all the possible numbers are divisble by 6, we can conclude that when M is divided by 6, the remainder is 0.

The answer is therefore C.

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

(1) M is not odd. - so M is even, however, the remainder can be either 0, 2, 4 - insufficient

(2) x + y = 12

possible x,y = (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3) - each of which does not have the same remainder when divided by 6
for example, 39/6 remainder = 3, while 48/6 remainder = 0 - insufficient

Taken statement (1) + (2) together,
the only possible number is 48, 66, 84 - all of them when divided by 6 has 0 as a remainder. So, statement (1) + (2) is sufficient

Therefore, C is the correct answer choice.

10≤M=XY≤99…1+0≤X+Y≤9+9…1≤X+Y≤18

(1) M is not odd: insufic.

M=EVEN; M=32/6…R=2; M=36=6…R=0

(2) x + y = 12: insufic.

x+y=12; (xy)=(39,93,48,84,57,75,66)…R=(3,3,0,0,3,3,0)

(1&2) sufic.

M=Even; x+y=12; (xy)=(48,84,66)…R=(0,0,0)

Answer (C)

Imo. C

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

M = 10, 11, 12,,,,,,99

(1) M is not odd.
M= 10, 10/6=4 (remainder), M = 12/6=0 (remainder)

(2) x + y = 12
9 +3=12, M = 93/39, Remainder =3
8+4 =12, M = 84/48, Remainder =0,

1 +2,
M= 84/48/66 , Remainder =0
Hence, sufficient.

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

(1) M is not odd.

(2) x + y = 12

we need to fine M/6=? and M is a two digit number

1) states that M is even can be 24 or 28 if it is 24 then 24/6=0 remainder and 28/6 = 4 remainder. No definite answer hence eliminated options A and D
2) X+Y= 12 in this scenario X could be 5 and y could be 7 57/6=3 remainder and X=6 and Y=6 then 66/6=11

If we combine both the statements that M is not odd we still have different remainders 66/6=11 and 84/6=14.

IMO E

M= xy
#1
not odd
12,14,16,10 insufficeint
as remainder varies
#2
x+y=12
57,75,66,48,84 insufficient
again
from 1 &2
even no; 66,48,84 remainder always 0
IMO C sufficient

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

(1) M is not odd.

(2) x + y = 12

M — two digit number=xy
xy/6 — remainder???

(Statement1) M is not odd.
—> M could end with 0, 2,4,6, or 8.

—> If xy is equal to 20, then remainder will be 2.
—> if xy is equal to 22, then remainder will be 4.
Insufficient

(Statement2): x+y= 12
xy could be 93, 84, 66 and so on.
—> Remainders will be different from one another.
Insufficient

Taken together 1&2,
—> xy ends with 0,2,4,6 or 8 ( that means all xy numbers is divisible by 2.
—> x+y=12 (any xy number is divisible by 3)

(For a number to be divisible by 6, it must also be divisible by 2 and by 3.)
—> xy is divisible by 2 and by 3.
The remainder will always be ZERO.
Sufficient

The answer is C

Posted from my mobile device

(1) M is not odd.....I.e his even....Without knowing about x we cannot conclude anything....insuff

(2) x + y = 12.....Clearly insufficient since we get many combinations.


Combining both....Values for y=0,2,4,6,8

For x+y=12. Y cannot be 0,2 since x is single digit no...And also cannot be 6 because both,x,y cannot be same values ....


So if y=4 then x=8...M=48
if y=8 then y=4....M=84


Above both values are divisible by 6...


OA:C

Posted from my mobile device

(1) M can be 21,23,25,27, and so on, so remainder on division of M by 6 will be 3,5,1,3, and so on respectively.
Therefore, we don't get a unique value of remainder M on division by 6.
Thus, insufficient.

(2) Thus statement implies that M can be 93,39,84,48,57,75, and 66, so remainder on division of M by 6 will be 3,3,0,0,3,3, and 0 respectively.
Therefore, we don't get a unique value of remainder on M on division by 6.
Thus, insufficient.

From (1) and (2) combined, we get that M can be 93,39,57, and 75, so remainder on division of M by 6 in each case is 3.
Thus, sufficient.

Therefore, the correct answer is option C.

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

M = 10.x + y. The possible remainders when a two digit number is divided by 6 are 4, 5, 0 1, 2 and 3. Here remainders are even when M is even and odd when M is odd i.e. each has three different remainders.

(1) M is not odd.
Three different remainders are possible.

INSUFFICIENT.

(2) x + y = 12
Possible values of x and y are
a) 9 and 3: Number possible are 93 and 39 with both having 3 as remainder.
b) 8 and 4: Number possible are 84 and 48 with both having 0 as remainder.
c) 7 and 5: Number possible are 75 and 57 with both having 3 as remainder.
d) 6 and 6: Number possible 66 with 0 as remainder.

Thus, two possibilities exist.

INSUFFICIENT.

Together 1 and 2

Case a) and b) suffice the requirement.

SUFFICIENT.

Answer C.

Statement 1: M is not odd. When M = 13, the remainder after dividing it with 6 is 1. When M= 23, the remainder is 5. NOT SUFFICIENT.

Statement 2: Sum of the 2 digits of M, x and y is 12. M can be 39, 57, 84, 66, 93 or 75. The remainder is 3, 3, 0, 0, 3, 3 respectively. We can see the remainder is 0 when M is even and 3 when M is odd. NOT SUFFICIENT.

Adding both statement, we can say the remainder is 3.

Answer C.

If the two-digit number M contains the digits x and y, what is the remainder when M is divided by 6?

(1) M is not odd.
if M is even M can be 12 which R is 0
but M can also be 14 and R is 2
therefore insufficient

(2) x + y = 12
x can be 3 and y can be 9 39 divided by 6 result in R3; 93 divided by 6 result in R3
x can be 4 and y can be 8 48 divided by 6 result in R0; 84 divided by 6 result in R0
therefore insufficient

together if
x is 6 and y is 6, 66 divided by 6 R 0
and 8 with 4 yield same result of R0
therefore sufficient
C

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