2 digit number

[nidhis.1408]

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

[user123321]

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

1) a+b = 11
=> the possible numbers are 29,38,47,......83,92
and each of this number when div by 9 gives rem 2. hence sufficient.

2) x+7 is div by 9
so x when div by 9 leaves remainder 2. hence sufficient.

IMO D

user123321

[neelgandham]

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

Let X = 10a + b = 9a + a + b = 9a + 11 = 9a + 9 + 2 = 9(a+1) + 2
Let X = 10b + a = 9b + b + a = 9b + 11 = 9b + 9 + 2 = 9(b+1) + 2

If X is divided by 9 then the remainder is 2.

(2) X + 7 is divisible by 9

X + 7 is divisible by 9, remainder is 2.

IMO Option D

[shankar.ashwin]

For the number to be divisible by 9, a+b should be multiple of 9.

(1) a+b = 11 which is not a multiple of 9.

(2) X + 7 is divisible by 9, then X will not be divisible by 9.

D

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

[neelgandham]

For the number to be divisible by 9, a+b should be multiple of 9.

(1) a+b = 11 which is not a multiple of 9.

(2) X + 7 is divisible by 9, then X will not be divisible by 9.

D

I think it is just not sufficient to answer the question 'Is n a multiple of 9?' or 'Is n divisible by 9 ?' because the question posed here is 'What is the Remainder?' and An answer-NO to the question Is n divisible by 9 ? results in 8 unique remainders.

Please correct me if I am wrong.

[shankar.ashwin]

Oh! right..

I didn't see the remainder, i thought it asked if it was divisible by 9. luckily answer was right thanks btw.

For the number to be divisible by 9, a+b should be multiple of 9.

(1) a+b = 11 which is not a multiple of 9.

(2) X + 7 is divisible by 9, then X will not be divisible by 9.

D

I think it is just not sufficient to answer the question 'Is n a multiple of 9?' or 'Is n divisible by 9 ?' because the question posed here is 'What is the Remainder?' and An answer-NO to the question Is n divisible by 9 ? results in 8 unique remainders.

Please correct me if I am wrong.

[pemdas]

this q. tests divisibility rule (by 9)
st(1) is Sufficient as we are suggested about the sum of digits a+b=11 their division will always return remainder of 2
st(2) is also Sufficient, because X+7 can be divisible by 9 only if (a+b+7)/9 and (a+b)/9 + 7/9
(a+b) will bear two additional units (after 9) onto 7/9 for clean divisibility

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

[pemdas]

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

Guys, I have just realized there's a flaw - either in our logic or the q. itself
X can be (-) negative number too, in such case the answer tends to be A (st.1 is Sufficient Only)

example X=-34 and -34+7=-27 which is divisible by 9, BUT -34/9 will have the remainder of 7

whereas for all positive numbers X we have the remainder of 2

[shankar.ashwin]

Pemdas, negative remainders works opposite to positive remainders.

Here is the remainder is "-7" , the actual remainder is still 9-7 = 2.

I am guessing topic of negative remainder is outside the scope of the GMAT, but just incase you're interested.

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

Guys, I have just realized there's a flaw - either in our logic or the q. itself
X can be (-) negative number too, in such case the answer tends to be A (st.1 is Sufficient Only)

example X=-34 and -34+7=-27 which is divisible by 9, BUT -34/9 will have the remainder of 7

whereas for all positive numbers X we have the remainder of 2

[pemdas]

certainly there's a flaw, i know they are opposite
the q. doesn specify +ve/-ve remainder, just remainder
also -7 is remainder when q=am+p and not when q/m=a+p/q, the latter has remainder in the numerator and cannot be assigned +ve/-ve sign. Here though it's -7, you're right as we divide by 9 and not -9.
this q. is not from GMAT series rather some made up one

Pemdas, negative remainders works opposite to positive remainders.

Here is the remainder is "-7" , the actual remainder is still 9-7 = 2.

I am guessing topic of negative remainder is outside the scope of the GMAT, but just incase you're interested.

If a and b are the digits of the two-digit number X, what is the remainder when X is divided by 9?

(1) a + b = 11

(2) X + 7 is divisible by 9

Guys, I have just realized there's a flaw - either in our logic or the q. itself
X can be (-) negative number too, in such case the answer tends to be A (st.1 is Sufficient Only)

example X=-34 and -34+7=-27 which is divisible by 9, BUT -34/9 will have the remainder of 7

whereas for all positive numbers X we have the remainder of 2