the original question states:
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
My answer is A and the solution answer is D
Here is the solution provided, which I found is inaccurate with the question, for the question did not mention about the reminder 2: [spoiler]There is no useful rephrasing that can be done for this question. However, we can keep in mind that for a number to be divisible by 9, the sum of its digits must be divisible by 9.
(1) SUFFICIENT: The sum of the digits a and b here is not divisible by 9, so X is not divisible by 9. It turns out, however, that the sum of the digits here can also be used to find the remainder. Since the sum of the digits here has a remainder of 2 when divided by 9, the number itself has a remainder of 2 when divided by 9.
We can use a few values for a and b to show that this is the case:
When a = 5 and b = 6, 56 divided by 9 has a remainder of 56 – 54 = 2
When a = 7 and b = 4, 74 divided by 9 has a remainder of 74 – 72 = 2
(2) SUFFICIENT: If X + 7 is divisible by 9, X – 2 would also be divisible by 9 (X – 2 + 9 = X + 7). If X – 2 is divisible by 9, then X itself has a remainder of 2 when divided by 9.
Again we could use numbers to prove this:
If X + 7 = 27, then X = 20, which has a remainder of 2 when divided by 9
If X + 7 = 18, then X = 11, which has a remainder of 2 when divided by 9
The correct answer is D. [/spoiler]
I would like to know if the error is in the solution or did I miss anything, thanks
a question from MGMAT CAT online, I think there's a mistake
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I think the solution is corect.
Stmt I
For any 2 digit number in which the 2 digits (is unique) since they add up to 11 the remainder is 2
SUFF-> One defnite answer for remainder
Stmt II
x+7 / 9 = Some integer k
x = 9k - 7
Any integer of this form will always leave a remainder of 2 when divided by 7
Think of it this way If x+7 is divisble by 9 , x must have been 2 away from a number that was divisible by 9 so the remainder will always be a 2.
SUFF
Choose D)
Stmt I
For any 2 digit number in which the 2 digits (is unique) since they add up to 11 the remainder is 2
SUFF-> One defnite answer for remainder
Stmt II
x+7 / 9 = Some integer k
x = 9k - 7
Any integer of this form will always leave a remainder of 2 when divided by 7
Think of it this way If x+7 is divisble by 9 , x must have been 2 away from a number that was divisible by 9 so the remainder will always be a 2.
SUFF
Choose D)
Last edited by cramya on Mon Dec 29, 2008 4:01 am, edited 1 time in total.
got it, i confused myself, read the st. incorrectly, thankscramya wrote:I think the solution is corect.
Stmt I
For any 2 digit number in which the 2 digits (is unique) since they add up to 11 the remainder is 2
SUFF-> One defnite answer for remainder
Stmt II
x+7 / 9 = Some integer k
x = 9k - 7
Any integer of this form will always leave a remainder of 2 when divided by 7
Think of it this way If x+7 is divisble by 7 x must have been 2 away from a number that was divisible by 9 so the remainder will always be a 2.
SUFF
Choose D)
- ronniecoleman
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IMO D
X + 7 is divisible by 9
X + 7 = 9K
X = 9K - 7
X = 9K -9 + 2
X = 9(K-1) + 2
Hence we have remainder 2
X + 7 is divisible by 9
X + 7 = 9K
X = 9K - 7
X = 9K -9 + 2
X = 9(K-1) + 2
Hence we have remainder 2
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