If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an

If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an integer?

It's given that x and y are two-digit integers. Any two-digit integer can be represented as 10a + b (where a and b are single digit integers), for example 37 = 3*10 + 7, 88 = 8*10 + 8, etc.

(1) x and y have the same two digits, but in reverse order.

So, given that x = 10a + b and y = 10b + a;

\(\frac{(x - y)}{9} = \frac{(10a+b-10b-a)}{9}=a-b=integer\). Sufficient.

(2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

If x = 31 and y = 13, then (x - y)/9 = 2 = integer, but if x = 31 and y = 24, then (x - y)/9 = 7/9 = not an integer. Not sufficient.

Answer: A.

Hope it's clear.
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Statement 1: x and y have the same two digits, but in reverse order.
Let the digits be ab and ba
x-y = 10a+b - 10b -a = 9a - 9b = 9*(a-b)

therefore, (x-y)/9 = a-b = Integer. Sufficient

Statement 2 : The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

Let say the numbers are 42 and 24, then (x-y)/9 = (42 - 24)/9 = 18/9 = 3 (integer). Therefore, (x - y)/9 an integer
Let say the numbers are 42 and 46, (x-y)/9 = (42 - 46)/9 = -4/9 (not an integer). Therefore , (x - y)/9 is not an integer
Hence insufficient

Therefore, A) should be the answer

Similar to Bunuel's method, but more algebraic:

The question is asking if (P-Q) is divisible by 9. We don't care which one is bigger, because negative numbers can still be integers.

1. \(P = 10a+b, Q = 10b +a\)
\((P-Q) = 10a+b - 10b -a\)
\(= a(10-1) - b(10 -1)\)
\(= 9(a-b)\)obviously divisible by 9. Sufficient

2. \(P = 10(a+2) +a\)
\(Q = 10(b-2) + b\)
\((P-Q) = 10(a+2) +a - 10(b-2) - b\)
\(= 10a + 20 + a -10b + 20 - b = 11a -11b +40 = 11(a-b) +40\), which is a linear equation of (a-b) and not always divisible by 9. Insufficient.


Answer: A

Alt Rephrase - Is (x-y) a multiple of 9?

Statement 1:

x = 10a + b
y = 10b - a

x-y = 9a - 9b
x - y = 9 (a-b)
As such, x-y is a multiple of 9.

S1 is sufficient

Statement 2:
I utilized the test case approach, and I looked to prove insufficiency.

Test 1:
x is 53, y is 13
(x-y) in this case (40) is not a multiple of 9

Test 2:
x is 42, y is 24
(x-y) in this case (18) is a multiple of 9

As such, only Statement 1 is sufficient.

Solution: Whether (x - y) is a multiple of 9

Statement 1: x and y have the same two digits, but in reverse order
Say x= ab i.e. x=10a + b
y=ba i.e. y = 10b + a

(x-y) = 9(a-b) -- multiple of 9 -- Sufficient

Statement 2: The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit
No relationship is defined between x and y - Insufficient

Answer A

Bunuel VeritasKarishma chetan2u
For Statement(2): (2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

If we approach analytically, Lets assume units digit of X is x...SO tens digit of X is x+2...X=10(x+2)+x=11x+20...(1)
Lets assume units digit of Y is y...SO tens digit of Y is y-2... Y=10(y-2)+y =11y-20...(2)
From (1) and (2)...X-Y=11(x-y)+40...
so (X-Y)/9= [11(x-y)+40]/9....
40/9 gives a remainder of 4.....So if 11(x-y)/9 gives a remainder of 5 or -4.....only then will the entire expression be divisible by 9...
(x-y) will range from 0 to 9 (+/-)....so 11(x-y) can have values 0,11,22,33,..99.....which will never give a remainder of 5 when divided by 9....
So this entire expression will never be divisible by 9....

Plz explain where did I go wrong..

Thanks

Considering x and y to be single digits as you defined them,
x and (x+2) need to be single digits so the value of x can vary from 0 to 7.
y and (y-2) need to be single digits such that (y-2) is not 0 so the value of y can vary from 3 to 9

Value of (x -y) can go from -9, -8, ... -3, -2, -1, 0, 1, 2, 3, 4

When (x-y) = -2, 11*(x - y) = -22 which gives remainder 5.

In other words, when say x = 1 and y = 3, the two two-digit numbers X and Y are 31 and 13.
(31 - 13)/9 = 18/9 = 2 is an integer.
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I got confused because of the language of the question, instead of asking "is expression an integer" they have written in statement form with a question mark at the end.

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Fixed the typos. Thank you!
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If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an integer?

(1) x and y have the same two digits but in reverse order.
(2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

Let's gather all the info in the question as well as analyze the Q. stem.

x and y are integers between 10 and 99, inclusive: x and y are 2 digit positive integers.

Q. Stem : is (x - y)/9 an integer?

A normal Yes/No type DS question, where you need to find Is (x- y ) exactly divisible by 9 or not?

Statement 1: x and y have the same two digits but in reverse order.

Whenever you deal with a question based on a two-digit number and its reverse, the most effective method would be to represent them in an algebraic expression.

Let's assume that x is a two-digit number 'ab' and y is the reverse of it i.e 'ba'.

We can represent x as 10a + b and y as 10b + a.

x- y = 10 a + b - ( 10b + a ) = 10a + b -10b -a = 9a - 9b = 9(a-b)

We can conclude that x-y is a multiple of 9.

Hence, (x - y)/9 will be an integer.
Statement 1 alone is sufficient.

(2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.

There are 2 ways to tackle this statement, either by plugging values or using algebraic expressions.

#Approach 1: Plugin values

The tens’ digit of x is 2 more than the units digit, Let's assume x= 42
The tens digit of y is 2 less than the units digit, y = 13

x - y = 42 - 13 = 29
Is x - y /9 is an integer? No, 29/9 is not an integer.

This doesn't mean that it's a NO in all cases. If you plugin values in yes/No type DS Questions, you should always try to get a yes as well as a NO as answer, else you will fall for the GMAT traps.

The next question is how do you get the values for x and y that give a YES to the Q.stem. Are there any such values?
These are ambiguities you might face when you plugin values in DS questions and it will cost some of your time as well.

Instead of randomly plugin values, if you could apply some logic when you choose numbers then the task would be a lot easier.

For eg: in St2, we can see that difference between ten's and unit digit of x is 2 . Also the difference between unit and ten's digit of y is 2.

Already in St1, we found that the difference between 2 digit number and its reverse is divisible by 9.

Why don't we use this in St2 for our advantage to find out the values?

Let's say x = 31 . Assume any 2 digit number where ten's digit is greater than unit digit by 2.

For y, you take the reverse of x i.e 13. The ten's digit will be less than the unit digit by 2.

Since y is the reverse of x, x-y should be a multiple of 9 as explained in St 1.
x-y = 31-13 = 18
18 is exactly divisible by 9.

Is x - y /9 is an integer? Yes

Another plugin options : ( 42,24) (53,35) (63,36)..

Since you are getting Yes as well as No as answers, Statement 2 alone is not sufficient.

Option A is the correct answer.

Thanks,
Clifin J Francis,
GMAT Quant Expert

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