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

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

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New post 12 Dec 2014, 07:17
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 12 Dec 2014, 10:03
1
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
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 12 Dec 2014, 17:21
1
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.
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 19 Dec 2014, 12:51
1
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
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 19 Dec 2014, 12:54
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 08 Dec 2016, 02:44
kinghyts wrote:
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



how did you get 10a+b?
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 03 Jan 2017, 06:52
nishaadnak wrote:
kinghyts wrote:
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



how did you get 10a+b?


Hi nishaadnak,

Any two digit number can be written in the form of 10a+b .

for example 45 = 10*4 + 5, a =4 and b =5.

Similarly three digit number can be written as 100a+ 10b + c.

451 = 100*4 + 10*5 + 1; a=4, b =5 , and c=1

Thanks.
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 03 Jan 2017, 17:42
Hello Team,

I am a bit confused.... but why we are not considering a=b as a possibility in 9(a-b) assumption ( As it have never been mentioned x=!y) in stmt 01.

If we consider that x could be equal to y, then stmt 1 together with stmt 2 will take us to the conclusion.

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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 03 Jan 2017, 19:01
ushapdatta wrote:
Hello Team,

I am a bit confused.... but why we are not considering a=b as a possibility in 9(a-b) assumption ( As it have never been mentioned x=!y) in stmt 01.

If we consider that x could be equal to y, then stmt 1 together with stmt 2 will take us to the conclusion.

Regards


Hi ushapdatta,

You are right, it's not mentioned that x != y, and we don't have to assume this either.

If x = y, then also statement 1 alone is sufficient, you don't need statement 2 to get the unique answer.

x = y => x-y = 0 => (x-y)/9 = 0/9 = 0 ( please note that 0 is an integer). Hence statement 1 alone is sufficinet.

If you consider x = y, that will be a special case of the above-discussed solution.

Thanks.
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 07 Mar 2017, 10:37
kinghyts wrote:
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


I understand how it is 10a+b, but how is it 10b - a?
Shouldn't that too be 10b + a?

Thank you!
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 21 Mar 2017, 04:29
menontion wrote:
kinghyts wrote:
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


I understand how it is 10a+b, but how is it 10b - a?
Shouldn't that too be 10b + a?

Thank you!

Hi menontion,

Please note that y is \(10b + a\) (as you have mentioned).

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

Hope this helps.
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 26 Feb 2019, 23:06
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
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 27 Feb 2019, 01:24
Debashis Roy wrote:
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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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 27 Feb 2019, 02:58
VeritasKarishma

Thanks!....Gross mistake on my part to wrongly assess the range of x and y...
Thanks again...you have been very helpful..
Cheers!
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an  [#permalink]

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New post 01 Mar 2019, 10:19
Bunuel wrote:

Tough and Tricky questions: Number Properties.



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.



Kudos for a correct solution.


#1
x & y have same 2 digit ; reverse order
x=ab and y = ba
10a+b=10b+a
9a=9b
9(a-b)=0
yes sufficeint
#2
x=42 , y = 40
42-40/9 ; no
x=42,y=24
yes
in sufficient
IMO A
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Re: If x and y are integers between 10 and 99, inclusive, is (x - y)/9 an   [#permalink] 01 Mar 2019, 10:19
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