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A and B are positive two-digit integers such that the tens digit in A

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A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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25 Jun 2015, 02:32
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A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following CANNOT be the difference of A and B ?

A 72
B 45
C 33
D 27
E 9
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Posts: 59587
Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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25 Jun 2015, 02:35
1
1
RudeyboyZ wrote:
A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following CANNOT be the difference of A and B ?

A 72
B 45
C 33
D 27
E 9

A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B:

A = 10x + y
B = 10y + x

A - B = (10x + y) - (10y + x) = 9(x - y). As you can see the difference must be a multiple of 9. Only option C is not a multiple of 9.

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Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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25 Jun 2015, 02:36
1
Bunuel wrote:
RudeyboyZ wrote:
A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following CANNOT be the difference of A and B ?

A 72
B 45
C 33
D 27
E 9

A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B:

A = 10x + y
B = 10y + x

A - B = (10x + y) - (10y + x) = 9(x - y). As you can see the difference must be a multiple of 9. Only option C is not a multiple of 9.

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Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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25 Jun 2015, 18:43
1
Hi RudeyboyZ,

This question is based on an issue that many Accountants have to deal with (and it's something that you might have learned about in an Accounting Class).

When digits are 'transposed' incorrectly (eg. You mean to type 23, but you type 32 instead), the difference between the original two numbers will ALWAYS be a multiple of 9.

Here, we're asked which of the following is NOT a possible difference....You might notice that the 4 wrong answers are ALL multiples of 9, so they are all possible differences.

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A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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02 Nov 2018, 04:27
Bunuel wrote:
RudeyboyZ wrote:
A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following CANNOT be the difference of A and B ?

A 72
B 45
C 33
D 27
E 9

A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B:

A = 10x + y
B = 10y + x

A - B = (10x + y) - (10y + x) = 9(x - y). As you can see the difference must be a multiple of 9. Only option C is not a multiple of 9.

i didnt quite get the logic of highlighted part. Bunuel can please you give an example with real numbers
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Posts: 59587
Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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02 Nov 2018, 04:31
1
dave13 wrote:
Bunuel wrote:
RudeyboyZ wrote:
A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following CANNOT be the difference of A and B ?

A 72
B 45
C 33
D 27
E 9

A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B:

A = 10x + y
B = 10y + x

A - B = (10x + y) - (10y + x) = 9(x - y). As you can see the difference must be a multiple of 9. Only option C is not a multiple of 9.

i didnt quite get the logic of highlighted part. Bunuel can please you give an example with real numbers

We got that A - B = 9*(x - y) = 9*integer = {a multiple of 9}.

For example, 91 - 19 = 72 = {a multiple of 9}.
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A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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02 Nov 2018, 05:00
1
Hi dave13

We can try examples such as the below

91 - 19 = 72
82 - 28 = 54
73 - 37 = 36
64 - 46 = 18
10 - 01 = 9

So all these are multiples of 9

Or algebraically as explained by Bunuel

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A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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02 Nov 2018, 05:05
Salsanousi wrote:
Hi dave13

We can try examples such as the below

91 - 19 = 72
82 - 28 = 54
73 - 37 = 36
64 - 46 = 18
10 - 01 = 9

So all these are multiples of 9

Or algebraically as explained by Bunuel

Posted from my mobile device

MASHALLAH ! THANK YOU
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Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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02 Nov 2018, 05:10
dave13 you’re welcome

You could also do it 91 - 19, 92 - 29 etc..

The difference will be 72,63,54,45,36,27,18,9

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Re: A and B are positive two-digit integers such that the tens digit in A  [#permalink]

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25 Nov 2019, 04:48
2 integers XY & YX can be written as (10Y + X) & (10X + Y)

=> difference = 9Y - 9X = 9(X-Y)

=> The answer has to be a multiple of 9 and 33 is not.

Smash C
Re: A and B are positive two-digit integers such that the tens digit in A   [#permalink] 25 Nov 2019, 04:48
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