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If a and b are both two digit integers, is a + b a multiple of 11?

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If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd
[Reveal] Spoiler: OA

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Re: If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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New post 29 Feb 2016, 03:38
Bunuel wrote:
If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd

HI,

Info from Q--
A and b are 2-digit numbers

lets see the statements:-

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
Means xy = yx.. or
a=10x+y, then b=10y+x..
so a+b= 10x+y+10y+x=11x+11y..
Ans is YES a+b is div by 11
Suff

(2) Both a and b are odd
not much info to work on..
Insuff

A

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Re: If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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New post 29 Feb 2016, 19:46
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd


In the original condition, there are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become a=10x+y=odd and b=10y+x=odd -> a+b=10x+y+10y+x=11(x+y), in which 11 is the factor. Thus, it is sufficient and C is the answer.
However, this is an integer question, which is one of the key questions. Apply the mistake type 4(A). Even just for 1), 11 is always the factor, which is yes and sufficient.
Thus, the answer is A.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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New post 09 Mar 2016, 20:50
Here the statement 1 is sufficient as when we interchange the digits the the form is => 10n +m + 10m+n = 11(m+n) m,n being the unit and tens digits..
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Re: If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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New post 10 Mar 2016, 00:56
Bunuel wrote:
If a and b are both two digit integers, is a + b a multiple of 11?

(1) The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
(2) Both a and b are odd


Required: Is a + b a multiple of 11?

Statement 1: The tens digit of a is equal to the units digit of b, and the tens digit of b is equal to the units digit of a
This means the numbers will be of the form xy and yx.

On adding, we get (10x + y) + (10y + x) = 11x + 11y
This clearly is divisible by 11
SUFFICIENT

Statement 2: Both a and b are odd
The different values of a and b can be:
a = 23 and b = 35
Here a + b = 58. Not a multiple of 11

a = 11 and b = 33
a + b = 44. Multiple of 11
INSUFFICIENT

Option A

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Re: If a and b are both two digit integers, is a + b a multiple of 11? [#permalink]

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New post 06 Dec 2017, 14:58
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Re: If a and b are both two digit integers, is a + b a multiple of 11?   [#permalink] 06 Dec 2017, 14:58
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