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If the two digit integers M and N are positive and have the

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If the two digit integers M and N are positive and have the [#permalink]

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If the two digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N?

A) 181
B) 165
C) 121
D) 99
E) 44
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Aug 2015, 23:56, edited 3 times in total.
Edited the typo in B: 164 was incorrect it should be 165

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 04 Jan 2006, 18:32
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Let the first digit be xy, then the second digit would be yx.

xy = 10x + y
yx = 10y + x

sum = 11 (x+y)

so the sum has to be multiple of 11. A is not.

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 04 Jan 2006, 19:15
I believe the answer is A and B. Both cannot be expressed as a sum of 2 digit numbers whose digits are the same but interchanged.

Let M= AB(A.. B = 0…9) and N = BA where A and B are the digits 0-9.

- One deducted rule is the last digit of the sum should be the sum of A+B . To illustrate further take for eg 44 the last digit is 4 which can be expressed as a sum of 2+2 , 1+3,0+4. (which means First digit is A and the second B) = 22+22 = 44, Similarly 13+31 = 44, 04+40 = 44. Therefore 44 can be expressed as per above rule. Applying the same rule for 99 = 9 can be expressed as sum of 1+8, 9+0,7+2, 5+4,6+3 all of which can yield 99. Therefore 99 is eliminated. For numbers less than 100 the sum of the last 2 digits cannot exceed 9.
- For numbers greater than 100 the last digit can be expressed as a sum of numbers from 0-9 with one of the numbers > 5. To illustrate this let us analyze 121. Last digit is 1. Since the number is > than 100 we need to express it as a sum of single digit number whose total is 11 therefore it can be 9+2,8+3,7+4,6+5, if we apply any of this we find 92+29 = 121, 83+38 = 121, 74+47+121, 65+56=121. Therefore 121 is eliminated.

164 = 4 can be expressed as 8+6, 9+5,7+7, = 86+68=154,95+59=154,77+77=154. Therefore 164 is one of the choices

Let us look at 181 = 1 can be expressed as 9+2,7+4,8+3,6+5, none of which will yield 181 . Therefore according to me there are 2 answers A or B.

I do not know whether this is the correct way. If someone can elaborate further that will be great.
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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 04 Jan 2006, 23:02
ellisje22 wrote:
If the two digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N?

A) 181
B) 164
C) 121
D) 99
E) 44



The official answer is A...though I don't know how they came up with it.


The original number is 10X+Y
The reversed number is 10Y+X
Sum = 11X+11Y
= 11(X+Y) so in the answer choices we need to find the number that is not a multiple of 11. In this case it happens to be A and B.

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 04 Jan 2006, 23:15
Agree with A and B. My working is the same - resulting in the sum of M and N being a multiple of 11.

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 05 Jan 2006, 00:38
M = 10x + y
N = 10y + x

M + N = 11x + 11y = 11(x+y)

So sum must be a multiple of 11. A and B both can not be sum.

I think there is a typo in B. It may be 154.
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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 05 May 2015, 10:57
10x + y = M
10y + x = N

M+N = 11x + 11y so it must be divisible by 11, only 181 is not divisible by 11 hence it is the answer

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 18 Aug 2015, 05:42
You can get 165 by using 9 and 6.

96+69=165.

It's better to use the equations as mentioned in the other comments. The question is perfectly fine. Answer should be A.

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 18 Aug 2015, 05:48
ellisje22 wrote:
If the two digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N?

A) 181
B) 165
C) 121
D) 99
E) 44



The official answer is A...though I don't know how they came up with it.


let the number be 10x+y
the reverse of the number is 10y+x
now, M+N=11x+11Y=11(x+y)
the sum is divisible by 11
the only option is A which is not divisible by 11

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Re: If the two digit integers M and N are positive and have the [#permalink]

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New post 18 Aug 2015, 21:21
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Hi All,

This type of question can be solved with algebra or "brute force"; here's how you use "brute force" to figure out what's possible and what's not.

We're told that M and N are each two-digit numbers with the two digits in reverse order (e.g. 23 and 32). We're asked which of the 5 answers CANNOT BE the sum of the M and N.

Let's see if we can sum to the given answers (starting with the easiest; keep in mind that there might be more than one way to "hit" each answer):

44? 13 + 31
99? 45 + 54
121? 56 + 65
165? 78 + 87

Since we can sum to each of those 4 answers, there's only one answer left...

Final Answer:
[Reveal] Spoiler:
A


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Re: If the two digit integers M and N are positive and have the [#permalink]

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ellisje22 wrote:
If the two digit integers M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of M and N?

A) 181
B) 165
C) 121
D) 99
E) 44


Remember this as a formula/theorem. This helps me solve all the \(m & n\) \(2-digit\) \(reversed\) \(problems\)

If M is a 2 digit number and n is obtained by reversing the digits of m, then:

1. The sum is a multiple of 11
2. Difference is a multiple of 9


Back to the question. Since the question asks us about the sum, the ans would be the option that will not be divisible by 11

[Reveal] Spoiler:
Ans: A

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Re: If the two digit integers M and N are positive and have the [#permalink]

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Re: If the two digit integers M and N are positive and have the   [#permalink] 19 Dec 2017, 13:50
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