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

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.

simple solution:
M=10x+y
N=10y+x
M+N=11x+11y=11(x+y)
In other words the answer is a multiple of 11.
Now the question becomes " which of the following is NOT a multiple of 11?"
answer----->181----->choice A

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.

madn800 is the best solution in my opinion.. :D

But, just to give another solution I've just saw: think of M = AB and N = BA.
(1) If A+B < 10, then:

AB
BA +
-----
B B ---> sum will result in a two digit number
+ +
A A

The sum of A + B will be less than 10, so the summed number digit's will be the same, so D and E satisfy. Now... if
2) A+B > 10, then:

AB
BA
----
1 B ---> sum will result in a three digit number
+ +
B A
+
A

Since B+A in this case is greater than 10, you should add a plus 1 in the sum of the next digit, then, this next digit will be the first digit + 1. In a 3 digit number __ ___ ___, the first two digits will be __ A+B+1 A+B , so B and C satisfy, leaving A as the only choice left.

Hi All,

Since the question asks for the answer that CANNOT be the sum of M and N, and the answers are numbers, we can use a combination of TESTing VALUES and TESTing THE ANSWERS to eliminate the possible values and find the answer to the question.

We're told that M and N are two-digit positive integers and have the SAME DIGITS but in REVERSE ORDER. We're asked which of the 5 answers CANNOT be the SUM of M and N.

Let's start with the 'easiest' answer first:

44. Can we get to 44 in the manner described?
Yes, if the numbers are 13 and 31.....13+31 = 44. Eliminate Answer E

Now let's work through the rest of the list....

Can we get to 99 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 18 and 81.....18+81 = 99. Eliminate Answer D

Can we get to 121 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 38 and 83.....38+83 = 121. Eliminate Answer C

Can we get to 165 in the manner described?
Yes, there are a couple of ways to do it. For example, if the numbers are 78 and 87.....78+87 = 165. Eliminate Answer B

There's only one answer left....

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

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:

GMAT assassins aren't born, they're made,
Rich
_________________

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

Bunuel

Same question as: https://gmatclub.com/forum/if-the-two-d ... 87908.html
Probably retire?
_________________

Cleaned both topics and merged. Thank you.
_________________

First digit :
10x + y
Second digit :
10y+x

Adding both : 11x + 11 y = 11 (x+y)
Only option A is not a multiple of 11, rest all are,

IMO A

We know that both a and b integers must have the same digits: that's the crucial piece of information!

You can get it as a=11 and b=11 or 22: they'll equal 22 or 33, which can be multiplied by 11: this can go all the way for all numbers except 181: answer A, findable in 50 seconds when u really wanna analyze the instructions

Hello,
Experts,
VeritasKarishma, Bunuel, IanStewart, EMPOWERgmatRichC, ArvindCrackVerbal, AaronPond, GMATinsight, JeffTargetTestPrep, ccooley
Is there any chance to take both M and N as 33?
Thanks in advance..
_________________

Hi Asad,

Yes, you could use a pair of two-digit numbers that have the same 2 digits - but that would only help you to eliminate one of the answer choices (Answer A; if the numbers were 22 and 22, then the sum would be 44). With all of the other possibilities, you will end up with an even number (re: 11 + 11 = 22, 33 + 33 = 66, etc.) - and since the remaining 4 answers are all ODD numbers, you'll have to find other combinations of numbers to eliminate the 3 answers that can be eliminated.

GMAT assassins aren't born, they're made,
Rich
_________________

Asad

Yes you can take such values but here in this case such values do NOT eliminate options while the entire objective in "can be" or "can NOT be" or "must be " type questions is to eliminate options by choosing smart numbers which eliminate at least a few options to begin with.
_________________

Hello Asad,

When you add a two-digit number and its reverse, the sum will always be divisible by 11. That is, if two two-digit numbers AB and BA are added, their sum will always be divisible by 11.
This is the concept on which this question needs to be solved. If you apply this concept, you will not have to test numbers at all. You will realise that 181 is the only number that is not divisible by 11.

In the above rule, note that there is no constraint saying A should not be equal to B. If A = B, it actually helps you prove faster that the sum will be divisible by 11, since two digit numbers with the same digits are multiples of 11.

However, taking a number like 33 does not help in THIS question. Can you take 33? Yes, why not! But does it help you? It doesn’t.
So, the better question you need to ask yourself – should I take 33? The answer would be a clear NO.

Asad, in a lot of questions, you will hit this fork in the road where you are unsure of what to do. At that stage, “Should I do this?” is a better question to ask yourself THAN “Can I do this?”. That’s because when you ask yourself the second question, you will mostly answer it with a yes and proceed further; but it may not be the best approach to the solution. Whereas, the answer to the first question will always make you think about your approach and then decide the best one.

Hope that helps!

I have seen this type of question (11A +11B) so many times that I instantly knew it had to be a non-multiple of 11

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