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OG 182

by nkaur » Fri Apr 22, 2011 7:03 am
Hi Guys,

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

I dont understand why the solution given is right, because they choose to take (10t+u)+(10u+t) and then get 11t+11u=11(t+u) and therefore reason that is has to be that 181 is no multiple of 11.

However if i take (11+u)+(11u+t) I get another solution. Where does it say that one has to say and if i take any other two-digit integer, will I get the same solution? Can somebody explain this please?
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by Brent@GMATPrepNow » Fri Apr 22, 2011 7:16 am
nkaur wrote:Hi Guys,

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

I dont understand why the solution given is right, because they choose to take (10t+u)+(10u+t) and then get 11t+11u=11(t+u) and therefore reason that is has to be that 181 is no multiple of 11.

However if i take (11+u)+(11u+t) I get another solution. Where does it say that one has to say and if i take any other two-digit integer, will I get the same solution? Can somebody explain this please?
This question relies on our ability to determine the value of any 2-digit number.
For example, what is the value of 83? For most of us, it has been a very long time since we examined this (we learned this when we we 5 or 6). Most of us just say that 83 has a value of 83, but 83 is really just a "recipe" for some value.
83 is equal to 8 tens plus 3 ones.
Similarly, 76 is equal to 7 tens plus 6 ones.

In general, if tu represents a 2-digit number (the t stands for the digit in the tens position and the u stands for the digit in the units position), then the value of tu is 10t + u

This is where we got the "10t + u" part in the original solution.

So, if M = tu, then N = ut
The value of M is 10t + u and the value of N us 10u + t
This means the value of M+N is (10t + u) + (10u + t) = 11t + 11u = 11(t + u)

So, we can see that M+N must be a multiple of 11
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by Anurag@Gurome » Fri Apr 22, 2011 7:18 am
nkaur wrote:Hi Guys,

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

I dont understand why the solution given is right, because they choose to take (10t+u)+(10u+t) and then get 11t+11u=11(t+u) and therefore reason that is has to be that 181 is no multiple of 11.

However if i take (11+u)+(11u+t) I get another solution. Where does it say that one has to say and if i take any other two-digit integer, will I get the same solution? Can somebody explain this please?
Here the 2-digit integer is tu, which can be written as M = tu = 10*t + u (Take an example: 39 = 3*10 + 9)
So, the other 2-digit integer with the same digits means we have to interchange the digits, and the reversed 2-digit integer is N = ut = 10*u + t (Continuing the above example, reverse digit of 39 is 93).
So, M + N = (10*t + u) + (10*u + t) = 11t + 11u = 11(t + u), which means that the sum should be a multiple of 11.
Of the given choices, only 181 is not a multiple of 11.

The correct answer is A.
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by manpsingh87 » Fri Apr 22, 2011 7:19 am
nkaur wrote:Hi Guys,

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

I dont understand why the solution given is right, because they choose to take (10t+u)+(10u+t) and then get 11t+11u=11(t+u) and therefore reason that is has to be that 181 is no multiple of 11.

However if i take (11+u)+(11u+t) I get another solution. Where does it say that one has to say and if i take any other two-digit integer, will I get the same solution? Can somebody explain this please?
any two digit number can be represented in the form of 10t+u ; where t and u can be any number from 1 to 9; for example consider 13; this can be written as 10+3; here t is 1 and u is 3, also consider 99; 10*9+9; here t is 9 and u is 9;

i hope that clears your doubt that why its written as 10t+u and not as 11u+t;
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