The following addition operation shows the sum of the two-digit positi

\(X,Y\,\,\, \in \,\,\,\,\left\{ {1,2, \ldots ,9} \right\}\)
\(Z \in \,\,\,\,\left\{ {0,1,2, \ldots ,9} \right\}\)
\(X,Y,Z\,\,\,{\text{distinct}}\)

\(? = Z\)


\(\left\langle {XXZ} \right\rangle = \left\langle {XY} \right\rangle + \left\langle {YX} \right\rangle < 99 + 99 < 200\,\,\,\,\,\,\, \Rightarrow \,\,\,\,X = 1\)

\(\left\langle {1Y} \right\rangle + \left\langle {Y1} \right\rangle = \left\langle {11Z} \right\rangle \,\,\,\,\, \Rightarrow \,\,\,Y = 9\,\,\,\,\,\,\left( {18 + 81 < 100} \right)\)

\(\left\langle {11Z} \right\rangle = 19 + 91 = 110\,\,\,\,\, \Rightarrow \,\,\,Z = 0\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Considering the stem, the sum of two-digit integer must result in a three-digit integer with the first 2 digits of this integer being similar.
The only case possible is for X to equal one (for example no two digit integer can sum above 198), hence we now have 1Y+Y1 = 11Z
Which leave us with the only option possible for Y to yield a three digit number : 9 (for example 8 for Y would yield : 18+81= 99)
Finally : 19+91=110

Can you explain how u got the below plzz?

⟨XXZ⟩=⟨XY⟩+⟨YX⟩<99+99<200⇒X=1⟨XXZ⟩=⟨XY⟩+⟨YX⟩<99+99<200⇒X=1???

⟨1Y⟩+⟨Y1⟩=⟨11Z⟩⇒Y=9???(18+81<100)

Hi, ias882000 !

Thank you for your interest in my solution.

1st doubt: <XXZ> is a positive 3-digit number less than 200, therefore X must be 1.

2nd doubt: If Y is 8 or less, ⟨1Y⟩+⟨Y1⟩ would be 18+81 or less, hence it would not be a three-digit number.
It must be, to be equal to ⟨11Z⟩.

Regards,
Fabio.

Dear Moderator,
This same problem has been duplicated in the link below, you may wish to merge the same, Thank you.

https://gmatclub.com/forum/in-the-corre ... ml#p384737

Now with hit and trial, we can see that the given nos. are 19 and 91.
19 + 91 = 110 [XXZ]; Z=0
Hence E

Kudos if helpful

only possible pair of XY ; 91 and YX ; 19
we get 110 ; which is XXZ
hence Z= 0
option E

How can we just assume that it has to be the multiple of 11?

because
XY
+YX
can be written as X+Y+10*X+10*Y = 11*X + 11*Y
I hope it is clear.

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