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

Question

The following addition operation shows the sum of the two-digit positive integers XY and YX. If the three-digit integer XXZ, and X, Y , and Z are different digits, what is the value of the integer Z?

XY
+YX
_____
XXZ

(A) 8
(B) 7
(C) 2
(D) 1
(E) 0

gracie · Accepted Answer

XXZ has to be a multiple of 11
with a 100s digit of 1
and a matching tens digit
the only possibility is 110
0
E

CounterSniper · Answer

Given the options , we want the units digit of XXZ 
consider 10 
10 can be written as 19, 91 
19+91= 110

Therefore E

fskilnik · Answer

X,Y\,\,\, \in \,\,\,\,\left\{ {1,2, \ldots ,9} ight\} 
 Z \in \,\,\,\,\left\{ {0,1,2, \ldots ,9} ight\} 
 X,Y,Z\,\,\,{	ext{distinct}}

? = Z

\left\langle {XXZ} ightangle = \left\langle {XY} ightangle + \left\langle {YX} ightangle < 99 + 99 < 200\,\,\,\,\,\,\, \Rightarrow \,\,\,\,X = 1

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

\left\langle {11Z} ightangle = 19 + 91 = 110\,\,\,\,\, \Rightarrow \,\,\,Z = 0\,\,\,

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

Regards, 
Fabio.

loneo3006 · Answer

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

ias882000 · Answer

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)

fskilnik · Answer

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.

stne · Answer

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

HSogani · Answer

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

Kudos if helpful

Archit3110 · Answer

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

aash932 · Answer

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

Zied123 · Answer

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

bumpbot · Answer

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