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

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The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 24 Sep 2018, 00:17
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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

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The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 24 Sep 2018, 03:59
1
Bunuel wrote:
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



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

Therefore E
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Re: The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 24 Sep 2018, 09:59
Bunuel wrote:
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

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

\(? = Z\)


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

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

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


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

Regards,
Fabio.
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The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 24 Sep 2018, 12:20
1
Bunuel wrote:
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


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
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Re: The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 05 Oct 2018, 09:16
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
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Re: The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 06 Oct 2018, 08:51
fskilnik wrote:
Bunuel wrote:
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

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

\(? = Z\)


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

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

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


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

Regards,
Fabio.



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)
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Re: The following addition operation shows the sum of the two-digit positi  [#permalink]

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New post 06 Oct 2018, 19:13
1
ias882000 wrote:
fskilnik wrote:
Bunuel wrote:
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

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

\(? = Z\)


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

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

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


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

Regards,
Fabio.



Can you explain how u got the below plzz?

⟨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.
_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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Re: The following addition operation shows the sum of the two-digit positi &nbs [#permalink] 06 Oct 2018, 19:13
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