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06 Feb 2019, 15:40
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GMATH practice exercise (Quant Class 16)
If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
KanishkM
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 769
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Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
06 Feb 2019, 23:46
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fskilnik
GMATH practice exercise (Quant Class 16)
If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
Key word : \(x\) and \(y\) are integers, can be -ive or +ive
x^2 - y^2 = x-y x+y
So now the product of 2 expressions to be odd, both expressions need to be odd, said that
from 1) x + y is odd
lets put back in the question, this means that x-y will be odd as well, Even + odd = Odd, Even - Odd = Odd
from 2) x - y is odd
lets put back in the question, this means that x+y will be odd as well, Even - odd = Odd, odd + even= Odd
D
07 Feb 2019, 06:18
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fskilnik
GMATH practice exercise (Quant Class 16)
If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
\(x,y\,\,{\rm{ints}}\,\,\,\,\,\left( * \right)\)If \(x\) and \(y\) are integers, is \(\,{x^2} - {y^2}\,\) odd?
\(\left( 1 \right)\,\,\,x + y\,\,\) is odd
\(\left( 2 \right)\,\,\,x - y\,\,\) is odd
\({x^2} - {y^2}\,\,\mathop = \limits^? \,\,{\rm{odd}}\)
First Approach: ("the smart way")
\(\left( 1 \right)\,\,x + y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)\)
\(\left( 2 \right)\,\,x - y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)\)
\(\left( {**} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
Second Approach: ("the elegant way")
\(\left( 1 \right)\,\,\,x + y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y = \underbrace {x + y}_{{\rm{odd}}} - \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)\)
\(\left( 2 \right)\,\,\,x - y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y = \underbrace {x - y}_{{\rm{odd}}} + \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)\)
\(\left( {***} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
The correct answer is therefore (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
