Gre question related to finding unknown number
$begingroup$
If $displaystylefrac{x^2}{4}$ is an integer greater than 50, then what is the smallest possible value of $x^2$ ?
IMO the answer should be $204$, the next small integer after $50$ is $51$, therefore, $displaystyle frac{x^2}4=51$ thus $x^2=204$. But the answer given is $256$, the question's answers considers the below constraint which I do not understand since my answer satisfies the constraints specifically given in the question.
The textbook answer states that given $x^2$ is divisible by $4$, so $x$ must be divisible by $2$ and even, thus answer given is $16cdot16=256$. I understand that it should be divisible by $2$ but where does question state $x$ should also be an integer?
integers gre-exam
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
$endgroup$
add a comment |
$begingroup$
If $displaystylefrac{x^2}{4}$ is an integer greater than 50, then what is the smallest possible value of $x^2$ ?
IMO the answer should be $204$, the next small integer after $50$ is $51$, therefore, $displaystyle frac{x^2}4=51$ thus $x^2=204$. But the answer given is $256$, the question's answers considers the below constraint which I do not understand since my answer satisfies the constraints specifically given in the question.
The textbook answer states that given $x^2$ is divisible by $4$, so $x$ must be divisible by $2$ and even, thus answer given is $16cdot16=256$. I understand that it should be divisible by $2$ but where does question state $x$ should also be an integer?
integers gre-exam
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
$endgroup$
$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
2
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19
add a comment |
$begingroup$
If $displaystylefrac{x^2}{4}$ is an integer greater than 50, then what is the smallest possible value of $x^2$ ?
IMO the answer should be $204$, the next small integer after $50$ is $51$, therefore, $displaystyle frac{x^2}4=51$ thus $x^2=204$. But the answer given is $256$, the question's answers considers the below constraint which I do not understand since my answer satisfies the constraints specifically given in the question.
The textbook answer states that given $x^2$ is divisible by $4$, so $x$ must be divisible by $2$ and even, thus answer given is $16cdot16=256$. I understand that it should be divisible by $2$ but where does question state $x$ should also be an integer?
integers gre-exam
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
$endgroup$
If $displaystylefrac{x^2}{4}$ is an integer greater than 50, then what is the smallest possible value of $x^2$ ?
IMO the answer should be $204$, the next small integer after $50$ is $51$, therefore, $displaystyle frac{x^2}4=51$ thus $x^2=204$. But the answer given is $256$, the question's answers considers the below constraint which I do not understand since my answer satisfies the constraints specifically given in the question.
The textbook answer states that given $x^2$ is divisible by $4$, so $x$ must be divisible by $2$ and even, thus answer given is $16cdot16=256$. I understand that it should be divisible by $2$ but where does question state $x$ should also be an integer?
integers gre-exam
integers gre-exam
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
edited Dec 16 '18 at 3:28
Richard Ambler
1,298515
1,298515
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
asked Nov 5 '18 at 19:51
LoveWithMathsLoveWithMaths
417
417
$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
2
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19
add a comment |
$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
2
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19
$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
2
2
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19
add a comment |
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$begingroup$
Does $x$ need to be an integer itself?
$endgroup$
– Josh B.
Nov 5 '18 at 19:57
2
$begingroup$
It sounds to me like you are correct.
$endgroup$
– John Douma
Nov 5 '18 at 20:15
$begingroup$
I think you’re right. It seems that the GRE test expected the reader to assume $x$ is an integer without stating it. Your solution, however, meets all the information given in the question and is, of course, a smaller integer than 256. Too bad that you’d be penalized for being correct!
$endgroup$
– Richard Ambler
Dec 16 '18 at 1:41
$begingroup$
ETS, which administers both the GRE and the SAT, sometimes gets the answer wrong.
$endgroup$
– David K
Dec 16 '18 at 14:19