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Bunuel
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If a and b are integers and 250a = b^2, all of the following values 27 Apr 2016, 00:07
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60% (01:58) correct 40% (02:08) wrong based on 249 sessions

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If a and b are integers and 250a = b^2, all of the following values must divide evenly into b^2, except

A. 20
B. 40
C. 250
D. 625
E. 1,250
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B
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chetan2u
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GMAT Focus 1: 735 Q90 V89 DI81
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If a and b are integers and 250a = b^2, all of the following values 27 Apr 2016, 07:17
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Bunuel
If a and b are integers and 250a = b^2, all of the following values must divide evenly into b^2, except

A. 20
B. 40
C. 250
D. 625
E. 1,250
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Hi,

\(250a=b^2.\)..
this means 250a should be perfect square..
and Since it is MUST, we will assign a value to a, just enough to make 250a a PERFECT square..
\(250a = 2*5^3 a.\).. so a MUST be atleast 2*5..
so \(b^2= 250*10 = 2500\)..

lets see the choices..
20, 250, 625 and 1250 all are factors of 2500 so OK

but 40 contains 8, whereas 2500 is evenly divided ONLY by 4.. our ANSWER
B
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If a and b are integers and 250a = b^2, all of the following values 27 Apr 2016, 08:00
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Same approach as chetan2u and same answer

Just wondering can there be any other way of solving this Question? Anyone with alternative solutions welcome..
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If a and b are integers and 250a = b^2, all of the following values Updated on: 10 May 2025, 22:50
Originally posted by anirchat on 10 May 2025, 21:35.
Last edited by anirchat on 10 May 2025, 22:50, edited 1 time in total.
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What if a = 2^3*5 = 40, even then it is a perfect square 10,000. And hence. b^2/a = 250, which is also true.
So how is 40 not a solution.
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If a and b are integers and 250a = b^2, all of the following values 10 May 2025, 22:06
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anirchat
What if a = 2^2*5 = 40, even then it is a perfect square 10,000. And hence. b^2/a = 250, which is also true.
So how is 40 not a solution.
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Could that be because 2^2*5 is 20?
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If a and b are integers and 250a = b^2, all of the following values 10 May 2025, 22:51
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Hey Bunuel. Sorry I have edited. 40 =2^3*5, which still keeps the RHS a perfect square.
bb
anirchat
What if a = 2^2*5 = 40, even then it is a perfect square 10,000. And hence. b^2/a = 250, which is also true.
So how is 40 not a solution.

Could that be because 2^2*5 is 20?
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If a and b are integers and 250a = b^2, all of the following values 11 May 2025, 01:40
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anirchat
If a and b are integers and 250a = b^2, all of the following values must divide evenly into b^2, except

A. 20
B. 40
C. 250
D. 625
E. 1,250

What if a = 2^3*5 = 40, even then it is a perfect square 10,000. And hence. b^2/a = 250, which is also true.
So how is 40 not a solution.
Show more

The question asks which of the following MUST divide b^2, not which COULD divide b^2. That's why we check for the SMALLEST possible value of b^2: if something divides the smallest possible b^2, then it will for sure divide all other valid values of b^2. However, the reverse is not true; if something divides larger possible values of b^2, it does not necessarily divide the smallest one.

The smallest possible value of b^2 is 2,500, for a = 10. In this case, each option is a factor of b^2 except B. Therefore, while ANY number could divide different valid b^2, 40 is the one, from the options, that will not always divide b^2.
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