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Re: If x and y are integers, then [#permalink]
amorphous wrote:
take \(y = 10\) and for consecutive integers take some small numbers such as \(1, 2, 3 or 2, 3, 4 or 3, 4, 5\)
doing so will invalidate all of the options from A to D because non of the final result will be an integer
Leaving only E for ans



Can you please elaborate your explanation.
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Re: If x and y are integers, then [#permalink]
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When we assume \(x, x+1\) and \(x+2\) as \(1,2,3\) or \(2,3,4\) or \(3,4,5\)
while replacing these values in the given expression we can cancel of digits.
In the first two cases \(2\) and \(3\) are cancelled off and we are left with \(\frac{1}{5^y}\) and \(\frac{4}{5^y}\)
Take \(y\) as a large number because we are working with a \(must be\) type questions so the ans must be valid in all circumstances.
In both the case the ans will not be an integer unless y = 0.
For the final case \(3\) gets cancelled of and we are left with \(\frac{10}{5^y}\). For this we can still get an integer if \(y\)= odd i.e \(1\) but it is not the case always as we have found out earlier.
therefore best way to answer this question is to chose a large value for \(y\) and make sure we can eliminate as many options as possible.
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Re: If x and y are integers, then [#permalink]
amorphous wrote:
When we assume \(x, x+1\) and \(x+2\) as \(1,2,3\) or \(2,3,4\) or \(3,4,5\)
while replacing these values in the given expression we can cancel of digits.
In the first two cases \(2\) and \(3\) are cancelled off and we are left with \(\frac{1}{5^y}\) and \(\frac{4}{5^y}\)
Take \(y\) as a large number because we are working with a \(must be\) type questions so the ans must be valid in all circumstances.
In both the case the ans will not be an integer unless y = 0.
For the final case \(3\) gets cancelled of and we are left with \(\frac{10}{5^y}\). For this we can still get an integer if \(y\)= odd i.e \(1\) but it is not the case always as we have found out earlier.
therefore best way to answer this question is to chose a large value for \(y\) and make sure we can eliminate as many options as possible.



So for this types of questions, there is no general rule? It's trial and error approach?
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Re: If x and y are integers, then [#permalink]
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Sometimes the only way to tackle a question is trial and error, even though is time consuming.

However, amorphous above provided a pretty fast and elegant solution to you

Quote:
take \(y = 10\) and for consecutive integers take some small numbers such as \(1, 2, 3 or 2, 3, 4 or 3, 4, 5\)
doing so will invalidate all of the options from A to D because non of the final result will be an integer
Leaving the only E for ans


To speed up, even more, the process you could think about two things:

- first off: when a question asks you "which of the following" start always from the bottom of the answer choices. More often than not the correct answer is from the bottom-up:
- secondly: the numerator of the fraction is 3 consecutive integers.

if x is even = 2 AND y= 0, which means that \(5^0 = 1\) then you do have

\(\frac{2*3*4}{2*3*1}\)

2 and 3 cancel out and you do have \(\frac{4}{1} = 4\).

Now, the question has only one correct answer and it is also a must be true question. As such, E must be the correct answer without bothering you to check the others.

Hope now is clear to you.

Regards
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Re: If x and y are integers, then [#permalink]
2

Easy rule: N consecutive integers are always divisible by N!


Since the numerator is 3 consecutive integers, it's always divisible by 3!=6.
If a number is divisible by 6, it is also divisible by 2 and 3.

There are 2 and 3 in the denominator already, so the only choice that preserves this is E.
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Re: If x and y are integers, then [#permalink]
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