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Bunuel
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

Is \(\frac{y}{x}=integer\)?

Is \(\frac{y}{y^{a-2}}=integer\)?

Is \(y^{3-a}=integer\)?

For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer. So, the question basically asks whether a is an integer and \(3-a \geq 0\)? Or which is the same: is a an integer and \(a \leq 3\)?

(1) a is an integer. Not sufficient.

(2) a < 4. Not sufficient.

(1)+(2) Sufficient.

Answer: C.

If anyone wants to know the step by step on how to get \(y^{3-a}\):

using the exponent rule: \(\frac{x^{a}}{x^{b}}\);
so, \(\frac{y}{y^{a-2}}\) is the same as \(\frac{y^1}{y^{a-2}}\) which becomes, \(y^{1-(a-2)}\) --> \(y^{1-a+2}\) finally, \(y^{-a+3}\).
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Bunuel
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

Is \(\frac{y}{x}=integer\)?

Is \(\frac{y}{y^{a-2}}=integer\)?

Is \(y^{3-a}=integer\)?

For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer. So, the question basically asks whether a is an integer and \(3-a \geq 0\)? Or which is the same: is a an integer and \(a \leq 3\)?

(1) a is an integer. Not sufficient.

(2) a < 4. Not sufficient.

(1)+(2) Sufficient.

Answer: C.


I just need to ask one doubt.
In the question it says, x and y are positive integers and x = y^(a-2)
Now if a is not an integer or a < 2, x will not be an integer.
So it is already given that a is an integer and it is greater than 2.
Thus A is not at all required.
Please correct me if there is any mistake
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Bunuel
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

Is \(\frac{y}{x}=integer\)?

Is \(\frac{y}{y^{a-2}}=integer\)?

Is \(y^{3-a}=integer\)?

For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer. So, the question basically asks whether a is an integer and \(3-a \geq 0\)? Or which is the same: is a an integer and \(a \leq 3\)?

(1) a is an integer. Not sufficient.

(2) a < 4. Not sufficient.

(1)+(2) Sufficient.

Answer: C.


I just need to ask one doubt.
In the question it says, x and y are positive integers and x = y^(a-2)
Now if a is not an integer or a < 2, x will not be an integer.
So it is already given that a is an integer and it is greater than 2.
Thus A is not at all required.
Please correct me if there is any mistake

First of all, if x = y = 1, then a could be any number, not necessarily an integer less than 2 for \(x=y^{a-2}\) to hold true.

Next, if a > 3, say if a = 10 and y is not 1, then \(y^{3-a}=integer\) won't be true.
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Hi Bunuel,
Need a clarification. We assume a is less than or equal to 3. How can we??? When we don't know y/x is an integer???
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Kezia9
Hi Bunuel,
Need a clarification. We assume a is less than or equal to 3. How can we??? When we don't know y/x is an integer???

The question became: Is \(y^{3-a}=integer\)? Now, for any y ≠ 1, if a is not an integer, for example if a = 1/2, then \(y^{\frac{5}{2}} \neq integer\). Also, if \(a > 3\), for example if a = 4, then \(y^{-1}=\frac{1}{y} \neq integer\).

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Understood after posting. Thanks!!!!
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Bunuel


For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer.

Answer: C.

Could you please tell me the reason? For example, if y=9, x=3 then a=5/2, because x=y^(a-2); so (3-a) = 1/2 is not non-negative integer. So, (3-a) doesn't need to be non-negative integer, it can be a fraction.
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Bunuel


For \(y^{3-a}=integer\) to be true for ANY value of y, the exponent must be a non-negative integer.

Answer: C.

Could you please tell me the reason? For example, if y=9, x=3 then a=5/2, because x=y^(a-2); so (3-a) = 1/2 is not non-negative integer. So, (3-a) doesn't need to be non-negative integer, it can be a fraction.

Pay attention to the highlighted part. \(y^{3-a}=integer\) will hold true for ANY integer y, if the exponent is a non-negative integer.
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abansal1805
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

(1) a is an integer
(2) a < 4

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (\(x\), \(y\) and \(a\)) and 1 equations, C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2)
\(a = 3\): \(x = y\) \(⇒ x\) is a factor or \(y\).
\(a = 2\): \(x = y^0 = 1 ⇒ x\) is a factor or \(y\) whatever integer \(y\) is.
\(a ≤ 1\): \(x = 1 ⇒ x\) is a factor or y whatever integer \(y\) is.
Both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
\(x = 2, y = 2, a = 3\): Yes
\(x = 4, y = 2, a = 4\): No
The condition 1) only is not sufficient.

Condition 2)
\(x = 2, y = 2, a = 3\): Yes
\(x = 8, y = 4, a = 7/2\): No
The condition 2) only is not sufficient.

Therefore, C is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Bunuel
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

Is \(\frac{y}{x}=integer\)?

Is \(\frac{y}{y^{a-2}}=integer\)?

Is \(y^{3-a}=integer\)?

For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer. So, the question basically asks whether a is an integer and \(3-a \geq 0\)? Or which is the same: is a an integer and \(a \leq 3\)?

(1) a is an integer. Not sufficient.

(2) a < 4. Not sufficient.

(1)+(2) Sufficient.

Answer: C.

Dear Bunuel,

If a = 1.........x = y^-1 = 1/y.........if y = 2....then x = 1/2 ...... 1/2 is Not factor of 2.........So a can't equal 1 and hence does not hold true that \(a \leq 3\)

what do you think?

Thanks
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Bunuel
If x and y are positive integers and \(x=y^{a-2}\), is x a factor of y ?

Is \(\frac{y}{x}=integer\)?

Is \(\frac{y}{y^{a-2}}=integer\)?

Is \(y^{3-a}=integer\)?

For \(y^{3-a}=integer\) to be true for any value of y, the exponent must be a non-negative integer. So, the question basically asks whether a is an integer and \(3-a \geq 0\)? Or which is the same: is a an integer and \(a \leq 3\)?

(1) a is an integer. Not sufficient.

(2) a < 4. Not sufficient.

(1)+(2) Sufficient.

Answer: C.

Dear Bunuel,

If a = 1.........x = y^-1 = 1/y.........if y = 2....then x = 1/2 ...... 1/2 is Not factor of 2.........So a can't equal 1 and hence does not hold true that \(a \leq 3\)

what do you think?

Thanks

We are given that x and y are positive integers, your example does not satisfy this.
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