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# If x and y are positive integers and x = y^(a -2), is x a factor of y?

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Manager
Joined: 03 Jul 2017
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If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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Updated on: 24 Nov 2017, 00:35
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Question Stats:

32% (01:54) correct 68% (02:03) wrong based on 380 sessions

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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

Originally posted by abansal1805 on 24 Nov 2017, 00:31.
Last edited by Bunuel on 24 Nov 2017, 00:35, edited 1 time in total.
Renamed the topic and edited the question.
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Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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24 Nov 2017, 00:45
11
7
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.
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Manager
Joined: 03 Jul 2017
Posts: 68
Location: India
Concentration: Finance, Economics
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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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24 Nov 2017, 01:19
Thank Bunuel for the awesome explanation!
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Joined: 24 Nov 2016
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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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24 Nov 2017, 07:03
1
Bunuel wrote:
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, $$y^{-a+3}$$.
Manager
Joined: 17 Mar 2015
Posts: 86
If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Jan 2018, 12:52
Bunuel wrote:
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
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Jan 2018, 13:01
prince00113 wrote:
Bunuel wrote:
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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If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Jan 2018, 19:35
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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Joined: 02 Sep 2009
Posts: 53066
Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Jan 2018, 23:41
Kezia9 wrote:
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

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Hope it helps.
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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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02 Jan 2018, 07:56
Understood after posting. Thanks!!!!
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Joined: 13 Jan 2014
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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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12 Jan 2018, 08:47
Bunuel wrote:

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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Posts: 53066
Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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12 Jan 2018, 08:51
bka2 wrote:
Bunuel wrote:

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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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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14 Jan 2018, 21:43
1
abansal1805 wrote:
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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If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Mar 2018, 13:33
Bunuel wrote:
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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Joined: 02 Sep 2009
Posts: 53066
Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?  [#permalink]

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01 Mar 2018, 20:09
Mo2men wrote:
Bunuel wrote:
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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Re: If x and y are positive integers and x = y^(a -2), is x a factor of y?   [#permalink] 01 Mar 2018, 20:09
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