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# If x and y are positive integers and 5^x

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If x and y are positive integers and 5^x [#permalink]

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06 Apr 2012, 08:42
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If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12
[Reveal] Spoiler: OA
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If x and y are positive integers and 5^x [#permalink]

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06 Apr 2012, 13:01
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BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$;

$$5^x-2^{y-1}*5^{x-1}=5^y$$;

$$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$;

$$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$:

$$5^x(10-2^3)=2*5^{3+1}$$;

$$2*5^x=2*5^4$$;

$$x=4$$ --> $$xy=12$$.

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Re: If x and y are positive integers and 5^x [#permalink]

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24 Sep 2013, 14:38
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$ --> $$5^x-2^{y-1}*5^{x-1}=5^y$$ --> $$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$ --> $$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From $$5^x-5^y=2^{y-1}*5^{x-1}$$ TO $$5^x-2^{y-1}*5^{x-1}=5^y$$

Thanks!
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Re: If x and y are positive integers and 5^x [#permalink]

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24 Sep 2013, 22:04
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danzig wrote:
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$ --> $$5^x-2^{y-1}*5^{x-1}=5^y$$ --> $$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$ --> $$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From $$5^x-5^y=2^{y-1}*5^{x-1}$$ TO $$5^x-2^{y-1}*5^{x-1}=5^y$$

Thanks!

Not directed at me, However you can re-arrange it another way.

We have $$5^x-5^y = 2^{y-1}*5^{x-1}$$, Dividing on both sides by $$5^{x-1}$$, we have
$$5-5^{y+1-x} = 2^{y-1} \to 5 = 5^{y+1-x} + 2^{y-1}$$. Now, as x and y are positive integers, the only value which $$2^{y-1}$$ can take is 4.Thus, y-1 = 2, y = 3. Again, the value of $$5^{y+1-x}$$ has to be 1, thus, y+1-x = 0 $$\to$$ x = y+1 = 3+1 = 4. Thus, x*y = 4*3 = 12.

Thus, I think you could re-arrange in any order, as long as you get a tangible logic.
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

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19 Feb 2014, 21:25
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MrWallSt wrote:
If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^(x-1), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

A little bit of observation can help you solve this question within a minute.

x and y are positive integers which means we will have clean numbers. On the right hand side, you have a 2 as a factor while it is not there on the left hand side. Can a 2 be generated on the left hand side by the subtraction? Here I am thinking that if we take 5^y common on the left hand side, I might be able to get a 2.

$$5^y (5^{x-y} - 1) = 2^{y-1}*5^{x-1}$$
Now I want only 2s and 5s on the left hand side. If x-y is 1, then $$(5^{x-y} - 1)$$ becomes 4 which is 2^2. If instead x - y is 2 or more, I will get factors such as 3, 13 too. So let me try putting x - y = 1 to get
$$5^y (2^2) = 2^{y-1}*5^{x-1}$$

This gives me y - 1 = 2
y = 3
x = 4
Check to see that the equations is satisfied with these values. Hence xy = 12

Note that it is obvious that y is less than x and that is the reason we took $$5^y$$ common. The reason it is obvious is that the right hand side is positive. So the left hand side must be positive too. This means $$5^x > 5^y$$ which means x > y.
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

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19 Feb 2014, 21:50
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Faster: First constrain the possible answers. We know $$x>y$$ since if $$x=y$$ then the left-hand side is 0 or if $$x<y$$ then the LHS is negative... but the RHS is always positive. Now act: divide both sides by $$5^{x-1}$$ to get $$5-5^{y-x+1} = 2^{y-1}$$.

Since the new RHS is a power of two, the LHS must equal 1, 2, or 4. The only power of 5 that gets us one of those is $$5-5^0=5-1=4$$. That means $$y=3$$ and thus $$x=4$$.

Oh, and BTW: The thread title is different than the equality in your post. (Title should have 5^{x-1}, not 5^x.)
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

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19 Feb 2014, 22:01
@Karishma, thanks again.

@SizeTrader, appreciate the solution and that reasoning was excellent. Also, the reason the title says 5^x is because I reached the character limit for the title and it got cut off
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Re: If x and y are positive integers and 5^x [#permalink]

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20 Feb 2014, 03:53
BN1989 wrote:
If x and y are positive integers and $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$, what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

by dividing both sides of the equation with 5^x we have 1- 5^(y-x) = 2^(y-1) * 5^(x-1-x)

1- 5^(y-x) = 2^(y-1) * 5^(-1)

5= 2^(y-1) + 5^(y-x+1)

now minimum value of 2^(y-1) =1, hence 2^(y-1) must be equal to 4 and 5^(y-x+1) must be equal to 1 for the R.H.S to become equal to L.H.S.

2^(y-1) = 4 for y=3
and 5^(y-x+1) =1 for x=4 (as y=3)

hence product of xy = 12
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Re: If x and y are positive integers and 5^x [#permalink]

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20 Feb 2014, 14:02
If x and y are positive integers and (5^x)-(5^y)=(2^{y-1})*(5^{x-1}), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

E

Protocol:Simplify expression

1. need to find X and Y but cant isolate X and Y directly so start by separating the bases:

divide both sides by (5^{x-1}

left and right side simplifies to 5-5^{y-x+1} = 2^{y-1}

2. after simplifying, analyze.

Right side must be positive integer ( y is at least 1 ) thus left side must be positive too.
==> Left side is 5 minus an expression so answer must be at max 4 and at minimum 1. Right side must also be equal to a power of 2. Thus 4 is only possible answer for left side. Thus Y = 3.

Note: From original expression it is clear that X> Y. However by the time you get to simplifying the expression, the possible number of answers is so constrained that this information isn't critical to arriving at a faster answer.
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

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24 Feb 2014, 22:57
Faster: First constrain the possible answers. We know $$x>y$$ since if $$x=y$$ then the left-hand side is 0 or if $$x<y$$ then the LHS is negative... but the RHS is always positive. Now act: divide both sides by $$5^{x-1}$$ to get $$5-5^{y-x+1} = 2^{y-1}$$.

Since the new RHS is a power of two, the LHS must equal 1, 2, or 4. The only power of 5 that gets us one of those is $$5-5^0=5-1=4$$. That means $$y=3$$ and thus $$x=4$$.

Oh, and BTW: The thread title is different than the equality in your post. (Title should have 5^{x-1}, not 5^x.)

Nicely explained.
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If x and y are positive integers and 5^x [#permalink]

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12 Feb 2015, 10:00
Hey Bunuel
Can you please guide me where I went wrong?

My take:

5^x( 1- (2^y)/10)=5^y

Cant we assume that x=y as we have same base 5.
Moreover,if i assume that x=y then in equation 5^x( 1- (2^y)/10)=5^y,

1-(2^y)/10 as to be positive then

1-(2^y)/10>0
then 1>(2^y)/10

so max value 10>2^y

so max value of y ha to be 3 and min value any negative value so any negative value<y<3

Now as x=y value of xy can be y^2,
now within the range any negative value<y<= 3,
we can assume y to be -6 ,
so xy can be 36.

I know I have done something terrible please guide me where I went wrong
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If x and y are positive integers and 5^x [#permalink]

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19 Aug 2015, 06:30
Bunuel wrote:
BN1989 wrote:
If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48
B. 36
C. 24
D. 18
E. 12

Notice that we are told that $$x$$ and $$y$$ are positive integers.

$$5^x-5^y=2^{y-1}*5^{x-1}$$;

$$5^x-2^{y-1}*5^{x-1}=5^y$$;

$$5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y$$;

$$5^x(10-2^y)=2*5^{y+1}$$.

Now, since the right hand side is always positive then the left hand side must also be positive, hence $$10-2^y$$ must be positive, which means that $$y$$ can take only 3 values: 1, 2 and 3.

By trial and error we can find that only $$y=3$$ gives integer value for $$x$$:

$$5^x(10-2^3)=2*5^{3+1}$$;

$$2*5^x=2*5^4$$;

$$x=4$$ --> $$xy=12$$.

i did another logic is it right?

$$(5^x)(1-(2^y-1)/5)=5^y$$

then $$(5^x-y-1)(5-(2^y-1))=1$$

which means this must be $$x-y-1= 0$$
and $$5-(2^y-1) = 1$$ means also $$2^y-1 = 4$$ then $$y-1=2$$
then y=3 replace y in old equation we get x =4 then finally xy =12
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Re: If x and y are positive integers and 5^x [#permalink]

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19 Aug 2015, 07:04
(5^x)-(5^y)=(2^(y-1))*(5^(x-1))
(5^x)(1-5^y-x) = (2^(y-1))*(5^(x-1))
5(1-5^y-x) = (2^(y-1))
10(1-5^y-x) = (2^(y))

We note that (2^(y)) is always positive. which translates to (1-5^y-x) >0 or y-x<0

So, 10(5^y-x)(5^(x-y)-1) = (2^(y))
5^(y-x+1)*2*(5^(x-y)-1) =(2^(y))

Now, RHS is 5^0, which means y-x+1 = 0, x-y =1

inputting values in above,

5^(0)*2*(5^(1)-1) =(2^(y))
2*4 = (2^(y))
implies y =3
x= 4

xy = 12
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Re: If x and y are positive integers and 5^x [#permalink]

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Re: If x and y are positive integers and 5^x [#permalink]

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21 Sep 2016, 01:31
Hi,
The explanations are great. However, I took a lot of time figuring out how to simplify the equation and get to some solution and then started exploring otherway around which worked faster for me..

Difference between any 2 powers of 5 would always yields an even number.,i.e., a multiple of 2. However, difference of only consecutive powers of 5 yields a number that is only multiple of 2 and 5. Check->(25-5), (125-25), (625-125). Also check(625-25), (625-5), etc. Then, I simply had to check the options which had consecutive integers as factors. Only 12 worked out with 3 and 4 as factors.

This is not a foolproof solution but just another way of thinking incase you feel trapped in a question.
Re: If x and y are positive integers and 5^x   [#permalink] 21 Sep 2016, 01:31
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