If x and y are positive integers and 5^x

Question

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

KarishmaB · Accepted Answer

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

Answer (E)

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.

Bunuel · Answer

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 .

Answer: E.

danzig · Answer

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!

mau5 · Answer

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} 	o 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  	o  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.

SizeTrader · Answer

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.)

MrWallSt · Answer

@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

manpreetsingh86 · Answer

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

clearmountain · Answer

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.

jimwild · Answer

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

HSidhu89 · Answer

(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

Sneha Jaiswal · Answer

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.

BrentGMATPrepNow · Answer

Given:  (5^x)-(5^y)=(2^{y-1})*(5^{x-1})

Divide both sides by  5^{x-1}  to get:  5^1 - 5^{y-x+1} = 2^{y-1}

Simplify:  5 - 5^{y-x+1} = 2^{y-1}

OBSERVE: Notice that the right side,  2^{y-1} , is POSITIVE for all values of y

Since y is a positive integer,  2^{y-1}  can equal 1, 2, 4, 8, 16 etc (powers of 2)

So, the left side,  5 - 5^{y-x+1} , must be equal 1, 2, 4, 8, 16 etc (powers of 2).

Since  5^{y-x+1}  is always positive, we can see that  5 - 5^{y-x+1}  cannot be greater than 5

So, the only possible values of  5 - 5^{y-x+1}  are 1, 2 or 4

In other words, it must be the case that: 
case a)  5 - 5^{y-x+1} = 2^{y-1} = 1 
case b)  5 - 5^{y-x+1} = 2^{y-1} = 2 
case c)  5 - 5^{y-x+1} = 2^{y-1} = 4

Let's test all 3 options.

case a)  5 - 5^{y-x+1} = 2^{y-1} = 1 
This means y = 1 (so that the right side evaluates to 1)
The left side,  5 - 5^{y-x+1} = 1 , when  5^{y-x+1} = 4 . Since x and y are positive integers, it's IMPOSSIBLE for  5^{y-x+1}  to equal 4
So, we can eliminate case a

case b)  5 - 5^{y-x+1} = 2^{y-1} = 2 
This means y = 2 (so that the right side evaluates to 2)
The left side,  5 - 5^{y-x+1} = 2 , when  5^{y-x+1} = 3 . Since x and y are positive integers, it's IMPOSSIBLE for  5^{y-x+1}  to equal 3
So, we can eliminate case b

case c)  5 - 5^{y-x+1} = 2^{y-1} = 4  
This means  y = 3  (so that the right side evaluates to 4)
The left side,  5 - 5^{y-x+1} = 4 , when  5^{y-x+1} = 1 . 
If  5^{y-x+1} = 1 , then  y-x+1 = 0 
In this case, y =  3 
So, we can write:  3  - x + 1 = 0, which mean  x = 4 
So, the only possible solution is  y = 3  and  x = 4 , which means xy = ( 4 )( 3 ) = 12

Cheers,
Brent

shashankism · Answer

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

5^{x-1} =/=0. So,  (5-5^{y-x+1})=(2^{y-1}) 
Since  (2^{y-1})  must be +ve. Also Y is +ve so, y-1>0 and hence  (2^{y-1})  will be an integer only. 
Hence y-x+1 can be 0 only.
y-x+1 = 0 
-> y-x = -1

also at y-x+1 = 0 
 5-1 = (2^{y-1})

4 = (2^{y-1}) 
y -1 = 2
y = 3

x= y+1 = 4
xy = 12

Answer E

EMPOWERgmatRichC · Answer

Hi All,

GMAT Quant questions can almost always be solved in a variety of ways, so if you find yourself not able to solve by doing complex-looking math, then you should look for other ways to get to the answer. Think about what's in the question; think about how the rules of math "work." This question is LOADED with Number Property clues - when combined with a bit of "brute force", you can answer this question by doing a lot of little steps.

Here are the Number Properties (and the deductions you can make as you work through them):

1) We're told that X and Y are POSITIVE INTEGERS, which is a great "restriction." 
2) We can calculate powers of 5 and powers of 2 rather easily:

5^0 = 1 
5^1 = 5 
5^2 = 25 
5^3 = 125 
5^4 = 625 
Etc.

2^0 = 1 
2^1 = 2 
2^2 = 4 
Etc.

3) The answer choices are ALL multiples of 3. Since we're asked for the value of XY, either X or Y (or both) MUST be a multiple of 3.

4) 
*The left side of the equation is a positive number MINUS a positive number. 
*The right side is the PRODUCT of two positive numbers, which is POSITIVE. 
*This means that 5^X > 5^Y, so X > Y.

5) 
*Notice how we have 5^X (on the left side) and 5^(X-1) on the right side; these are consecutive powers of 5, so the first is 5 TIMES bigger than the second. 
*On the left, we're subtracting a number from 5^X. 
*On the right, we're multiplying 2^(Y-1) times 5^(X-1). 
*2^(Y-1) has to equal 1, 2 or 4, since if it were any bigger, then multiplying by that value would make the right side of the equation TOO BIG (you'd have a product that was bigger than 5^X). 
*By extension, Y MUST equal 1, 2 or 3. It CANNOT be anything else.

6) Remember that at least one of the variables had to be a multiple of 3. What if Y = 3? Let's see what happens….

5^X - 5^3 = 2^2(5^(X-1))

5^X - 125 = 4(5^(X-1))

Remember that X > Y, so what if X = 4?…..

5^4 - 125 = 500 
4(5^3) = 500

The values MATCH. This means Y = 3 and X = 4. XY = 12

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

rahul16singh28 · Answer

Good Question.. I solved it this way..

Let  x-1 = a ,  y-1 = b .. So the equation becomes,

5^a - 5^b  =  \frac{5^a*2^b}{5}..  
Now, we have to find the value of  xy  i.e.  (a+1)(b+1) 
By looking at the options and the above equation only E satisfies.

ScottTargetTestPrep · Answer

We can simplify the given equation by dividing both sides by 5^(x - 1):

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

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

It’s not easy to solve an equation with two variables by algebraic means; however, since both variables are positive integers, we can try numbers for one of the variables and solve for the other.

If we let y = 1, then the right hand side (RHS) = 2^0 = 1 and thus the left hand side (LHS) is 5^(1 - x + 1) = 4. However, a power of 5 can’t be equal to 4 when the exponent is an integer.

Now, let’s let y = 2; then the RHS = 2^1 = 2 and thus, for the LHS, 5^(2 - x + 1) = 3. However, a power of 5 can’t be equal to 3 when the exponent is an integer.

Finally, let’s let y = 3; then the RHS = 2^2 = 4 and thus, for the LHS, 5^(3 - x + 1) = 1. We see that a power of 5 can be equal to 1 when the exponent is 0. Thus:

3 - x + 1 = 0

x = 4

We see that x = 4 and y = 3 and thus xy = 12.

Answer: E

standyonda · Answer

-- The simplest solution so far --

1/ Move the 5^(x-1) to the left side
2/ Move (1/2) from the right side 2^(y-1) = (2^y)*(1/2) to the left side 
3/ Now you have 10 - 2*5^(y-x+1) = 2^y
4/ Ask when this is possible? -- The right side is always >0 so the left side must be >0 and this is true only if (y-x+1)=0
5/ Now you are looking for a combination of numbers from the choices that is y+1=x
6/ 3*4=12 looks good!

GMATGuruNY · Answer

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

\frac{5^x}{5^{x-1}}-\frac{5^y}{5^{x-1}}=2^{y-1}

5-5^{y-x+1}=2^{y-1}

The blue equation implies the following:
5 - POWER OF 5 = POWER OF 2.
The only logical option is as follows:
 5 - 5^0 = 2^2 .

Since the right side of the blue equation is equal to  2^2 , we get:
 2^{y-1}=2^2 
 y-1=2 
 y=3 .

Since y=3 and the subtracted term on the left side is equal to  5^0 , we get:
 5^{3-x+1}=5^0 
 3-x+1=0 
 4=x .

Thus:
 xy = 4*3 = 12 .

E

sarphant · Answer

Clearly,
5^x - 5^y = 2^y-1 * 5^x-1
(5^x-5^y)/5^(x-1) = 2^(y-1)
5 - 5^(y-x+1) = 2^ (y-1)

Now,keeping conditions RHS always +ve .So LHS must always be a positive.
RHS , a factor of 2 can be made on the LHS by when 5^(y-x+1) = 1
So y-x+1 = 0
y-x=-1

Also y-1 =2
so y =3
x =4

which gives xy = 12.

If x and y are positive integers and 5^x

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If x and y are positive integers and 5^x

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