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If x and y are positive integers and 5^x
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06 Apr 2012, 08:42
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If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy? A. 48 B. 36 C. 24 D. 18 E. 12
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If x and y are positive integers and 5^x
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06 Apr 2012, 13:01
BN1989 wrote: If x and y are positive integers and (5^x)(5^y)=(2^(y1))*(5^(x1)), 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. \(5^x5^y=2^{y1}*5^{x1}\); \(5^x2^{y1}*5^{x1}=5^y\); \(5^x(1\frac{2^y}{2}*\frac{1}{5})=5^y\); \(5^x(102^y)=2*5^{y+1}\). Now, since the right hand side is always positive then the left hand side must also be positive, hence \(102^y\) 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 \(x\): \(5^x(102^3)=2*5^{3+1}\); \(2*5^x=2*5^4\); \(x=4\) > \(xy=12\). Answer: E.
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Re: If x and y are positive integers and 5^x  5^y = 2^(y1)*5^x
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19 Feb 2014, 21:50
Faster: First constrain the possible answers. We know \(x>y\) since if \(x=y\) then the lefthand side is 0 or if \(x<y\) then the LHS is negative... but the RHS is always positive. Now act: divide both sides by \(5^{x1}\) to get \(55^{yx+1} = 2^{y1}\).
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 \(55^0=51=4\). 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^{x1}, not 5^x.)




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Re: If x and y are positive integers and 5^x
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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^(y1))*(5^(x1)), 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. \(5^x5^y=2^{y1}*5^{x1}\) > \(5^x2^{y1}*5^{x1}=5^y\) > \(5^x(1\frac{2^y}{2}*\frac{1}{5})=5^y\) > \(5^x(102^y)=2*5^{y+1}\). Now, since the right hand side is always positive then the left hand side must also be positive, hence \(102^y\) 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 \(x\): \(5^x(102^3)=2*5^{3+1}\) > \(2*5^x=2*5^4\) > \(x=4\) > \(xy=12\). Answer: E. Bunuel, I have a question, how did you know that you had to reorganize the equation in this way? From \(5^x5^y=2^{y1}*5^{x1}\) TO \(5^x2^{y1}*5^{x1}=5^y\) Thanks!



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Re: If x and y are positive integers and 5^x
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24 Sep 2013, 22:04
danzig wrote: Bunuel wrote: BN1989 wrote: If x and y are positive integers and (5^x)(5^y)=(2^(y1))*(5^(x1)), 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. \(5^x5^y=2^{y1}*5^{x1}\) > \(5^x2^{y1}*5^{x1}=5^y\) > \(5^x(1\frac{2^y}{2}*\frac{1}{5})=5^y\) > \(5^x(102^y)=2*5^{y+1}\). Now, since the right hand side is always positive then the left hand side must also be positive, hence \(102^y\) 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 \(x\): \(5^x(102^3)=2*5^{3+1}\) > \(2*5^x=2*5^4\) > \(x=4\) > \(xy=12\). Answer: E. Bunuel, I have a question, how did you know that you had to reorganize the equation in this way? From \(5^x5^y=2^{y1}*5^{x1}\) TO \(5^x2^{y1}*5^{x1}=5^y\) Thanks! Not directed at me, However you can rearrange it another way. We have \(5^x5^y = 2^{y1}*5^{x1}\), Dividing on both sides by \(5^{x1}\), we have \(55^{y+1x} = 2^{y1} \to 5 = 5^{y+1x} + 2^{y1}\). Now, as x and y are positive integers, the only value which \(2^{y1}\) can take is 4.Thus, y1 = 2, y = 3. Again, the value of \(5^{y+1x}\) has to be 1, thus, y+1x = 0 \(\to\) x = y+1 = 3+1 = 4. Thus, x*y = 4*3 = 12. Thus, I think you could rearrange 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^(y1)*5^x
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19 Feb 2014, 21:25
MrWallSt wrote: If x and y are positive integers and 5^x  5^y = 2^(y1)*5^(x1), 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. \(5^y (5^{xy}  1) = 2^{y1}*5^{x1}\) Now I want only 2s and 5s on the left hand side. If xy is 1, then \((5^{xy}  1)\) 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 \(5^y (2^2) = 2^{y1}*5^{x1}\) 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.
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Re: If x and y are positive integers and 5^x  5^y = 2^(y1)*5^x
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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
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20 Feb 2014, 03:53
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), 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^(yx) = 2^(y1) * 5^(x1x) 1 5^(yx) = 2^(y1) * 5^(1) 5= 2^(y1) + 5^(yx+1) now minimum value of 2^(y1) =1, hence 2^(y1) must be equal to 4 and 5^(yx+1) must be equal to 1 for the R.H.S to become equal to L.H.S. 2^(y1) = 4 for y=3 and 5^(yx+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
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20 Feb 2014, 14:02
If x and y are positive integers and (5^x)(5^y)=(2^{y1})*(5^{x1}), 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^{x1}
left and right side simplifies to 55^{yx+1} = 2^{y1}
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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If x and y are positive integers and 5^x
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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^(y1))*(5^(x1)), 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. \(5^x5^y=2^{y1}*5^{x1}\); \(5^x2^{y1}*5^{x1}=5^y\); \(5^x(1\frac{2^y}{2}*\frac{1}{5})=5^y\); \(5^x(102^y)=2*5^{y+1}\). Now, since the right hand side is always positive then the left hand side must also be positive, hence \(102^y\) 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 \(x\): \(5^x(102^3)=2*5^{3+1}\); \(2*5^x=2*5^4\); \(x=4\) > \(xy=12\). Answer: E. i did another logic is it right? \((5^x)(1(2^y1)/5)=5^y\) then \((5^xy1)(5(2^y1))=1\) which means this must be \(xy1= 0\) and \(5(2^y1) = 1\) means also \(2^y1 = 4\) then \(y1=2\) 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
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19 Aug 2015, 07:04
(5^x)(5^y)=(2^(y1))*(5^(x1)) (5^x)(15^yx) = (2^(y1))*(5^(x1)) 5(15^yx) = (2^(y1)) 10(15^yx) = (2^(y))
We note that (2^(y)) is always positive. which translates to (15^yx) >0 or yx<0
So, 10(5^yx)(5^(xy)1) = (2^(y)) 5^(yx+1)*2*(5^(xy)1) =(2^(y))
Now, RHS is 5^0, which means yx+1 = 0, xy =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
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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>(255), (12525), (625125). Also check(62525), (6255), 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.



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Re: If x and y are positive integers and 5^x
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01 Aug 2017, 08:57
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy?
A. 48 B. 36 C. 24 D. 18 E. 12 Given: \((5^x)(5^y)=(2^{y1})*(5^{x1})\) Divide both sides by \(5^{x1}\) to get: \(5^1  5^{yx+1} = 2^{y1}\) Simplify: \(5  5^{yx+1} = 2^{y1}\) OBSERVE: Notice that the right side, \(2^{y1}\), is POSITIVE for all values of y Since y is a positive integer, \(2^{y1}\) can equal 1, 2, 4, 8, 16 etc (powers of 2) So, the left side, \(5  5^{yx+1}\), must be equal 1, 2, 4, 8, 16 etc (powers of 2). Since \(5^{yx+1}\) is always positive, we can see that \(5  5^{yx+1}\) cannot be greater than 5 So, the only possible values of \(5  5^{yx+1}\) are 1, 2 or 4 In other words, it must be the case that: case a) \(5  5^{yx+1} = 2^{y1} = 1\) case b) \(5  5^{yx+1} = 2^{y1} = 2\) case c) \(5  5^{yx+1} = 2^{y1} = 4\) Let's test all 3 options. case a) \(5  5^{yx+1} = 2^{y1} = 1\) This means y = 1 (so that the right side evaluates to 1) The left side, \(5  5^{yx+1} = 1\), when \(5^{yx+1} = 4\). Since x and y are positive integers, it's IMPOSSIBLE for \(5^{yx+1}\) to equal 4 So, we can eliminate case a case b) \(5  5^{yx+1} = 2^{y1} = 2\) This means y = 2 (so that the right side evaluates to 2) The left side, \(5  5^{yx+1} = 2\), when \(5^{yx+1} = 3\). Since x and y are positive integers, it's IMPOSSIBLE for \(5^{yx+1}\) to equal 3 So, we can eliminate case b case c) \(5  5^{yx+1} = 2^{y1} = 4\) This means y = 3 (so that the right side evaluates to 4) The left side, \(5  5^{yx+1} = 4\), when \(5^{yx+1} = 1\). If \(5^{yx+1} = 1\), then \(yx+1 = 0\) In this case, y = 3So, we can write: 3  x + 1 = 0, which mean x = 4So, the only possible solution is y = 3 and x = 4, which means xy = ( 4)( 3) = 12 Cheers, Brent
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Re: If x and y are positive integers and 5^x
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07 Aug 2017, 21:37
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy?
A. 48 B. 36 C. 24 D. 18 E. 12 \((5^x)(5^y)=(2^{y1})*(5^{x1})\) =\((5^{x1})(55^{yx+1})=(2^{y1})*(5^{x1})\) 5^{x1} =/=0. So, \((55^{yx+1})=(2^{y1})\) Since \((2^{y1})\) must be +ve. Also Y is +ve so, y1>0 and hence \((2^{y1})\) will be an integer only. Hence yx+1 can be 0 only. yx+1 = 0 > yx = 1 also at yx+1 = 0 \(51 = (2^{y1})\) \(4 = (2^{y1})\) y 1 = 2 y = 3 x= y+1 = 4 xy = 12 Answer E



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Re: If x and y are positive integers and 5^x
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30 Jan 2018, 13:30
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 complexlooking 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^(X1) 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^(Y1) times 5^(X1). *2^(Y1) 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^(X1)) 5^X  125 = 4(5^(X1)) 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: GMAT assassins aren't born, they're made, Rich
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Re: If x and y are positive integers and 5^x
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31 Jan 2018, 01:10
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy?
A. 48 B. 36 C. 24 D. 18 E. 12 Good Question.. I solved it this way.. Let \(x1 = a\), \(y1 = b\).. 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.



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Re: If x and y are positive integers and 5^x
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02 Feb 2018, 11:09
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy?
A. 48 B. 36 C. 24 D. 18 E. 12 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^(y1) 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
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If x and y are positive integers and 5^x
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02 Jul 2018, 12:15
 The simplest solution so far 
1/ Move the 5^(x1) to the left side 2/ Move (1/2) from the right side 2^(y1) = (2^y)*(1/2) to the left side 3/ Now you have 10  2*5^(yx+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 (yx+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!



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If x and y are positive integers and 5^x
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02 Jul 2018, 15:03
BN1989 wrote: If x and y are positive integers and \((5^x)(5^y)=(2^{y1})*(5^{x1})\), what is the value of xy?
A. 48 B. 36 C. 24 D. 18 E. 12 \(5^x5^y=2^{y1}*5^{x1}\) \(\frac{5^x}{5^{x1}}\frac{5^y}{5^{x1}}=2^{y1}\) \(55^{yx+1}=2^{y1}\)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: \(2^{y1}=2^2\) \(y1=2\) \(y=3\). Since y=3 and the subtracted term on the left side is equal to \(5^0\), we get: \(5^{3x+1}=5^0\) \(3x+1=0\) \(4=x\). . Thus: \(xy = 4*3 = 12\).
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Re: If x and y are positive integers and 5^x
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10 Aug 2019, 08:00
Clearly, 5^x  5^y = 2^y1 * 5^x1 (5^x5^y)/5^(x1) = 2^(y1) 5  5^(yx+1) = 2^ (y1)
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^(yx+1) = 1 So yx+1 = 0 yx=1
Also y1 =2 so y =3 x =4
which gives xy = 12.




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