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If x and y are positive integers and 5^x [#permalink]
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06 Apr 2012, 09:42
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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 [#permalink]
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06 Apr 2012, 14:01
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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 [#permalink]
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24 Sep 2013, 15:38
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 [#permalink]
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24 Sep 2013, 23:04
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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 [#permalink]
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19 Feb 2014, 22:25
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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 [#permalink]
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19 Feb 2014, 22:50
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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  5^y = 2^(y1)*5^x [#permalink]
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19 Feb 2014, 23:01
@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, 04:53
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 [#permalink]
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20 Feb 2014, 15:02
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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Re: If x and y are positive integers and 5^x  5^y = 2^(y1)*5^x [#permalink]
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24 Feb 2014, 23:57
SizeTrader wrote: 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.) Nicely explained.
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If x and y are positive integers and 5^x [#permalink]
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12 Feb 2015, 11:00
Hey BunuelCan 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, 07:30
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 [#permalink]
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19 Aug 2015, 08:04
(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 [#permalink]
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Re: If x and y are positive integers and 5^x [#permalink]
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21 Sep 2016, 02:31
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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