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Updated on: 29 Aug 2014, 14:42
Originally posted by plaverbach on 29 Aug 2014, 13:19.
Last edited by Bunuel on 29 Aug 2014, 14:42, edited 1 time in total.
Last edited by Bunuel on 29 Aug 2014, 14:42, edited 1 time in total.
Edited the question
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B
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Difficulty:
85%
(hard)
Question Stats:
54% (02:40) correct
46%
(02:26)
wrong
based on 700
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Suppose x, y, and z are positive integers such that xy + yz = 29 and xz + yz = 81. Which of the following variables has exactly one unique solution?
(i) x
(ii) y
(iii) z
A. none
B. ii only
C. iii only
D. i and ii only
E. ii and iii only
(i) x
(ii) y
(iii) z
A. none
B. ii only
C. iii only
D. i and ii only
E. ii and iii only
Updated on: 05 Jan 2021, 05:17
Originally posted by plaverbach on 29 Aug 2014, 13:20.
Last edited by plaverbach on 05 Jan 2021, 05:17, edited 2 times in total.
Last edited by plaverbach on 05 Jan 2021, 05:17, edited 2 times in total.
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How can one solve in under 3 min?
Official answer below:
Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.
But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).
Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.
That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.
Official answer below:
Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.
But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).
Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.
That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.
30 Aug 2014, 12:41
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Here's another approach:
Once we figure out that y=1, we use that to simplify the second equation to z(x+1) = 81. So what pairs of numbers make 81? 1*81, 3*27, or 9*9. Looking back at our first equation with y=1, we get x+z=29. So what do we know about x and z?
They add up to 29, and when we add 1 to x, they multiply to yield 81. The only pair of factors that works is 3 and 27. So we could have x=2 and y=27, or x=26 and y=3.
Once we figure out that y=1, we use that to simplify the second equation to z(x+1) = 81. So what pairs of numbers make 81? 1*81, 3*27, or 9*9. Looking back at our first equation with y=1, we get x+z=29. So what do we know about x and z?
They add up to 29, and when we add 1 to x, they multiply to yield 81. The only pair of factors that works is 3 and 27. So we could have x=2 and y=27, or x=26 and y=3.
