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Tricky Number Properties and Algebra

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Tricky Number Properties and Algebra

by Cedagmat » Wed Nov 10, 2010 5:51 pm
I only have the OA to these but not the explanations. Help!

If x and у are none negative integer, what is the value of xy?
(1) 7^x = 13^y.
(2) 3^x = 9^y.

[spoiler]Answer: A[/spoiler]

s1: Since only 7^0 and and 13^0 can be equal to each other, I thought x and y would be 0. Sufficient.
s2: 3^2 = 9^1, thus x and y are 2 and 1. Sufficient.
But D is not the answer here, can someone explain?


If x and у are integers, what is the value of (x - y)^4?
(1) The product of x and у is 7.
(2) The sum of x and у is - 8

[spoiler]Answer: A
[/spoiler]
I choose answer C since it had all the elements that put together the value. However, that is incorrect. Does anyone know a better way to simplify the question stem so that it is more approachable?

If x^n = l, where x and n are both integers, what is the value of the x?
(1) n is a multiple of 5.
(2) n is an odd number.

[spoiler]Answer: B[/spoiler]
If x and n are both integers, then x^0 = 1 or 1^n = 1.
S1: Since n is a multiple of 5 and not 0, then x must be 1.
S2: Since n is an odd number and not 0, then x must be 1.
My answer was D, but it is incorrect. Can anyone explain?
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by beat_gmat_09 » Wed Nov 10, 2010 7:00 pm
Cedagmat wrote:I only have the OA to these but not the explanations. Help!

If x and у are none negative integer, what is the value of xy?
(1) 7^x = 13^y.
(2) 3^x = 9^y.

[spoiler]Answer: A[/spoiler]

s1: Since only 7^0 and and 13^0 can be equal to each other, I thought x and y would be 0. Sufficient.
s2: 3^2 = 9^1, thus x and y are 2 and 1. Sufficient.
But D is not the answer here, can someone explain?


If x and у are integers, what is the value of (x - y)^4?
(1) The product of x and у is 7.
(2) The sum of x and у is - 8

[spoiler]Answer: A
[/spoiler]
I choose answer C since it had all the elements that put together the value. However, that is incorrect. Does anyone know a better way to simplify the question stem so that it is more approachable?

If x^n = l, where x and n are both integers, what is the value of the x?
(1) n is a multiple of 5.
(2) n is an odd number.

[spoiler]Answer: B[/spoiler]
If x and n are both integers, then x^0 = 1 or 1^n = 1.
S1: Since n is a multiple of 5 and not 0, then x must be 1.
S2: Since n is an odd number and not 0, then x must be 1.
My answer was D, but it is incorrect. Can anyone explain?
I)
i) 7 and 13 are both prime numbers only way 7^x=13^y is when x and y both are 0. Therefore xy=0. sufficient
ii) 3^x=9^y , 3^x=3^2^y; by the indices rule x=2y; we cannot get the value of xy. Not sufficient

II)
i) The statement which provides value of x-y will be sufficient. Product x*y =7; As 7 is prime number, this means
that either x=7, y=1 or y=7, x=1; either way the difference will be 6 or -6 and raising this to even power will
be a positive number.
ii) x+y does not provide any information to get x-y. Not sufficient.

III)
i) x^n=1. n is multiple of 5. n can also be 0, 0*5=0, x can have any value, or n can be multiple of 5 and x=1.
There are more possibilities for value of x. Not sufficient.
ii) x^n=1, n is odd. This means that x has to be 1. gives a single value of x.
Therefore Sufficient.
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by Rahul@gurome » Wed Nov 10, 2010 7:05 pm
s1: Since only 7^0 and and 13^0 can be equal to each other, I thought x and y would be 0. Sufficient.
s2: 3^2 = 9^1, thus x and y are 2 and 1. Sufficient.
But D is not the answer here, can someone explain?
Solution:
Statement (1) alone is sufficient because obviously, x and y both have to be zero.
Next consider statement (2) alone.
If 3^x = 9^y, we get 3^x = 3^2y or x = 2y.
If x = 2, y = 1 and xy = 2.
If x = 4, y = 2 and xy = 8.
Since we do not get a unique value of xy, statement (2) alone is not sufficient.

The correct answer is hence (A).
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by Cedagmat » Wed Nov 10, 2010 8:19 pm
II)
i) The statement which provides value of x-y will be sufficient. Product x*y =7; As 7 is prime number, this means
that either x=7, y=1 or y=7, x=1; either way the difference will be 6 or -6 and raising this to even power will
be a positive number.
ii) x+y does not provide any information to get x-y. Not sufficient.

III)
i) x^n=1. n is multiple of 5. n can also be 0, 0*5=0, x can have any value, or n can be multiple of 5 and x=1.
There are more possibilities for value of x. Not sufficient.
ii) x^n=1, n is odd. This means that x has to be 1. gives a single value of x.
Therefore Sufficient.
Thank you! These two solutions are very elegant and I see what my errors are now.
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by gmatmachoman » Tue Nov 16, 2010 10:38 am
beat_gmat_09 wrote:
Cedagmat wrote:

III)
i) x^n=1. n is multiple of 5. n can also be 0, 0*5=0, x can have any value, or n can be multiple of 5 and x=1.
There are more possibilities for value of x. Not sufficient.
ii) x^n=1, n is odd. This means that x has to be 1. gives a single value of x.
Therefore Sufficient.
n cannot be zero. least value of n will be 5. so X has to be 1.

Pick D.
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by frank1 » Tue Nov 16, 2010 8:30 pm
Cedagmat wrote:I only have the OA to these but not the explanations. Help!

If x and у are none negative integer, what is the value of xy?
(1) 7^x = 13^y.
(2) 3^x = 9^y.

[spoiler]Answer: A[/spoiler]

s1: Since only 7^0 and and 13^0 can be equal to each other, I thought x and y would be 0. Sufficient.
s2: 3^2 = 9^1, thus x and y are 2 and 1. Sufficient.
But D is not the answer here, can someone explain?
There are some good explanation from other gurus here
but i wonder how did you forget the 'principles you used few seconds ago' while dealing with s2 (just query)
In statement 1 you used 0 for both
In st 2, you were able to come up with x=2 and y=1 in which xy=2
and x=0 and y=0 in which xy=0 still holds (thats what you used in st 1)
so double case
Insufficient...
so sometimes it can be tricky....

thanks
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