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Veritas Prep Challenge Question

by Ashley@VeritasPrep » Thu Jul 07, 2011 6:59 pm
As advertised on the main page of Beat the GMAT, Brian, David and I will be posting some original questions today and tomorrow with prizes for the first five correct responders who show their work!

Good Luck!


Because there's no such thing as too many multivariable exponent questions in a forum, here's another one for you all :) --



If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.


Go!
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Veritas Prep

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by swetha2 » Thu Jul 07, 2011 7:04 pm
If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.

Answer is C - require both statements to reach a conclusion

Statement 1 - Insufficient
Z can have a units digit of 4 if x=2 and y=2 or if x=4 and y=2 etc

Statement 2 - Insufficient - whether y is odd or even is alone insufficient in determining value of X

Statement 1 and 2 --> taken together, we notice that only for x=4 and y= any odd integer will the units digit of Z yield 4. Hence x=4
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by soumava » Thu Jul 07, 2011 7:27 pm
From Statement (1) we get: -
As z has a unit digit of 4, then plugging values for x,y we can have many solution for x. For example
Let x = 2, and y = 2. then we have z = 4, Again let x = 4 and y = 3 then z = 64. So Statement 1 is insufficient.

From Statement (2) we get: -
As y is odd, there is nothing said about x and z, so x can have infinite number of values
Statement 2 is also insufficient

Combining the two statements together we have z can only have 4 in its unit place if value of x is 4 and the value of y is odd. Therefore combining the two we can answer

Hence the Answer is C
z has a
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by Itzmeash » Thu Jul 07, 2011 7:33 pm
Statement 1 - Units digit is 4
z=x^y -> Value of Z cannot determined without value of y.. As,Z could be 2^2 or 8^2. X can take many values - So, Insufficient
Statement 2 - Y is odd - y is odd or even cannot give unique value for X
-Insufficient

Both taken together, x=4 and y=1 z=>4, x=4, y=3 =>64, x=4 y=5 => 1024
SO for any odd value of y x takes the value 4

Ans C) Both statements are required
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by anandc19 » Thu Jul 07, 2011 7:41 pm
z=x ^ y
option 1 Alone : The units digits of z is 4.
infinite solutions
option 2 alone : y is odd
infinite solutions

Option 1 + 2 : again infinite solutions
if x =4 , y =1 , units digit of Z is 4
if x = 34 ,y =3 units digit of Z is also 4


Hence both the statements together are also not sufficient
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by krishnasty » Thu Jul 07, 2011 7:41 pm
Z = x^y

Stmnt1 : unit digit of z is 4.
z can be 4(2^2) or 64(8^2)
insufficient

Stmnt2 : y is odd...
irrelevant

combining both,
plugging number, we get
z = 1024 = 4^5

hence, IMo C
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by saurabh2525_gupta » Thu Jul 07, 2011 7:58 pm
The answer is E. Both the statements together are insufficient. The onlt thing that we can derive from the two statements is that the unit's digit of X is 4.
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by amit2k9 » Thu Jul 07, 2011 8:15 pm
a x=2,4. y=2,1. not sufficient.

b x=1,2,3,4 y= 1,2,3,4. not sufficient.

a+b

x= 4,14 y=1,3 not sufficient.

E.
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by Nemesis09 » Thu Jul 07, 2011 8:23 pm
Answer: (E)

If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.

Statement 1:
--> The cyclicity for 2 is 2,4,8,6
--> The cyclicity for 8 is 8,4,2,6
--> The cyclicity for 4 is 4,6
So x could either be a number with unit digit 2,4,or 8 and y could be number yielding a remainder of 2 when it is divided by 4 or a remainder of 1 when it is divided by 2.
Stmnt1: Insufficient

Statement 2:
y is odd --> Can't determine the vaue of x.

Stmnt2: Insufficient

Stmnt1&2: y is odd, so unit digit of x is going to be 4 (4^1=4, 24^3=13824)

Answer (E)
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by Frankenstein » Thu Jul 07, 2011 8:51 pm
Hi,
From(1):
Let z = 64,then
x can be 2 and y =6
x can be 4 and y = 3
Not sufficient
From(2):
No info. about x and z
Not sufficient

From(1) and (2):
Every number with units digit 4 when raised to odd power gives 4 as units digit. So, x can be many numbers with units digit 4.
Example: if z= 64, x= 64, y=1 or x=4, y=3
Not sufficient

Hence, E
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by beyondgmatscore » Thu Jul 07, 2011 9:47 pm
If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.

Statement 1) says that units digit of z is 4. Plugging number we can derive that this is possible for a number of x and y combinations such as 4^3 or 2^6 so not sufficient

Statement 2) says y is odd - that doesn't tell us anything about z and x and hence there are infinite solutions possible here as well

Combining two statement, we know that z has 4 in units digits and that y is odd, but this is also achievable by multiple ways - For e.g., 4^3 = 64 and 4^5 = 1024 and hence there is no unique solution possible

Hence, Answer is E
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by srini1988 » Thu Jul 07, 2011 10:14 pm
If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.

From statement 1: unit digit of z is 4, which means z is even so x or y is even, but cannot get the value of x.

From statement 2: y is odd , cannot infer anything from this

combining 1 and 2: we come to know y is odd and x is even, but cannot find the value of X.

Hence answer is E.
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by rppala90 » Fri Jul 08, 2011 6:22 am
Z=x^y.

4^1 = 4.
4^3 = 64.

14^1 = 14.
14^3 = @@@@4. --> units is 4.

1) units digit of z is 4 BUT In both cases i.e x=4 and x=14 , units is 4. So (1) is not sufficient.
2) Y is odd BUT in both cases above y is odd. so (2) is not sufficient.

Both (1) and (2) together is not sufficient as X can be 4 or 14.

Hence answer is E.
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by mirantdon » Fri Jul 08, 2011 12:02 pm
+1 for C . .
Cyclicity series
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by ronaldramlan » Sat Jul 09, 2011 7:41 pm
Ashley@VeritasPrep wrote:As advertised on the main page of Beat the GMAT, Brian, David and I will be posting some original questions today and tomorrow with prizes for the first five correct responders who show their work!

Good Luck!


Because there's no such thing as too many multivariable exponent questions in a forum, here's another one for you all :) --



If x and y are positive integers and z = x^y, what is the value of x?

(1) The units digits of z is 4.
(2) y is odd.


Go!
The answer is E.

What we know :
x and y > 0 (int)
z = x^y -> which means that z > 0 (int)

What is being asked :
x = ???

Answer :
Statement 1 is INSUFFICIENT, because z can be either 4 (x = 2) or 64 (x = 8)
Statement 2 is INSUFFICIENT, because x can be any positive integers
Statement 1 and 2 combined is also INSUFFICIENT, because let's say y = 1, then x can come in many numbers (4, 14, 24, 34, ...)

Therefore, E is the answer.
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