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If x is a positive integer, what is the units digit of 567^24*x^y

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chetan2u
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If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  Updated on: 22 Sep 2016, 23:56
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If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


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Originally posted by chetan2u on 22 Sep 2016, 23:16.
Last edited by Bunuel on 22 Sep 2016, 23:56, edited 1 time in total.
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deepthit
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If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  Updated on: 23 Sep 2016, 00:26
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I go with B

Statement 1: -3<x<2 so x is 1
statement 2 : y=0, what ever the value of x, if y = 0 , the value of the term will be 1 so only term to consider to determine the value of entire expression is 7^(24)

Hence D

Originally posted by deepthit on 22 Sep 2016, 23:20.
Last edited by deepthit on 23 Sep 2016, 00:26, edited 1 time in total.
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NEELAGIRI
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If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  23 Sep 2016, 00:21
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0



statement1: x^2+x-6<0 so -3<x<2. given x is positive integer. so the value of x will be 1
the value of y is independent. SUFFICIENT

statement2: y=0. so what ever is the positive value of x, anything power 0 will be 1.
SUFFICIENT


SO D

#NEELAGIRI
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  23 Sep 2016, 11:07
Statement 1: X^2 - X < 6 for it to be true. Since x is a positive integer the only value for x is 1. Sufficient since the value or sign of y doesn't matter
Statement 2: states than x^y is 1 so sufficient.

Ans D
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  23 Sep 2016, 22:50
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


Self made


My take is D.

Before approaching, 567^24 follows a unit digit cycle of 3 : 567^1 = 7, 567^2 = 9, 567^3 = 3, and rem(24/3) = 0. That means the question gets reduced to 3*(x)^y.

Statement 1: X is a positive integer x=1 alone satisfies the equation and y can take any value less than 5

3(1)^y...since irrespective of value of y , the result is going to be 3 ---Sufficient

Statement 2:

Y=0....we know that 3(x)^0 = 3 as anything to the power of zero is 1 ---Sufficient
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  24 Sep 2016, 11:23
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


Self made


Guys is not 0 a positive integer?
Answer may be 0 or 1. why are you all not including the value x=0??
Ans is not D
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  24 Sep 2016, 11:34
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RatneshS wrote:
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


Self made


Guys is not 0 a positive integer?
Answer may be 0 or 1. why are you all not including the value x=0??
Ans is not D


0 is a non negative non positive integer. So, x cannot be 0.
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  24 Sep 2016, 11:40
RatneshS wrote:
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


Self made


Guys is not 0 a positive integer?
Answer may be 0 or 1. why are you all not including the value x=0??
Ans is not D


Hi
0 is considered as a neutral number. That's the reason it separates positive and negative in a graph.

So positive integer starts from 1

Hope this helps

Posted from my mobile device
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SanjayaGupta
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  24 Sep 2016, 23:52
The answer would be D.

As the statement for 1 is enough to get the unique ans.

The unit digit of 567^24 comes for 4 cycles of the unit digit 7 and which will result as 1.
x lies between -3 and 2 and its mentioned that the x is positive number hence we can take x as 1 . And 1^y will result as 1

And the statement 2 also satisfies the ans .as the y=0.

Hence option D
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  24 Nov 2016, 04:52
One of the Best Questions on Units digit concept.
Here we need the units digit of n=> (567)^24 * x^y
Applying the concept of unit digit ->
Units digit of (567)^24 is always 1
So we need the unit digit of x^y to be able to tell the unit digit of n.


Statement 1
Here x=> (-3,2) is the range of x
x>0 so x must be 1
and y is a integer <5

Basically y can be anything
as (one)^any integer = one
hence the unit digit of n will be 1*1 => 1

Statement 2
y=0
So in this case x can be any integer >0
Because x^0 will be 1
hence the unit digit of n will be 1*1 => 1


Hence D
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  07 Nov 2019, 12:24
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


x = positive integer

\(units.digit:567^{24}*x^y=units(7^{24}*x^y)\)
\(cycles(7)=[7,9,3,1]=4…units:7^{24}=24/4=integer=4th.cycle=[1]\)
\(find:units(1*x^y)=units(x^y)\)

(1) \(x^2+x-6<0\) and y is an integer less than 5. sufic.

\(x^2+x-6<0…(x+3)(x-2)<0…(less.than=inside.rng)…-3<x=positive.integer<2…x=1\)
\(units(x^y)=(1^y)=1\)

(2) y=0 sufic.

\(units(x^y)=(x^0)=1\)

Ans. (D)
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Re: If x is a positive integer, what is the units digit of 567^24*x^y [#permalink] New post  29 Nov 2019, 02:33
Balajikarthick1990 wrote:
chetan2u wrote:
If x is a positive integer, what is the units digit of \(567^{24}*x^y\)?

(1) \(x^2+x-6<0\) and y is an integer less than 5.
(2) y=0


Self made


My take is D.

Before approaching, 567^24 follows a unit digit cycle of 3 : 567^1 = 7, 567^2 = 9, 567^3 = 3, and rem(24/3) = 0. That means the question gets reduced to 3*(x)^y.

Statement 1: X is a positive integer x=1 alone satisfies the equation and y can take any value less than 5

3(1)^y...since irrespective of value of y , the result is going to be 3 ---Sufficient

Statement 2:

Y=0....we know that 3(x)^0 = 3 as anything to the power of zero is 1 ---Sufficient


Hi, just to correct the cyclicity piece. The unit digit cyclicity for 7 is 4 ( 7,9,3,1) and not 3. In this case it worked as 24 is divisible by both 3 and 4 but just something to keep in mind.

Thanks
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