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If x and y are positive integers, what is the remainder when x^y is

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If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 12 Jun 2015, 03:20
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 12 Jun 2015, 03:38
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Bunuel wrote:
If x and y are positive integers, what is the remainder when x^y is divided by 10?

(1) x = 26
(2) y^x = 1


Kudos for a correct solution.


CONCEPT: Remainder when a number is divided by 10 remains the unit digit of Number

Question Redefined: Find unit digit of x^y?

Statement 1: x = 26

Since y is a positive integer and the cyclicity of any number with unit digit 6 remain 6 always for any positive integer exponent of it

Therefore x^y = 26^y will have Unit digit 6 for any positive Integer value of y

Hence, SUFFICIENT

Statement 2: y^x = 1

i.e. y has unit digit 1 but since the value of x is unknown therefore unit digit of x^y can't be calculated

Hence, NOT SUFFICIENT

Answer: option
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 12 Jun 2015, 03:34
If x and y are positive integers, what is the remainder when x^y is divided by 10?
(1) x = 26
(2) y^x = 1

Stmt 1 - y is not know. In Sufficient.
Stmt 2 - There are two possible solutions available. Lets take x=y=1, then y^x=1.
Lets take x=0 and y=1, then y^x=1.
Two possible solutions, In sufficient.

1+2 -> x=26, then y must be 1 to validate 1^26=1. Remainder is 6. Sufficient.

Ans C.
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 12 Jun 2015, 03:42
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balamoon wrote:
If x and y are positive integers, what is the remainder when x^y is divided by 10?
(1) x = 26
(2) y^x = 1

Stmt 1 - y is not know. In Sufficient.
Stmt 2 - There are two possible solutions available. Lets take x=y=1, then y^x=1.
Lets take x=0 and y=1, then y^x=1.
Two possible solutions, In sufficient.

1+2 -> x=26, then y must be 1 to validate 1^26=1. Remainder is 6. Sufficient.

Ans C.


Hi Balamoon,

You have made a mistake in evaluating the first statement. The unit digit of x is 6 and y is a positive Integer and by the principle of cyclicity, the number with unit digit 6 will always have unit digit 6 for any positive Integer exponent of the number

Hence the first statement alone answers the question here.

I hope it clears your mistake part in above working.
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 13 Jun 2015, 03:15
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Bunuel wrote:
If x and y are positive integers, what is the remainder when x^y is divided by 10?

(1) x = 26
(2) y^x = 1


Kudos for a correct solution.



Both x and y are positive integers => Both x and y are not zero.
Keeping this in mind, let us approach the question.


(1) x = 26
Here, the units digit is 6 and 6^(any number) will result in 6 as the units digit.
So, x^y => 6 as units digit => Divided by 10 will give 6 as the remainder. Sufficient.

(2) y^x = 1

Now here we can have two possibilities:
(i) 1^x = 1, where x can be any number. So, x^y = x^1 = x.
We do not get a single remainder when x is divided by 10 as it is dependent on value of x. Not Sufficient
(ii) y^0 = 1 => 0^y = 0.
:!: Here x = 0 is not possible because both x and y are POSITIVE INTEGERS :!:
So option (ii) is not possible, and per (i) y^x = 1 is Not Sufficient


Many people, including me, make a mistake by assuming value of variables to be 0 even when the question explicitly states "positive intergers". DO NOT do that.

Answer is A.


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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 15 Jun 2015, 06:05
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Bunuel wrote:
If x and y are positive integers, what is the remainder when x^y is divided by 10?

(1) x = 26
(2) y^x = 1


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

(1) SUFFICIENT: The tendency is to deem statement (1) insufficient because we have no information about the value of y. But 26 has a units digit of 6, and remember that 6^(any positive integer) has a units digit of 6 (the pattern is a single-term repeat).

6^1 = 6
6^2 = 36
6^3 = 216
etc.

Thus, 26 raised to ANY positive integer power will also have a units digit of 6 and therefore a remainder of 6 when divided by 10.

(2) INSUFFICIENT: Given that y^x = 1, there are a few possible scenarios:
Attachment:
2015-06-15_1659.png
2015-06-15_1659.png [ 27.94 KiB | Viewed 2315 times ]

However, the question stem tells us that x and y are POSITIVE integers, so we eliminate the first and third scenarios.
Attachment:
2015-06-15_1700.png
2015-06-15_1700.png [ 44.81 KiB | Viewed 2313 times ]


The remaining scenario indicates that y = 1 and x = any positive integer. Without more information about x, we cannot determine the remainder when x^y is divided by 10.

Since statement (1) tells us the value of x and statement (2) indirectly tells us the value of y (y = 1), the temptation might be to combine the information to arrive at an answer of C. This is a common trap on difficult Data Sufficiency problems. It might seem that we need both statements, when one statement alone actually provides enough information.

The correct answer is A.
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 16 Jun 2015, 07:55
Bunuel wrote:
If x and y are positive integers, what is the remainder when x^y is divided by 10?

(1) x = 26
(2) y^x = 1


Kudos for a correct solution.


Given
X and Y both positive.

Statement 1 :
X = 26

power cycle of 6 is 6, hence x ^ Y will always have unit's digit as 6 irrespective of Y. Henc X^y/10 will always have remainder as 6.

Hence statement 1 is sufficient.

Statement 2 :
y^x = 1
Now X > 0, hence Y = 1.

if Y = 1 and X is anything, then X^Y can be anything
Say x = 10, X^y/10 will give reminder as Zero
say X = 15, X ^Y/10 will give reminder as Five.

Statement 2 is insufficient.

Hence option A
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 17 Jun 2015, 02:03
X and Y = pos int
x^y / 10 = ? REMAINDER

1) 26^1 = 26/20 -> R = 6
26^2 = 676 / 10 -> R=6
26^3 = A large number ending in 6 -> R=6

S.

2) y^x = 1

Y = 1, x could be anything.

I.

A.
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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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New post 07 Mar 2017, 10:50
Rule of cyclisity...
Unit digit is 6 so the power doesn't matter...n the reminder will remain 6.
So option a.

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Re: If x and y are positive integers, what is the remainder when x^y is  [#permalink]

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Re: If x and y are positive integers, what is the remainder when x^y is &nbs [#permalink] 16 Mar 2018, 02:51
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