Reminders

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Reminders

by vipulgoyal » Thu Aug 07, 2014 1:06 am
What is the remainder when X^4 + Y^4 is divided by 5

1. When X-Y is divided by 5 remainder is 1
2. When X+Y is divided by 5 remainder is 2

C

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by GMATGuruNY » Thu Aug 07, 2014 3:02 am
vipulgoyal wrote:What is the remainder when X^4 + Y^4 is divided by 5

1. When X-Y is divided by 5 remainder is 1
2. When X+Y is divided by 5 remainder is 2

C
When integer m is divided by 5, the remainder depends on the UNITS digit of m.
If the units digit of m is 0 or 5, dividing by 5 will yield a remainder of 0.
If the units digit of m is 1 or 6, dividing by 5 will yield a remainder of 1.
If the units digit of m is 2 or 7, dividing by 5 will yield a remainder of 2.
If the units digit of m is 3 or 8, dividing by 5 will yield a remainder of 3.
If the units digit of m is 4 or 9, dividing by 5 will yield a remainder of 4.

Statement 1:
In other words, x-y is a (MULTIPLE OF 5) + 1.
Thus:
x-y = 5a + 1 = 1, 6, 11, 16, 21...

Let x-y = 1.
If x=1 and y=0, then the units digit of x� + y� is 1, in which case dividing by 5 will yield a remainder of 1.
If x=2 and y=1, then the units digit of x� + y� is 7, in which case dividing by 5 will yield a remainder of 2.
Since the remainder can be different values, INSUFFICIENT.

Statement 2:
In other words, x+y is a (MULTIPLE OF 5) + 2.
Thus:
x+y = 5b + 2 = 2, 7, 12, 17, 22...

Let x+y = 2.
If x=2 and y=0, then the units digit of x� + y� is 6, in which case dividing by 5 will yield a remainder of 1.
If x=1 and y=1, then the units digit of x� + y� is 2, in which case dividing by 5 will yield a remainder of 2.
Since the remainder can be different values, INSUFFICIENT.

Statements combined:
Statement 2: x+y = 2, 7, 12, 17, 22...
Statement 1: x-y = 1, 6, 11, 16, 21...

If x+y=7 and x-y=1, then x=4 and y=3.
Here, the units digit of x� + y� is 7, in which case dividing by 5 will yield a remainder of 2.
If x+y=12 and x-y=6, then x=9 and y=3.
Here, the units digit of x� + y� is 2, in which case dividing by 5 will yield a remainder of 2.
If x+y=17 and x-y=1, then x=9 and y=8.
Here, the units digit of x� + y� is 7, in which case dividing by 5 will yield a remainder of 2.

In every case, dividing x� + y� by 5 yields a remainder of 2.
SUFFICIENT.

The correct answer is C.
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by GMATinsight » Thu Aug 07, 2014 6:26 am
vipulgoyal wrote:What is the remainder when X^4 + Y^4 is divided by 5

1. When X-Y is divided by 5 remainder is 1
2. When X+Y is divided by 5 remainder is 2

C
Question : Remainder when X^4 + Y^4 is divided by 5 ?

Statement 1) X - Y = 5a+1

i.e. X - Y = 6 or 11 or 16 and so on...
Case 1: X-Y = 6 i.e. X = 7, Y = 1 then X^4 + Y^4 = 7^4 + 1^4 = Unit digit (1+1) = Unit digit 2
i.e. Remainder when divided by 5 = 2

Case 2: X-Y = 6 i.e. X = 15, Y = 9 then X^4 + Y^4 = 5^4 + 9^4 = Unit digit (5+1) = Unit digit 6
i.e. Remainder when divided by 5 = 1
INCONSISTENT SOLUTIONS

INSUFFICIENT

Statement 2) X + Y = 5b+2
i.e. X+Y = 7 or 12 or 17 and so on...

Case 1: X+Y = 7 i.e. X = 5, Y = 2 then X^4 + Y^4 = 5^4 + 2^4 = Unit digit (5+6) = Unit digit 1
i.e. Remainder when divided by 5 = 1

Case 2: X+Y = 7 i.e. X = 6, Y = 1 then X^4 + Y^4 = 6^4 + 1^4 = Unit digit (6+1) = Unit digit 7
i.e. Remainder when divided by 5 = 2
INCONSISTENT SOLUTIONS

INSUFFICIENT

Combining the two Statements
X - Y = 5a+1 AND X + Y = 5b+2
Adding the two equation
2X = 5(a + b) + 3 => [for X to be an Integer]
i.e. X = 4 or 9 or 14... and so on

let's say X = 5c - 1


and Y = 5b+1 - X = 5b+1 - [5c-1] = 5 (b-c)+(2)
i.e. Y = (Multiple of 5)+(2)

let's say X = 5d + 2


i.e. X^4 + Y^4 = (5c-1)^4 + (5d+2)^4 when divided by 5 will always give us the remainder (1+6) i.e. Remainder = 2

SUFFICIENT

Answer: Option C
Last edited by GMATinsight on Thu Aug 07, 2014 9:55 am, edited 2 times in total.
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by GMATGuruNY » Thu Aug 07, 2014 8:42 am
GMATinsight wrote: Combining the two Statements
X - Y = 5a+1 AND X + Y = 5b+1
Adding the two equation
2X = 5a + 5b + 2
]i.e. X = Multiple of 5 + 1

let's say X = 5c+1

and Y = 5b+1 - X = 5b+1 - (5c+1) = 5 (b-c)
]i.e. Y = Multiple of 5

i.e. X^4 + Y^4 = (5c+1)^4 + (5d)^4 when divided by 5 will always give us the remainder 1

SUFFICIENT

Answer: Option C
Bhoopendra,

The equation in red should be X+Y = 5b + 2.
This typo seems to have lead to an erroneous conclusion.
When x� + y� is divided by 5, the remainder is not 1 but 2.
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by GMATinsight » Thu Aug 07, 2014 9:54 am
GMATGuruNY wrote:
Bhoopendra,

The equation in red should be X+Y = 5b + 2.
This typo seems to have lead to an erroneous conclusion.
When x� + y� is divided by 5, the remainder is not 1 but 2.
Thank you Mitch for pointing out. I have made the required correction. :)
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