Data Sufficiency

What is the remainder when X^4 + Y^4 divide by 5

A) X-Y divided by 5 gives remainder 1
B) X+Y divided by 5 gives remainder 2

Please suggest an easy method to solve the above question

it is neither A nor B.

Statement 1 says,
(x-y)/5 has R=1, so x-y could be any value from 1,6,11,16.... This is insufficient
Statement 2 says,
(x+y)/5 has R=2, so x+y could be any value from 2,7,12,17... This is also insufficient

Combining the two, we can see we don't have any common remainder for both (x+y) and (x-y), we are likely to get different results. So in my opinion it is E

zaarathelab wrote:What is the remainder when X^4 + Y^4 divide by 5

A) X-Y divided by 5 gives remainder 1
B) X+Y divided by 5 gives remainder 2

Please suggest an easy method to solve the above question

A general strategy for the following prompt:

When positive integer x is divided by the divisor D, the remainder is R.

-- Make a list of possible values for x.
-- The smallest possible value of x = the remainder R.
-- To determine the other possible values of x, just keep adding multiples of the divisor D.

Statement 1: X-Y divided by 5 gives a remainder of 1.
Smallest possible value = 1.
Keep adding multiples of 5:
x-y = 1,6,11,16...
Since x and y could be many different values, insufficient.

Statement 2: X+Y divided by 5 gives a remainder of 2.
Smallest possible value =2.
Keep adding multiples of 5:
x+y = 2,7,12,17...
Since x and y could be many different values, insufficient.

Statements 1 and 2 combined:
x+y = 2,7,12,17...
x-y = 1,6,11,16..
The units digit of x+y is 2 or 7.
The units digit of x-y is 1 or 6.
Try different combinations:
If x+y = 7 and x-y = 1, then x=4 and y=3.
If x+y = 22 and x-y = 6, then x=14 and y=8.
If x+y = 52 and x-y = 26, then x=39 and y=13.
If x+y = 27 and x-y = 11, then x=19 and y=8.

The units digit of x is 4 or 9.
Since 4^4 has a units digit of 6 and 9^4 has a units digit of 1, the units digit of x^4 is either 6 or 1.

The units digit of y is 3 or 8.
Since 8^4 has a units digit of 6 and 3^4 has a units digit of 1, the units digit of y^4 is 6 or 1.

Thus, the units digit of x^4 + y^4 could be:
6+1 = 7
6+6 = 12 (implying a units digit of 2)
1+6 = 7
1+1 = 2.

The units digit is either 2 or 7.
When an integer with a units digit of 2 or 7 is divided by 5, the remainder in each case will be 2.
Sufficient.

The correct answer is C.

GMATGuruNY wrote:

zaarathelab wrote:What is the remainder when X^4 + Y^4 divide by 5

A) X-Y divided by 5 gives remainder 1
B) X+Y divided by 5 gives remainder 2

Please suggest an easy method to solve the above question

A general strategy for the following prompt:

When positive integer x is divided by the divisor D, the remainder is R.

-- Make a list of possible values for x.
-- The smallest possible value of x = the remainder R.
-- To determine the other possible values of x, just keep adding multiples of the divisor D.

The question at hand: What is the remainder when x^4 + y^4 is divided by 5?

Statement 1: X-Y divided by 5 gives a remainder of 1.
x-y = 1,6,11,16...
Since x and y could be many different values, insufficient.

Statement 2: X+Y divided by 5 gives a remainder of 2.
x+y = 2,7,12,17...
Since x and y could be many different values, insufficient.

Statements 1 and 2 combined:
x+y = 2,7,12,17...
x-y = 1,6,11,16..
The units digit of x+y is 2 or 7.
The units digit of x-y is 1 or 6.
Try different combinations:
If x+y = 7 and x-y = 1, then x=4 and y=3.
If x+y = 37 and x-y = 11, then x=24 and y=13.
If x+y = 12 and x-y = 6, then x=9 and y=3.
If x+y = 52 and x-y = 26, then x=39 and y=13.

In each case, the units digit of x is 4 or 9.
Since 4^4 has a units digit of 6 and 9^4 has a units digit of 1, the units digit of x^4 is either 6 or 1.

In each case, the units digit of y is 3.
Since 3^4 = 81, the units digit of y^4 is 1.

Thus, the units digit of x^4 + y^4 is either 6+1 = 7 or 1+1 = 2.
When an integer with a units digit of 7 or 2 is divided by 5, the remainder in each case will be 2.
Sufficient.

The correct answer is C.

Mitch

Just to confirm Initially even i thought its C but certainly there are 2 numbers..Example

x= 9
y= 3
It satisfies both the statements and hence should be the answer but lets look at another example

x=19
y=8

again it satisfies both the statements but the ques is asking for a value which will vary from the above choices....

Hence the answer [spoiler]IMOE

Please correct me if my approach is incorrect.[/spoiler]

prateek_guy2004 wrote: Mitch

Just to confirm Initially even i thought its C but certainly there are 2 numbers..Example

x= 9
y= 3
It satisfies both the statements and hence should be the answer but lets look at another example

x=19
y=8

again it satisfies both the statements but the ques is asking for a value which will vary from the above choices....

Hence the answer [spoiler]IMOE

Please correct me if my approach is incorrect.[/spoiler]

The remainder will be 2 in every case.
Please check my amended post above.

The prompt does not state that x and y are integers, so there are cases for x+y and x-y where we get fractions for x and y. For exmaple, x+y=7 and x-y=6. In this scenario how can we establish a definite answer?

Can anyone give a more algebraic solution than plugging values.

sl750 wrote:The prompt does not state that x and y are integers, so there are cases for x+y and x-y where we get fractions for x and y. For exmaple, x+y=7 and x-y=6. In this scenario how can we establish a definite answer?

By asking for a remainder, the question stem implies -- but should state explicitly -- that x and y are positive integers. Generally, a remainder is defined as the integer value that "remains" when one positive integer is divided by another.

If this question is from GMATPrep, I doubt that it has been transcribed correctly. I can't imagine that an actual GMAT question would use the phrase "gives a remainder".