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What is the remainder when X^4 + Y^4 is divided by 5 1. When

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What is the remainder when X^4 + Y^4 is divided by 5 1. When [#permalink] New post 20 Jul 2010, 04:02
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

25% (01:03) correct 75% (01:30) wrong based on 8 sessions
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

This question has been treated in another thread, but I simply cannot filter
out the posts for reference.

The approach used in one of the post:
Each statement alone is not SUFFICIENT.
Let X - Y = 5A + 1
X + Y = 5B + 2
2X = 5(A+B) + 3
2Y = 5(B-A) + 1
16 (X4 + Y4) = {5(A+B)+3}4 + {5(B-A)+1}4
= (5P+3)4 + (5Q+1)4
= 81 + 1 + 5R
= 82 + 5R
X4 + Y4 = 5 + (5R+2)/16
Since L.H.S is multiple of 16, R.H.S. should also be multiple of 16(since R.H.S cannot be a fraction.).
Lowest value of R for which R.H.S. is an integer is R=6
X4 + Y4 = 82
Remainder of (X4 + Y4)/5 = 2

Hence C

Could anyone assist to especially with the highlighted part. Thanks
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Re: Remainder for X^4 + Y^4 divided by 5 [#permalink] New post 22 Jul 2010, 04:10
The method you posted seems to me to be a bit too contrived and lengthy, though it may work for some people. On the GMAT, it may be better to test cases and visualize possibilities rather than perform too much of algebra that you would be prone to make errors in under a time crunch. The following method may seem pretty contrived too, but in fact it's quite intuitive - you just need to test obvious cases. This approach will work for most DS questions and is something you can rely on for test-day.

x^4 + y^4

1. (x - y) R 5 = 1

i.e. x - y is some number ending in 1 or 6.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4

INSUFFICIENT

2. (x + y) R 5 = 2

i.e. x + y is some number ending in 2 or 7.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4

INSUFFICIENT


Both 1 and 2:

x + y is a number ending in 2 or 7 and x - y is a number ending in 1 or 6, i.e. there are 4 combinations we need to test to see if they are possible. i.e. (x+y, x-y) are numbers ending in (2,1) (2,6) (7,1) or (7,6). It shouldn't be too tough to come up with the following examples:

19 + 13 = 32 and 19 - 13 = 6 ... satisfies (2,6)
14 + 13 = 27 and 14 - 13 = 1 ... satisfies (7,1)
19 + 18 = 37 and 19 - 18 = 1 ... satisfies (7,1)

The other 2 cases of (2,1) and (7,6) are evidently contradictory (you won't be able to find two numbers that satisfy it) and are thus not possible. Now let's test our stem for the 3 possible examples:

1. 19^4 is a number ending in 1, and 13^4 is a number ending in 1 => the sum is a number ending in 2... remainder when divided by 5 will also be 2.

2. 14^4 is a number ending in 6, and 13^4 is a number ending in 1 => the sum is a number ending in 7 ... remainder when divided by 5 will be 2.

3. 19^4 is a number ending in 1 and 18^4 is a number ending in 6 => the sum is a number ending in 7 ... remainder when divided by 5 will be 2.

Three solid cases are usually sufficient on the GMAT and this is a safe bet.

SUFFICIENT

Pick C.
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Re: Remainder for X^4 + Y^4 divided by 5 [#permalink] New post 22 Jul 2010, 04:15
Nice explanation!!
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Re: Remainder for X^4 + Y^4 divided by 5 [#permalink] New post 13 Aug 2011, 11:35
AbhayPrasanna wrote:
The method you posted seems to me to be a bit too contrived and lengthy, though it may work for some people. On the GMAT, it may be better to test cases and visualize possibilities rather than perform too much of algebra that you would be prone to make errors in under a time crunch. The following method may seem pretty contrived too, but in fact it's quite intuitive - you just need to test obvious cases. This approach will work for most DS questions and is something you can rely on for test-day.

x^4 + y^4

1. (x - y) R 5 = 1

i.e. x - y is some number ending in 1 or 6.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4

INSUFFICIENT

2. (x + y) R 5 = 2

i.e. x + y is some number ending in 2 or 7.

There are various values of x and y that could yield such numbers, and would thus yield various different values of x^4 + y^4



Dear AbhayPrasanna

can you please tell what calues combination we can try

btw when i tries 3 combination remainder was always 2
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Re: Remainder for X^4 + Y^4 divided by 5   [#permalink] 13 Aug 2011, 11:35
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