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this ds killed me

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this ds killed me [#permalink] New post 17 Sep 2006, 06:07
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A
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What is the reminder when X^4 + Y^4 is divided by 5
A. When X-Y is divided by 5 reminder is 1
B. When X+Y is divided by 5 reminder is 2


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 [#permalink] New post 17 Sep 2006, 08:18
interesting Question
Each statement by itself is not sufficient
If you square both and then add them together it seems that 2*(X^2+Y^2) when divided by 5 will have reminder 0. Then X^2+Y^2 is divisible of 5. Do not think that we can get something about X^4+Y^4
think it is E)
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 [#permalink] New post 17 Sep 2006, 08:25
X^4 + Y^4

can be written as

1/2[(x^2+y^2)^2 + (x^2-y^2)^2]

from (1) we get the rem for X-Y when divided by 5

from (2) we get the rem for X+Y when divided by 5

from both we get the rems from (1) and (2)

1/2[(x^2+y^2)^2 + (x^2-y^2)^2] can be re written as

1/2[(x^2+y^2)^2 + [(x-y)(x+y)]^2]

1/2[((x+y)^2)^2 - 2*x^2*y^2+ [(x-y)(x+y)]^2]

we do not know anything about the rem for x^2*y^2

So E
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 [#permalink] New post 17 Sep 2006, 08:33
For the problem, we know S1 and S2 are not sufficient.

(Pick two different (x,y) pairs for S1 and S2 and you see that x^4+y^4 has different remainders.)

The problem is S1 & S2; I believe this will not be sufficient too, as you cannot find two integers such that their difference has a remainder of 1 and sum has a remainder of 2 when divided by 5.

i.e. x - y = {1, 6, 11, 16, ...}
& x +y = {2, 7, 12, 17, ...}

If x = 1.5 and y = 0.5 this is true
If x = 6.5 and y = 5.5 this is true again

However for each case x^4 + y^4 divided by 5 gives different remainders.

(3/2)^4 + (1/2)^4 = 27/16 + 1/16 = 1.75 (1 int. 3/4)

This divided by 5 gives a remainder of 1.75!! ??

I think this one is an E... I would guess and move on on the test.


On another aspect, I looked at the following property, but haven't been able to figure out how I can use it...

a^n+b^n is not divisible by (a+b) is n is even.
a^n+b^n is never disible by (a-b).

Maybe this can be applied for S1 & S2.
We know x^n + y^n has n = 4 even

Therefore x^n+y^n is not divisible by x+y.

(x^y+y^)/(x+y) not an integer

i.e. (x^n+y^n)/5 x 5/(x+y) not an integer

Given x+y leaves a remainder of 1... don't know where I am going with this.
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 [#permalink] New post 17 Sep 2006, 08:53
What is the reminder when X^4 + Y^4 is divided by 5
A. When X-Y is divided by 5 reminder is 1
B. When X+Y is divided by 5 reminder is 2

X^4 + Y^4 COULD BE REPHRASED TO

(X^2 - XY SQRT 2 + Y^2) (X^2 +XY SQRT 2 + Y^2)

FROM ONE (X-Y)^2 =X^2 -2XY+Y^2
FROM TWO (X+Y)^2=X^2 +2XY+Y^2

MULTIPLY THEM TOGETHER

WE END UP NO WHERE EXACTLY AS I DID

SO JUST GUESS IT IS TIME CONSUMING AND WE VE TO GUESS AND FLY AWAY TRYING TO FORGET THIS BAD MEMORY :lol: :lol:

STILL IF WE PLUGGED ANY TWO NUMBERS IN X^4 + Y^4

A REMAINDER OF TWO WILL ALWAYS REMAIN WN DEVIDING BY 5

I GIVE UP HERE.......Help ..anyone
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 [#permalink] New post 17 Sep 2006, 11:20
What is the reminder when X^4 + Y^4 is divided by 5
A. When X-Y is divided by 5 reminder is 1
B. When X+Y is divided by 5 reminder is 2


after a second thought

from A (X,Y) COULD BE (7,1) , (8,2) , (9,3) , (10,4)

IF WE RAISED ALL OF THOSE BEAUTIFULL COUPLES TO THE POWER OF 4

AND ADDED

(7,1)...... 7^4+1 = 2401+1 =2402 .. REMAINDER 2

(8,2) ..... 8^4 + 2^4 = 4132 REMAINDER IS 2

(10,4)....10000+ 256 = 10256 REMAINDER IS 1

A IS INSUFF

FROM B
USING THE SAME CONCEPT

(1,6) .....1297 REMAINDER IS 2

(8,4) 4352.....REMAINDER IS 2

(10,7) 12401 REMAINDER IS 1

B INSUFF

we could never find intigers that satisfy both statments

i say it is E
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 [#permalink] New post 17 Sep 2006, 15:56
I get (C)

As X and Y are real numbers, I prefer fraction not decimal.

Stat 1:
X-Y remainder 1

Lets try : X=4/3 and Y=1/3 and X=1 and Y=0

The result is that the first case has no remainder and the second has 1 for reminder.

Stat 2:
X+Y remainder 2

Lets try : X=5/3 and Y=1/3 and X=2 and Y=0

The result is that the first case has no remainder and the second has 1 for reminder.

(1) & (2) :
X=4 and Y=3 > X^4+Y^4 = 337
or
X=9 and Y=3 > X^4+Y^4 = 6642

The remainders are equal to 2.
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 [#permalink] New post 17 Sep 2006, 21:12
I think it is C.

from bother, we can get x2 - y2.
and we can get x2 + y2 if you multiply both 1 and 2 and add them together.

then we square both of them again and add them together again. then we can get the term x4 + y4. it will equal to a term.

i will guess it is C. but i won't work it out in the real test.
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Re: this ds killed me [#permalink] New post 18 Sep 2006, 22:18
ashuspark wrote:
What is the reminder when X^4 + Y^4 is divided by 5
A. When X-Y is divided by 5 reminder is 1
B. When X+Y is divided by 5 reminder is 2


rate this also


Hey folks, this is again simple if u know a basic concept of numbers.

So it's time again to concentrate..........don't worry i will not eat too much of ur brain as i have done in my earlier posts.

Let's say that when N is divided by D the remiander is R.
Now when k x N is divided by D the remainder will be k x R ,provided k x R is less than D.

If k x R is > D, it is further divided by D and the remainder is given.

Let me take an example here.

92 when divided by 7 gives remainder 1
Now 2x92 when divided by 7 will give 2x1=2 as remainder .
Now 3x92 when divided by 7 will give 3x1=3 as remainder.

Let's look at one more example.

96 when divided by 7 leaves a remainder 5
Now 2x92 when divided by 7 should give 2x5=10 as remainder.
But, since 10 > 7, we further divide it by 7 and hence the remainder will be 3.

Similarly 3x96 /7 the remainder will be 3x5 =15.
This 15/7 leaves 1 as the remainder.

So i think this funda is clear to all of u.

Now let's take our question.
I think it's clear to everyone that each statement alone is not sufficient.
Now combining both...........

From the first statement X-Y = 5p+1
From the second statement X+Y=5q+2

So 2X = 5(p+q)+3
2Y = 5(q-p)+1

Now each term in the expansion of (2X)^4 will contain 5 other than 3^4
So 16(X)^4 when divided by 5 the remainder will be the remainder obtained when 3^4 is divided by 5 ie 1

Similary 16(Y)^4 when divided by 5 the remainder will be the remainder obtained when 1^4 is divided by 5 ie 1.

So the remainder of 16(X^4) +16(Y^4) when divided by 5 is 1+1 =2
So the remainder of 16 (X^4+Y^4) when divided by 5 is 2.

Now apply the above said rule;

(X^4 +Y^4)/5 the possible remainders are 0, 1, 2, 3, 4
16(X^4+Y^4)/5 the possibel remainders are 16x0,16x1,16x2,16x3,16x4
i.e 0, 1, 2, 3, 4
Since we know that the remainder of 16(X^4+Y^4) is 2,
the corresponding remainder of X^4 + Y^4 ie 2 will be the remainder for (X^4+Y^4)/5

So the answer is C.

Is this clear guys,..........
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Last edited by cicerone on 25 Sep 2008, 00:25, edited 1 time in total.
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 [#permalink] New post 19 Sep 2006, 10:20
You lost me at:
So 16(X)^4 when divided by 5 the remainder will be the remainder obtained when 3^4 is divided by 5 ie 1

Similary 16(Y)^4 when divided by 5 the remainder will be the remainder obtained when 1^4 is divided by 5 ie 1.

What I think you were doing was saying (5(p+q)+3)^4 must end in the same digit as 3^5, but this is not true because 5(p+q) could have a units digit of 5, and so (5(p+q)+3)^4 would have a units digit the same as 8^4.

I think it is E, but I haven't gotten the proof together. If I saw this on the test, I would freak.
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 [#permalink] New post 19 Sep 2006, 11:22
dabots wrote:
You lost me at:
So 16(X)^4 when divided by 5 the remainder will be the remainder obtained when 3^4 is divided by 5 ie 1

Similary 16(Y)^4 when divided by 5 the remainder will be the remainder obtained when 1^4 is divided by 5 ie 1.

What I think you were doing was saying (5(p+q)+3)^4 must end in the same digit as 3^5, but this is not true because 5(p+q) could have a units digit of 5, and so (5(p+q)+3)^4 would have a units digit the same as 8^4.

I think it is E, but I haven't gotten the proof together. If I saw this on the test, I would freak.


Hey dabots, when I expand (5(p+q) +3)^4.
what we get is [(5(p+q)+3)^2]^2
ie [25(p+q)^2+3^2+2x5(p+q)x3]^2
this is again of the form (a+b+c)^2
where a=25(p+q)^2, b= 3^2 and c= 2x5(p+q)x3.
Now a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ac
Clearly a^2 is divisible by 5 exactly
c^2 is divisible by 5 exactly as well as
2ab+2bc+2ac is also divisible by 5 exactly.

Finallly the remainder is nothing but b^2 ie 3^4= 81 when divided by 5 leaves remainder 1.

So the answer must be C.

Try to understand the above concept clearly.
Ofcourse I can say the answer is C by using completely another logic, but i want to answer this using basic fundamentals of numbers......
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 [#permalink] New post 19 Sep 2006, 12:08
OK, cicerone I think you are right.

Please let me know if this is correct:

x-y = 5p + 1
x+y = 5q + 2

2x = 5(p+q) + 3
2y = 5(q-p) + 1

Now, p+q, and q-p, must both be odd, or both be even. So:

Either p+q is even and:

5(p+q) + 3 has a unit digit of 3, and 5(q-p) + 1 has a unit digit of 1

Or p+q is odd and:

5(p+q) + 3 has a unit digit of 8, and 5(q-p) + 1 has a unit digit of 6

We know it cannot be the first case, because those are not divisible by 2. So it must be the second. The units digits of x and y are 8/2 and 6/2, that is 4, and 3.

4^4 has a units digit of 6
3^4 has a units digit of 1

x^4+y^4 therefore has a units digit of 7

and so the remainder for (x^4+y^4)/5 is 2
  [#permalink] 19 Sep 2006, 12:08
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