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Remainders

Expert replies
by Stockmoose16 » Wed Nov 12, 2008 8:54 pm
If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+q is divided by 5 is 1
The remainder when p-q is divided by 3 is 1

Anyone know how to solve this problem? I'm trying to use the remainders formula: N=QD+R

It mentions N in the question stem, then never mentions it again, so I'm confused.
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Source: — Data Sufficiency |

Re: Remainders

by logitech » Wed Nov 12, 2008 9:29 pm
SM16, where is this question from bro ?
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by vishubn » Wed Nov 12, 2008 9:31 pm
WOWW!!
No wayss @! this king of all typos!! no way one cal solve this

Vishu
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
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by logitech » Wed Nov 12, 2008 9:35 pm
vishubn wrote:WOWW!!
No wayss @! this king of all typos!! no way one cal solve this

Vishu
okay lets make our own problem then, what the hell!

If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+n is divided by 5 is 1
The remainder when p-n is divided by 3 is 1


alright lets go!
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
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by cramya » Wed Nov 12, 2008 9:39 pm
It mentions N in the question stem, then never mentions it again, so I'm confused
Either all n's needs to be replaced by q or vice versa.

I picked numbers and still got E)

Whats the OA?

Stmt I

p=4 q=2
Remainder for p^2-q^2/15 will be 12

p=10 q=6
Remainder for p^2-q^2/ 15 will be 4

INSUFF

Stmt II

p=8 q=4

Remainder for p^2-q^2/ 15 will be 3

p=11 q=4

Remainder for p^2-q^2 / 15 will be 0

INSUFF

Stmt I amd I together

lcm of 5 and 3 is 15 so pick 16 and 31 (1 more than the multiples of their lcm)

p=10 q=6

Remainder for p^2-q^2 / 15 will be 4

p=19 q=12

Remainder for p^2-q^2 / 15 will be 7

E)
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by logitech » Wed Nov 12, 2008 9:49 pm
logitech wrote:
vishubn wrote:WOWW!!
No wayss @! this king of all typos!! no way one cal solve this

Vishu
okay lets make our own problem then, what the hell!

If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+n is divided by 5 is 1
The remainder when p-n is divided by 3 is 1


alright lets go!
p^2-n^2 = (p+n) (p-n)

So we are trying to find the remainder when it is divided by 15

1 and 2 insuf they talk either p+n or p-n

1+2)

p+n = 5x+1
p-n= 3y+1

(5x+1)*(3y+1)/3*5

(x,y) = 0,0 --> R=1
(x,y) = 1,0 --> R=6

Hence, E
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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by vishubn » Wed Nov 12, 2008 10:34 pm
A) so adding up p+q is always one numebr greatre than multiple of 5

now p=11, q=5

and stem /15 give remainder of 1

again p=16 q=5 gives u the diffe remainder
Inssuff

B)p=12,n=3

Isufficent Again

A and C togetrjer also Insufficent

OA E
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
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by stop@800 » Thu Nov 13, 2008 9:52 am
IMO E

solution same as done by logitech

(5x+1)(3y+1) / 15

(15xy+5x+3y+1)/15

remainder when 5x+3y+1 is divided 15 can not be determined
so E
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