If p and n are positive integers

[ES354]

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

1) The remainder when p + n is divided by 5 is 1.
2) The remainder when p - n is dividided by 3 is 1.

Answer is E.

Here's what I did:
Took the original statement, used the difference of squares formula to change it to: what is the remainder when (p + n) (p - n) / (5*3)?

1) tells us that p + n divided by 5 is 1, but tells us nothing about the relationship of p - n and 3
2) tells us the relationship of p - n and 3, but nothing about p + n and 5
Together: they do give us relationships of 2 parts of the division, but i'm getting stuck on how to mix them all together to determine that when combined they are still not sufficient. I realized i was getting close to 2 minutes, guessed C over E, and moved on.

[Binit]

Hi Eric,
I too approached like u but while thinking about 1 & 2 together: p+n = 6, 11, 16, 21 etc. & p-n = 4, 7, 10, 13 etc.
Now we can assume paired values for p+n and p-n where both should be odd or both even (since their sum has to be even: (p+n)+(p-n)=2p,even)
(6,4) -> remainder= 9;
(11,7)-> r=2;
(6,10) -> r=0. Not sufficient. (E).

[RonPurewal]

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

1) The remainder when p + n is divided by 5 is 1.
2) The remainder when p - n is dividided by 3 is 1.

Answer is E.

Here's what I did:
Took the original statement, used the difference of squares formula to change it to: what is the remainder when (p + n) (p - n) / (5*3)?

1) tells us that p + n divided by 5 is 1, but tells us nothing about the relationship of p - n and 3
2) tells us the relationship of p - n and 3, but nothing about p + n and 5
Together: they do give us relationships of 2 parts of the division, but i'm getting stuck on how to mix them all together to determine that when combined they are still not sufficient. I realized i was getting close to 2 minutes, guessed C over E, and moved on.

as soon as you get to the blue thing--literally, the very second you get to the blue thing--IMMEDIATELY abandon the theoretical approach, and just start bombarding the problem with specific numbers (i.e., testing cases).

you'll probably have to make a few guesses to come up with the right numbers, but you'll eventually come up with them.

do not have patience!
do not persevere!
if something is not working, bail out and try something else.

[slaughterGMAT]

(y+z) = (u+v)^2 --- (1)
(z-x)/y = y/(z+x) => z^2-x^2 =y^2 =>
(z-y)(z+y)= 121 ( since x=11)
so now either 121 can be factored like 121*1 or 11*11
Case1: z+y =121 and z-y = 1 => z=61 and y =60.
Case2: z+y = 11 and z-y = 11 => z=11 and y =0.

As per the info given in a) - if y =60 then we can know that z =61.
As per info given in b) u=6 =>x = u^2 - v^2 => v = +/-(5) which gives us the value of z.

So both conditions independently are sufficient.

[tim]

I think you may have posted in the wrong thread. What you've done here doesn't seem to have any relevance to the problem being discussed.

[RonPurewal]

(y+z) = (u+v)^2 --- (1)
(z-x)/y = y/(z+x) => z^2-x^2 =y^2 =>
(z-y)(z+y)= 121 ( since x=11)
so now either 121 can be factored like 121*1 or 11*11
Case1: z+y =121 and z-y = 1 => z=61 and y =60.
Case2: z+y = 11 and z-y = 11 => z=11 and y =0.

As per the info given in a) - if y =60 then we can know that z =61.
As per info given in b) u=6 =>x = u^2 - v^2 => v = +/-(5) which gives us the value of z.

So both conditions independently are sufficient.

i think you meant to post this here:
https://www.manhattanprep.com/gmat/foru ... 12316.html

please re-post (in the correct place). thanks.