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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Tue Jan 26, 2010 8:31 pm

what is remainder when p^2-n^2 is divided by 15

1, remainder when p+n is divided by 5 is 1

2, remainder when p-n is divided by 3 is 1

I do not know how to solve remainder problem. pls, tell me your experience

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Source: Beat The GMAT — Data Sufficiency | Tue Jan 26, 2010 8:31 pm

thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Tue Jan 26, 2010 10:00 pm

IMO C

i just tried by plugging sme selected no's( and a bit of assumptions)

to meet s1) we need p+n = 5k+1(where k is int)
there can be n no. of combination hence
s1) alone insuff

to meet s2) we need p-n=3K+1

same as s1) insuff...

combining
u will come to knw that no,s like p=5 and n=1; p=20 and n=1are satisfying bth the statements
and remn is 9 for bth the cases

now time to assume that it holds true for all such no's

hence C

sorry for such a vague reply
but on exam such tricks will play a role of crocin to keep your headache away

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Tue Jan 26, 2010 10:28 pm

thephoenix wrote:IMO C

i just tried by plugging sme selected no's( and a bit of assumptions)

to meet s1) we need p+n = 5k+1(where k is int)
there can be n no. of combination hence
s1) alone insuff

to meet s2) we need p-n=3K+1

same as s1) insuff...

combining
u will come to knw that no,s like p=5 and n=1; p=20 and n=1are satisfying bth the statements
and remn is 9 for bth the cases

now time to assume that it holds true for all such no's

hence C

sorry for such a vague reply
but on exam such tricks will play a role of crocin to keep your headache away

What if p+n = 11
p-n = 7

ie p=9 and n =2

p^2-n2^2 = 77

and the remainder when divided by 15 is 2

So it is not universal

I think we have just about enough evidence to go with E

Always borrow money from a pessimist, he doesn't expect to be paid back.

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dmitriyaleyev: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Thu May 21, 2009 5:53 pm; Thanked: 1 times

by dmitriyaleyev » Wed Jan 27, 2010 8:11 am

in a nutshell,
once u realize neither a nor b is a right answer,
try to expand the equasion as ((p - n)(p+n))/3*5 >>>>>>>>>>>>> then plug (6+5)/5 r 1 and (6-5)/3 r 1, then, plug (4+2)/5 r 1 and (4-2) r 2.
Since there is no consistency - E

Please let me know if there is something faulty in this approach. Thanks

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Thu Jan 28, 2010 1:08 am

dmitriyaleyev wrote:in a nutshell,
once u realize neither a nor b is a right answer,
try to expand the equasion as ((p - n)(p+n))/3*5 >>>>>>>>>>>>> then plug (6+5)/5 r 1 and (6-5)/3 r 1, then, plug (4+2)/5 r 1 and (4-2) r 2.
Since there is no consistency - E

Please let me know if there is something faulty in this approach. Thanks

Here plugging in numbers will serve the purpose.

Case 1 :

(6+5)/5 r 1 and (6-5)/3 r 1

Case 2:

(11+5)/5: remainder 1

(11-5)/3: remainder:0

There are inconsistencies in the reaminder values for different p & n values. So Neither of the given conditions an help us to figure out the solution of the problem. Hence E

R&D on GMAT algorithm:

https://www.beatthegmat.com/r-d-on-gmat- ... tml#305739

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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Thu Jan 28, 2010 2:56 am

hard yes no question require algebra change before pluging number''

for 1) and 2), plugging number because it is easy to do so

now consider both 1 and 2

make algebra change before plugging number

(5k+1)(3m+1)= 15km+5k+3m+1

now we pluging number to see that 5k+3m+1 =remainder is not fixed

E is correct

very hard, I can not do it in 2 minutes

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