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# If p and q are two different prime numbers and n is the

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If p and q are two different prime numbers and n is the [#permalink]  12 Apr 2007, 06:18
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If p and q are two different prime numbers and n is the difference between their product and their sum, what is the remainder when n is divided by 4?

(1) Both p and q have a units digit of 1.
(2) p-q= 10
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Statement (1) Both p and q have a units digit of 1.
31, 41, 61, 71 ... So not sufficient

Statement (2) p-q= 10
23-13 = 10
41-31 = 10
71-61 = 10
So not sufficient

edit...
Statements 1 and 2 together
n=31*41-(31+41) = 1199
or
n=61*71-(61+71) = 4199

Reminder is 3 when n is devided by 4 for both the values of n.

So Answer should be C. (Please correct me if I am wrong.)

Last edited by newgmater on 12 Apr 2007, 06:47, edited 2 times in total.
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Hi
Stmnt 1 is not SUFF
N would be an integer that ends with 9. But a n integer is divisible by 4 if last 2 digits are divisible by 4. If N is 9 rem is 1, If N is 19 rem is 3 so Stmnt 1 is NOT suff

sTMNT 2 is SUFF

ANS B
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B

St1: Clearly INSUFF
St2:
If p-q = 10
pq-(p+q) can be written as q^2+8q-10
or (q^2 - 1) + (8q-9)
8q-9 will give a remainder of 3 when divided by 4
now lets see q^2 - 1
q^2 - 1 = (q-1) (q+1)
since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4.
Final remainder = 3: SUFF
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ps_dahiya. I am missing something. Could you please explain how you are writing pq - (p+q) = q^2+8q-10 (provided p-q=10)

Thank you.
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newgmater wrote:
ps_dahiya. I am missing something. Could you please explain how you are writing pq - (p+q) = q^2+8q-10 (provided p-q=10)

Thank you.

n = pq-(p+q)
substitute p= q+10 in above equation and you get q^2+8q-10

hope this helps...
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ps_dahiya wrote:
B

St1: Clearly INSUFF
St2:
If p-q = 10
pq-(p+q) can be written as q^2+8q-10
or (q^2 - 1) + (8q-9)
8q-9 will give a remainder of 3 when divided by 4
now lets see q^2 - 1
q^2 - 1 = (q-1) (q+1)
since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4.
Final remainder = 3: SUFF

Why is (1) clearly insufficient? Shouldn't you know me better by now?
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kevincan wrote:
ps_dahiya wrote:
B

St1: Clearly INSUFF
St2:
If p-q = 10
pq-(p+q) can be written as q^2+8q-10
or (q^2 - 1) + (8q-9)
8q-9 will give a remainder of 3 when divided by 4
now lets see q^2 - 1
q^2 - 1 = (q-1) (q+1)
since q is odd both (q-1) and (q+1) are divisible by 2 hence it is divisible by 4.
Final remainder = 3: SUFF

Why is (1) clearly insufficient? Shouldn't you know me better by now?

Oops...
St1:
p = 10x +1
q = 10y+1
where x and y are positive integers

Now pq-(p+q) can be written as
(10x+1) (10y+1) - 10x-1-10y-1
100xy +10x+10y +1 - 10x-1-10y-1
100xy - 1 so the remainder will always be 3: SUFF

A little out of practice...
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Very nice! Also note that for prime numbers p and q, n=pq-p-q=(p-1)(q-1)-1
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My first post.

Sorry, how did you get the reminder 3?

Thanx
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jaspetrovic wrote:
My first post.

Sorry, how did you get the reminder 3?

Thanx

If something is divisible by 4 and then you subtract 1 from that then we need to subtract 3 more from that to again make it divisible by 4. Hence remainder = 3

Same way if something is divisible by 4 and then you subtract 9 from that then we need to subtract 3 more from that to again make it divisible by 4. Hence remainder = 3
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