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little confused

by stufigol » Thu Feb 04, 2010 9:20 pm
From OG
If n=4p, where p is a prime number greater than 2, how many different positive divisors does n have, including n?
A/2
B/3
C/4
D/6
E/8
I UNDERSTAND THE ANSWER IS 2,4,2p and 4p, but it says including n so it might be 5 different positive divisors not only4.
correct me if i am wrong which is mostly possible.
thank you
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by Osirus@VeritasPrep » Thu Feb 04, 2010 9:26 pm
n = 4p so by counting 4p and n you would be counting the same number twice.

The factor box would look like this

n
|------------------------------
| 2, 2, p
|
|

The different factors are 2, 4, 2p, and 4p.
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by money9111 » Thu Feb 04, 2010 10:10 pm
i liked this one hehe
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by thephoenix » Thu Feb 04, 2010 10:18 pm
stufigol wrote:From OG
If n=4p, where p is a prime number greater than 2, how many different positive divisors does n have, including n?
A/2
B/3
C/4
D/6
E/8
I UNDERSTAND THE ANSWER IS 2,4,2p and 4p, but it says including n so it might be 5 different positive divisors not only4.
correct me if i am wrong which is mostly possible.
thank you
divisors will be
2,4,p,n,2p

makes it 5
but i am surprised y 1 is not counted here as divisors(which will make it 6 diff divisor)
on the D day i wud have marked D as ans
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by ajith » Thu Feb 04, 2010 11:29 pm
stufigol wrote:From OG
If n=4p, where p is a prime number greater than 2, how many different positive divisors does n have, including n?
A/2
B/3
C/4
D/6
E/8
I UNDERSTAND THE ANSWER IS 2,4,2p and 4p, but it says including n so it might be 5 different positive divisors not only4.
correct me if i am wrong which is mostly possible.
thank you
n= 2^2*p^1

total num of factors = (2+1)(1+1) =6

[If a number can be expressed as n = a^x*b^y*c^z.....

where a,b,c... are prime and x,y,z are integers, number of factors n has is, (x+1)(y+1)(z+1).....]
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