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GMATPrep - Q3 - Function prime

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GMATPrep - Q3 - Function prime

by California4jx » Tue Aug 26, 2008 2:49 pm
I need a better explanation. I got this right intuitively but want to ask if someone has better explanation. Someone tried to explain this in past threads but it was not very clear .... thanks.
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by nikhilagrawal » Wed Aug 27, 2008 2:46 am
no clue :(
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by amitansu » Wed Aug 27, 2008 3:07 am
The q is asking the no. of positive numbers those satisfy the given function.

I will take 5 as an example prime no., then according to choices given :

Is the total no. of positive integers going to be p-1=5-1=4
p-2=5-2=3; (p+1)/2=(5+1)/2=3 (incidentally same as earlier) or (p-1)/2=(5-1)/2=2 or 2 as the final choice.

According to q : if p(5) then, all no. which are satisfying the function would be 1,2,3,4.Because they are less than 5 and each of these has no positive factor/s in common with 5 except 1.

So p-1 satisfies here.

Take 7 or 11 (as prime nos.)and so on.. all of them satisfy the function because the total no. of the function would be one less than the prime no. itself to satisfy the function.


Amit
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by nervesofsteel » Wed Aug 27, 2008 1:31 pm
A is the answer

let us take a p as a prime number 7...
acc to question f(p) = total number of numbers less than 7 and don't have a common factor with 7 except 1

Every prime number has two factors 1 and the number itself...

So every number less tha a prime number will not have any factor common with prime number..

thus the solution is 1,2,3,4,5,6... = 7-1 = 6 digits.. or we can say p-1 numbers..

Hope it helps...
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