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lnaik
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GMATPrep 2 05 Oct 2007, 06:41
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Please solve this and explain.



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GMATPrep 2 05 Oct 2007, 06:54
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I got this one last night too. Pick any prime number. Start with 2 to make it easy. 1 is less than 2 and has no other factor in common with 2 than 1. Pick another prime, just to be sure - let's say 5. Positive integers less than 5 are 4 3 2 1. Luckily for us, the very first answer choice is the correct one - just plug in the prime values you chose. If p is 2, then p-1 is 1, which is in fact the number of positive integers less than 2 that has no other factor in common with 2 than 1. If p is 5, p-1 is 4, and once again, that's the correct # of positive integers less than that prime and having no other factor in common with it than 1.

A it is.
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StartupAddict
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GMATPrep 2 05 Oct 2007, 17:31
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This one is easy.

If p is prime, by definition the only numbers that divide it are 1 and itself. Hence all numbers before p are coprime with p, of which there are p-1 of them.

P-1 or A
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GMATBLACKBELT
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GMATPrep 2 06 Oct 2007, 08:59
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lnaik
Please solve this and explain.
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This one is actually pretty easy once you get through the convoluted wording of the problem.

Took me bout 1:30min just to read it, but only but 10sec to answer it.

f(n)= all the numbers that are only in common w/ n b/c of 1 and they are less than n. Thus they are not factors of n.


p is prime. f(p) then are all the numbers below p and only in share 1 as a factor.

Try out a prime. (also note this f(n) or f(p) applys to ONLY positive integers)

Lets try 11. Every number before 11 is only has factor 1 in common w/ 11.

Try 13. Same thing every number b/f 13 has only 1 as a factor in common w/ 13.

So A fits this criteria. 13-1 =12 #'s in common w/ 13 only b/c of 1.

11-1= 10. Same thing.


A it is.



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