og math 249

Enginpasa1 wrote:if n is a positive integer less than 200 and 14n/60 is an integer, then n has how many different positive prime factors.

a. 2
b.3
c.5
d.6
e.8



this question is actually very easy and belongs as one of the first questions. However, the trick to it is that 14n/60=7n/30. this makes n as 30. However, how in the world could you see then original fraction can be split like that. its this type of wit that gets you through the exam and I need more of that type of mathematical vision. any help?

The easiest way to do most of these wacky division questions is by using prime factors.

Let's break the expression down to primes:

14n/60

2*7*n/2*2*3*5

Since we know that the expression is an integer, we know that every prime factor on the bottom has a matching prime factor on the top.

One of the 2s from the denominator can come from the 2 in the numerator. However, the 2*3*5 remaining all have to come from n.

Therefore, we know that n is a multiple of 2*3*5 = 30 and that n has at least 3 prime factors.

Strategically, we already know that the answer has to be 3, because even if we're not sure why n can't have any other primes, it doesn't have to have any others.

For a more complete solution, we can show why n can't have any other different prime factors: we know that n < 200 and that 2, 3 and 5 are already factors of n. Therefore, the next candidate is 7. However, if 7 were a prime factor of n then the minimum value for n would be 30*7 = 210, which violates the condition that n<200.

I am learning from this conversation that the best way to break things down to bear minimum is prime facotorization.