stonecold wrote:

if k is a positive Integer,how many unique Prime Factors Does 14k have ?

(1) k^4 is divisible by 100

(2)50*k has 2 Prime Factors

Give me an algebraic solution. And i will tell you where you went wrong

(1) \(k^4\) is divisible by \(100=10^2\), we have \(k^4=10^2 k'\).

Since the left side is a double square of \(k\) and the right side contains only a square of 10, we must have \(k'=10^2 \times k"^4\).

So \(k^4=10^4 \times k"^4 \implies k=10k"\) or \(k\) is divisible by 10.

This means that \(k\) has

at least 2 prime factors 2 and 5, so \(14k=2 \times 7 \times k\) has

at least 3 prime factors 2, 5, and 7.

However, we cannot know the extract number of prime factors that \(14k\) has. Insufficient.

(2) \(50k=2 \times 5^2 \times k\). Since \(50=2 \times 5^2\) has 2 prime factors 2 and 5 and \(50k\) has

only 2 prime factors, it is deduced that \(k\) has

at most 2 prime factors 2 and 5.

If \(k\) has only one prime factors 2, \(14k\) has 2 prime factors 2 and 7.

If \(k\) has two one prime factors 2 and 5, \(14k\) has 3 prime factors 2, 5 and 7.

Insufficient.

Combine (1) & (2):

(1): \(k\) has at least 2 prime factors 2 and 5

(2): \(k\) has at most 2 prime factors 2 and 5

(1)+(2): \(k\) does have 2 prime factors 2 and 5.

So \(14k\) has 3 prime factors 2, 5, and 7. Sufficient.

The answer is C

ccooley wrote:

It depends on whether you're referring to prime factors or unique prime factors. I assume from the wording of statement 2 that it's the latter, but on the GMAT they'd have to be more specific.

Please correct me, but I see no diffirence in these two sentences:

(1) 14k has 3 prime factors.

(2) 14k has 3 unique prime factors.

I think i can answer that.

n=2*3^2 => two unique prime factors and 3 Total prime factors.

They refer to total Prime factors as "length" where we just have to add the exponents.

It just Creates confusion if you ask me since in GMAT/general mathematics we always write prime factors as to unique prime factors.