If k is a positive integer, then 20k is divisible by how man

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.


Back to the original question:

If k is a positive integer, then 20k is divisible by how many different positive integers?

\(20k=2^2*5*k\).

(1) k is prime. If \(k=2\), then \(20k=2^3*5\) --> the # of factors = \((3+1)(1+1)=8\) but if \(k=5\), then \(20k=2^2*5^2\) --> the # of factors = \((2+1)(2+1)=9\). Not sufficient.

(2) k = 7 --> \(20k=2^2*5*7\) --> the # of factors = \((2+1)(1+1)(1+1)=12\). Sufficient.

Answer: B.

Hope it's clear.
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What you are missing in F.S 1, is that we don't know the value of k.

Scenario I: k=2, the total no of factors for 20k = \(2^2*5*2 = 2^3*5 = (3+1)*(1+1) = 8\)

Scenario II: k=3, the total no of factors for 20k = \(2^2*5*3 = (2+1)*(1+1)*(1+1) = 12.\)
Hence, 2 different answers, thus, Insufficient.
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Thanks, the outcome is clear but this method of picking random numbers to test with makes me very uneasy. If you get lucky and the 2 numbers you pick yield different results, then all is fine. But if they yield the same result, you don't know anything. Do you pick a 3rd candidate? A 4th?

Here Statement ! is not sufficient as the prime may be 2 or 5 or any other prime.
But statement 2 is sufficient as we Number of divisors now = 2*3*2= 12
Hence B
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Very simple, no tricks here or anything. If we know the value of K then we obviously know the answer- though in the case of K being a prime number- what if K is 2? Then there is of course some overlap- do not forget 2 is a prime number.

K= 5 x 2 x 2 x (2 x 1)

Thus
B

20k = 2^2 * 5 * k

We already have 2 prime factors of 20k (2 and 5). The total number of factors depends on k.

(1) k is prime

k could be 2 or 5 or any other prime number. In each case, the number of factors would be different.
If k = 2, \(20k = 2^3 * 5\) has 4*2 = 8 factors
If k = 5, \(20k = 2^2 * 5^2\) has 3*3 = 9 factors
If k is any other prime number such as 11, \(20k = 2^2 * 5 * 11\) has 3*2*2 = 12 factors
Not sufficient

(2) k = 7
If k = 7,
20k = 2^2 * 5 * 7
Total number of factors = 3*2*2 = 12
Sufficient

Answer (B)
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It’s easy to think that statement (1) is sufficient by thinking, “Well, I can figure out quickly that 20 has six factors (1, 2, 4, 5, 10, 20), and if k is a new prime number, then 20k will have six MORE factors than 20 has, because we can just take the new prime number and multiply it by each of the original six factors. For instance, if k were 3, then the factors of 20k would be each of the six factors of 20 PLUS six new ones - 3, 6, 12, 15, 30, 60. So statement (1) is sufficient.”

Statement (1) is a trap, however, because we aren’t told that k is a NEW prime number!!! It may be a prime that is ALREADY a factor of 20 – i.e., k could be 2 or 5. Suppose k were 2. If that were the case, then the factors of 20k would be 1, 2, 4, 5, 8, 10, 20 and 40 – i.e., we would be adding only two new factors, 8 and 40, to the original list of six. Statement (1) is therefore insufficient.

So everyone here agrees that the answer is "B." When I answered "B" on the OG online practice test the program said my answer was wrong and the answer is actually "D." Do we think the OG is wrong?

Was the question on the test slightly different? Can you put up a screenshot?
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As soon as I posted this I saw that I read the correct answer wrong...ha! Thank you for the response, though!

If k is a positive integer, then 20k is divisible by how many different positive integers?

(1) k is prime ns as K can be 5 or any other prime no ,ans will be diff for both
(2) k = 7 sufficient

Dang it I missed 2! Arrgghh..

The question stem says that to find the no of DIFFERENT factors.
Statement 1 says k is a prime number.
So, 20k will always yield 4 factors (1,2,3 and k). The powers of each factor may be different for different values of k. But the question asks for the number of different positive factors. According to my reasoning, the question should have clearly asked for the total number of factors (NOT the number of different factors). And that's why I chose option D.

Need some expert opinions on this. cc: Bunuel VeritasKarishma

hopefulmba2021

"No. of different factors" just means "no. of factors". The question says "different" to clarify that when you prime factorise and get
20 = 2^2 * 5, don't consider the two 2s as two factors. The word "different" is added to give extra clarity. Don't get confused. The question is not asking for "prime factors". You need total number of factors only. That depends on value of k.

If k = 2, 20k = 2^3 * 5 so total number of factors = (3+1)(1+1)
If k = 5, 20k = 2^2 * 5^2 so total number of factors = (2+1)(2+1)
If k = some other prime, 20k = 2^2*5*k so total number of factors = (2+1)(1+1)(1+1)
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This is a quite simple one:

What we need to do is to find out how many different factors does 20k have.
Here are my steps:

--We know that 20 has 6 different factors: 1, 2, 4, 5, 10, 20
Among the six factors, 2 and 5 are prime (hope you still remember what a prime number is)

(1)K is prime.
In this case, K only has 2 factors: 1 and itself.
Thus there are two possibilities:
--K = 2 or 5
--K ≠ 2 or 5
If k = 2 or 5, then K itself is one of the 6 factors of 20, thus K doesn't provide a new factor for 20K. (The question is to find how many 'distinct' factors 20K has! )
So it's clear that 20K has only 6 factors(namely the 6 factors of 20)
If K≠ 2 or 5, then K will provide one new factor to 20K (namely itself). So 20K has 6+1=7 factors.

See, (1) is not sufficient because 6 and 7 factors are both possible in this case.

(2) K is 7.
In this case, K is a prime AND K ≠ 2 or 5.
So according to the reasoning in (1), 20K will have 7 factors. SUFFICIENT