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22 Nov 2023, 02:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
45% (02:37) correct
55%
(02:30)
wrong
based on 110
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Find the number of factors of 180 that are in the form (2*k + 2), where k is a positive integer?
A) 6
B) 7
C) 11
D) 12
E) 18
A) 6
B) 7
C) 11
D) 12
E) 18
22 Nov 2023, 08:41
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Bunuel
OA is C
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22 Nov 2023, 21:23
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Q. Number of factors of 180 that are in the form (2*k + 2), where k is a positive integer.
Now, we know that k is positive i.e. k>0 and k is integer.
For any value of k, 2k+2 should be even.
Using the formula for the number of factors using the power, we can write 180 as (2^2)*(3^2)*(5^1) which gives us a total of 18 factors.
Out of these 18, 6 are odd {(3^2) * (5^1)}.
Remember, we need to exclude 2 because k>0.
So, the total becomes 18-6-1=11
IMO, C
Now, we know that k is positive i.e. k>0 and k is integer.
For any value of k, 2k+2 should be even.
Using the formula for the number of factors using the power, we can write 180 as (2^2)*(3^2)*(5^1) which gives us a total of 18 factors.
Out of these 18, 6 are odd {(3^2) * (5^1)}.
Remember, we need to exclude 2 because k>0.
So, the total becomes 18-6-1=11
IMO, C
Bunuel
\(180 = 2^2*3^2*5\)
In a nutshell, question is asking to calculate (Total even factors - 1) = 2*3*2 - 1 = 11
IMO: C
For detailed explanation on the concept tested =>
This problem is testing the prime factors concept. For any number \(N = prime1^x*prime2^y*prime3^z*...\)
Total number of factors would be (x + 1) * (y + 1) * (z + 1) *.....
Generally, each prime with a power n contributes (n + 1) times in the total number of factors, which means that in any one of the resulting factor that prime number would have a power between 0 and n.
Note: + 1 denotes power 0 which means that prime number is not present in the resulting factor
Now an important thing to note here is that when when you have 2 as a prime number, all the factors having atleast 1 power of 2 would be an even number.
ie for \(60 = 2^2*3^1*5^1\)
Refer below prime factorization to understand breakdown of primes in each factor of 60
| Prime Factorization | Factor |
| \(2^0*3^0*5^0\) | 1 |
| \(2^0*3^1*5^0\) | 3 |
| \(2^0*3^0*5^1\) | 5 |
| \(2^0*3^1*5^1\) | 15 |
| \(2^1*3^0*5^0\) | 2 |
| \(2^1*3^1*5^0\) | 6 |
| \(2^1*3^0*5^1\) | 10 |
| \(2^1*3^1*5^1\) | 30 |
| \(2^2*3^0*5^0\) | 4 |
| \(2^2*3^1*5^0\) | 12 |
| \(2^2*3^0*5^1\) | 20 |
| \(2^2*3^1*5^1\) | 60 |
Total factors would be 3*2*2 = 12 and number of even factors would be 2*2*2 = 8
We are removing all factors having \(2^0\) and keeping all the factors having atleast 1 power of 2, which is nothing but instead of multiplying '(n + 1)' in our above formula we multiply 'n' to calculate no. of factors where that prime number is atleast present once.
Now with this knowledge, let's try solving the question
\(180 = 2^2*3^2*5\)
We are told to calculate factors with format 2*k + 2 = 2*(k + 1) ie even factors
Number of even factors = 2*3*2 = 12
Refer below prime factorization of all these even factors of 180 to better understand power dynamics.
| Prime Factorization | Factor |
| \(2^1*3^0*5^0\) | 2 |
| \(2^1*3^1*5^0\) | 6 |
| \(2^1*3^2*5^0\) | 18 |
| \(2^1*3^0*5^1\) | 10 |
| \(2^1*3^1*5^1\) | 30 |
| \(2^1*3^2*5^1\) | 90 |
| \(2^2*3^0*5^0\) | 4 |
| \(2^2*3^1*5^0\) | 12 |
| \(2^2*3^2*5^0\) | 36 |
| \(2^2*3^0*5^1\) | 20 |
| \(2^2*3^1*5^1\) | 60 |
| \(2^2*3^2*5^1\) | 180 |
Now, before we mark the answer choice let's just validate that this is what we are asked or did we miss anything?
Ohh right! k is a positive integer, which means k > 0 ie. 2*(k + 1) > 2 and hence we cannot have 2 as a factor in our final list.
Let's check if we have '2' in the list we calculated... (You can mentally check this, by keeping power of 2 as 1 and power of rest of the primes as 0 in the initial prime factorization. I have added list of factors just for reference here, and you shouldn't calculate this full list in your test as that would be too time consuming)
\(2^1*3^0*5^0 = 2\) => Indeed we have, so we need to remove one factor from our total
12 - 1 = 11
Answer: C
