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Bunuel
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Find the number of factors of 180 that are in the form (2*k + 2), wher 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% (02:35) correct 55% (02:25) wrong based on 106 sessions

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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

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C
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Find the number of factors of 180 that are in the form (2*k + 2), wher 22 Nov 2023, 08:41
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Bunuel
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

Show more

OA is C

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Find the number of factors of 180 that are in the form (2*k + 2), wher 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
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Find the number of factors of 180 that are in the form (2*k + 2), wher Updated on: 28 Dec 2024, 02:24
Originally posted by hr1212 on 26 Dec 2024, 07:56.
Last edited by hr1212 on 28 Dec 2024, 02:24, edited 7 times in total.
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Bunuel
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

Show more
\(180 = 2^2*3^2*5\)

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
Show Spoiler

Prime FactorizationFactor
\(2^0*3^0*5^0\) 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.
Show Spoiler

Prime FactorizationFactor
\(2^1*3^0*5^0\) 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
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Find the number of factors of 180 that are in the form (2*k + 2), wher 27 Dec 2024, 15:38
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hr1212
Bunuel
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

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, where +1 denotes power 0 which means that prime 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 \(60 = 2^2*3^1*5^1\)

Total factors would be 3*2*2 = 12 and number of even factors would be 2*2*2 = 8 ie. we need to remove +1 from the power multiplier of 2, as we need atleast one power of 2 for the number to be even.

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

Now, before we mark the answer choice let's just validate that this is the only think asked or did we miss anything?
Ohh right! k is a positive integer, which means k != 0, so we cannot include 2 as our factor in our final list.

Let's check if we have a factor of 2?

\(2^1*3^0*5^0\) => Indeed we have, so we need to remove one factor from our total

12 - 1 = 11

Answer: C
Show more

the removal of factors part is confusing
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Find the number of factors of 180 that are in the form (2*k + 2), wher 28 Dec 2024, 01:51
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hr1212
Bunuel
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

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, where +1 denotes power 0 which means that prime 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 \(60 = 2^2*3^1*5^1\)

Total factors would be 3*2*2 = 12 and number of even factors would be 2*2*2 = 8 ie. we need to remove +1 from the power multiplier of 2, as we need atleast one power of 2 for the number to be even.

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

Now, before we mark the answer choice let's just validate that this is the only think asked or did we miss anything?
Ohh right! k is a positive integer, which means k != 0, so we cannot include 2 as our factor in our final list.

Let's check if we have a factor of 2?

\(2^1*3^0*5^0\) => Indeed we have, so we need to remove one factor from our total

12 - 1 = 11

Answer: C

the removal of factors part is confusing
Show more
Have added some more details. Hope that clarifies the confusion.
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