How many of the factors of 72 are divisible by 2?

How many positive factors of 72 are divisible by 2?

A. 4
B. 5
C. 6
D. 8
E. 9


Finding the Number of Factors of an Integer

First, make the 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 respective 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: \(450 = 2^1 * 3^2 * 5^2\)

The 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:

Using the method above, since \(72=2^3*3^2\), the total number of factors of 72 is \((3+1)(2+1) = 12\). Of these, only three are odd: 1, 3, and 9. This is because the only factors of 72 that are odd arise from powers of 3. The number of odd factors for \(3^2\) is \((2 + 1) = 3\), specifically - 1, 3, and 9. Thus, the other \(12-3=9\) factors are even.

OR: Since \(72=2^3*3^2\), even factors MUST have 2 raised to either the power of 1, 2, or 3, providing 3 options, and 3 raised to the power of 0, 1, or 2, again giving 3 options. This results in a total of \(3*3=9\) even factors.


Answer: E

Hope it helps.
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How many of the factors of 72 are divisible by 2?
A. 4
B. 5
C. 6
D. 8
E. 9

I got it right, but I would like to know if my method is efficient.

72 = 2*2*2*3*3
therefore, different numbers that can be found from the above = 5!/3!*2! = 10
Out of these 10, only one number (3*3) is odd....hence, the answer is 10-1 = 9.

you were lucky there I'm afraid the method is incorrect.

Sol:

The number of factors of 72 will be 12 and not 10. The best way to find out is if X= a^b * c^d then number of factors are (b+1) * (d+1)

here 72= 2^3*3^2 so number of factors will be (3+1) * (2+1) = 12 --> this includes 1 and the number itself.

so for finding factors not divisible by 2, remove all the 2s from the prime factorization. You will be left with 3^2.

so factors will be 3 and 3^2(=9). Also, we need to include the number 1 to this list, as it is odd and not divisible by 2.

so 12-3=9

the flaw in your approach:
1.) 5!/2!*3! will give you the number of ways you can arrange (permute) 22233. essentially it gives you the following list:
22233,22323,33222,32322 etc. As you can see this is a mere representation of how you can write three 2s and two 3s. This does NOT give you the list of factors of 72.
2.) not only 3*3 is odd but 3 and 3*3 both are odd factors. Include 1 to this list and you get the 3 odd factors.


hope this helps.

I'm with the poster above - think you were lucky on this one.

A quick check (good thing listing factors of 72 doesn't take long):
Factors - (12) 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Factors NOT divisible by 2 - (3) 1, 3, 9

Therefore answer is 12 - 3 = 9.

BTW - jumsumtak actually shows you how to find the number of factors for a number, and will be a time saver in the exam. Note the example here works with only two prime factors. For another example with three prime factors: "How many factors in 360?", 360 = 5 x 8 x 9 --> (1+1) x (3+1) x (2+1) = 24.

That is correct. This works with every number not with just 2 prime factors. you can have 'n' PRIME factors and it will still hold true.

360 = 5 x 8 x 9 = 5 x 2^3 x 3^2. So the factors will be (1+1) x (3+1) x ( 2+1) = 2 x 4 x 3= 24

number of factors and prime factors is fine.... but out of those number of factors... how did you conclude on as only 3 being odd?

(Will try to explain this using an easier but slightly slower approach since many have difficulties grasping perms & combs)

Using the number 72 from the original question;
[1] Find number of factors:
1. 72 = 2^3 x 3^2;
2. therefore number of factors = (3+1) x (2+1) = 12

[2] Find number of ODD factors:
*Number property, N1: We know that all primes, except 2, are odd.
*Number property, N2: We know that ODD x ODD = ODD.
*Number property, N3: Multiplying any number by 2 (an Even Number) will yield an EVEN number.

1. Recalling from [1], we have identified 2 and 3 as the prime factors of 72.
2. We ignore the "2" remembering N3.
3. We can construct factors that consist of prime factor 3 only:3, 3 x 3 (since there are only two "3"s we stop here).
4. Let's not forget that "1" is also a non-even factor.
5. Total of ODD factors is 3.

[3] Find number of EVEN factors: Total of factors - total of odd factors = total of even factors = 12 - 3 = 9.


One could directly use combinations of 2, 2, 2, 3, 3 to list all EVEN factors but I've found it faster to find ODD factors first.

For example, in my explanation for counting factors for a number with three distinct primes:
1. 360 = 5 x 8 x 9 = 5^1 x 2^3 x 3^2
2. Number of primes = (1+1) x (3+1) x (2+1) = 24
3. Number of odd factors will be multiples of only up to 1 "5" and 2 "3"s.
4. List odd factors:
3
5
9 = 3 x 3
15 = 3 x 5
45 = 3 x 3 x 5
5. Do not forget that "1" is also a factor, therefore there are 6 ODD factors in 360.
6. Total of EVEN factors in 360 is 24 - 6 = 18.

*Quick Check with pairs indeed reveals 6 ODD factors:
1, 360 --> 1 is ODD
2, 180
3, 120 --> 3 is ODD
4, 90
5, 72 --> 5 is ODD
6, 60
8, 45 --> 45 is ODD
9, 40 --> 9 is ODD
10, 36
12, 30
15, 24 --> 15 is ODD
18, 20


Hope this clarifies things.
* Do note that I would expect 750+ questions to involve combinatorics that involve the use of perms & combs to solve within the time limit.

Bunuel,

how do we know the red part without writing all the factors of 72?

It's not hard to find the number of odd factors of 72 manually but if you want more systematic approach, refer to the red part above or consider the following:

Get rid of all the 2’s which give even factors in 72, so divide 72 by 2^3=8: 72/2^3=9=3^2. Now, 9 will have all the odd factors of 72 and won’t have its even factors. The number of factors of 9 is (2+1)=3.

So, we know that 72 has total of 12 factors out of which 3 are odd. Therefore 72 has 12-3=9 even factors.

Hope it's clear.
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Yes , it clear... thanks

Total no. of factors= (a+1).(b+1). and so on
These factors will include all the even factors, odd factors including 1, and the number itself.

Easier method can be: Total no. of factors-odd factors
72= 2^3. 3^2
Here a=3 and b=2

Therefore total no. of factors= (3+1).(2+1)=12
Total no. of odd factors (easy!! don't count power of 2)= 3

Thus the total no. of even factors are 12-3= 9

Bunuel
In the highlighted part why cant we have 2 in the power of 0 as we have 3 in the power of 0?

We are counting EVEN factors, so factors which have 2 in them. If 2 were in power of 0, the factor won't be even anymore (2^0 = 1).
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ANOTHER METHOD:-

72 can be written as 2*36

To find the number of factors of 36 is a straightforward application of number of factors formula:-
(p+1)(q+1)(r+1)... [where p,q,r are exponents of each prime factor]

Therefore 36 can be written as \(36=2^2*3^2\)
Therefore the number of factors of 36 are (2+1)(2+1) = 9 factors.

Therefore the no. of factors of 72 which are divisible by 2 are 9 factors.

Ans: Option E is correct!!
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72= 2^3 * 3^2
Total# Factors: 4*3
Total# Factors where power of 2 is atleast one: 3*3= 9

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