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08 Sep 2010, 22:58
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (01:18) correct
40%
(01:20)
wrong
based on 1966
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive factors of 72 are divisible by 2?
A. 4
B. 5
C. 6
D. 8
E. 9
A. 4
B. 5
C. 6
D. 8
E. 9
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Official Solution:
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
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
24 Nov 2012, 21:20
Kudos
Bookmarks
Amateur
(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.
I'm afraid the method is incorrect. 
