In the Arithmetic Session 2, while finding the even factors of 4500 we did the following:
First, I did PF
= 2^2 * 5^3 * 3^2
Total of 36 factors
Find # of odd factor
= 3 * 4 = 12
Even factors = 36 - 12 = (24)
2^2 = 2
2^2 * 3^2 = 3 * 3 = 9
2^2 * 5^3 = 3 * 4 = 12
2^2 * 5^3 * 3^2 = 1
This adds up to 24 factors which is correct.
But now in the case of 180
= 3^2 * 2^2 * 5
Total of 18 factors
Odd # of factors
= 3 * 2 = 6
Even
2^2 = 2
2^23^2 = 9
2^25^1 = 6
2^25^13^2 = 1
= 18 factors
Which now does not holdup.
Can anyone explain when this strategy works ? Or can identify mistake in my work
I would just approach it this way: when you find the ODD factors, you know there is not a two (just need to consider the 2^0 case). Similarly, to find the EVEN factors, you know that there is at least one two (just need to consider the cases of 2^1, 2^2, and all cases where the exponent of 2 > 0).
After prime factorization, and ONLY for the 2, instead of adding 1 to the exponent for the multiplication, just use the exponent of 2.
So to find the even factors of 4,500:
2^2 * 3^2 * 5^3 (prime factorization)
2 * 3 * 4 (exponents + 1, except for the two)
= 24 even factors.
To find the even factors of 180:
2^2 * 3^2 * 5^1 (prime factorization)
2 * 3 * 2 (exponents + 1, except for the two)
= 12 even factors
Can you explain how you are adding up the even factors? It is not clear to me what this section means. For each row, I was unable to identify a consistent rule as to what you are multiplying to get 2, 9, 12, and 1 factor:
- 2^2 = 2
- 2^2 * 3^2 = 3 * 3 = 9
- 2^2 * 5^3 = 3 * 4 = 12
- 2^2 * 5^3 * 3^2 = 1
I think this approach is too complex, but if you insist on breaking out into these scenarios, they would add up to 12:
- 2^2 = Exponent of 2 > 0, Exponents of 3 & 5 = 0
2 factors
- 2^2 * 3^2 = Exponents of 2 & 3 > 0, Exponent of 5 = 0
4 factors (2*2)
- 2^2 * 5^3 = Exponents of 2 & 5 > 0, Exponent of 3 = 0
6 factors (2*3)
- 2^2 * 5^3 * 3^2 = Exponents of 2 & 3 & 5 > 0
12 factors (232)