factors

[vaivish]

29. How many positive odd factors does 450 have?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[gowani]

break down 450 to prime factors...(5^2)(3^2)(2) then look at the powers and 1 to each and then sum it up

3 + 3 + 2 = 8

i'm saying C

[vaivish]

its D...i dont know how..

[gowani]

ok so i think i messed up on my math

420
/ \
20 21
/ \ / \
4 5 7 3
/ \
2 2

(2^2)(5^1)(7^1)(3^1)

2+1 + 1+1 + 1+1 + 1+1 = 9

[Ian Stewart]

break down 450 to prime factors...(5^2)(3^2)(2) then look at the powers and 1 to each and then sum it up

3 + 3 + 2 = 8

i'm saying C

The method starts off well, but there are a few mistakes at the end. If you want to work out how many positive divisors a number has:

-prime factorize
-look only at the powers
-add one to each power
-multiply what you get (don't add!)

So,
-450 = (2^1)*(3^2)*(5^2)
-the powers are 1, 2 and 2
-add one to each: 2, 3 and 3
-multiply: 2*3*3 = 18

450 has 18 different positive divisors, including 1 and itself.

Why does this work? Because any number that looks like (2^a)*(3^b)*(5^b) is a divisor of 450 = (2^1)*(3^2)*(5^2) as long as:

a = 0 or 1 (two choices)
b = 0, 1 or 2 (three choices)
c = 0, 1 or 2 (three choices)

and we multiply just as we would in any mathematical counting problem to work out the total number of choices for the exponents a, b and c.

Now, after all that, 18 is not the answer to the posted question. The question asks only for odd divisors. If a divisor is to be odd, it must not have a 2 in its prime factorization. Thus, the exponent on 2 must be 0:

a = 0 (one choice)
b = 0, 1 or 2 (three choices)
c = 0, 1 or 2 (three choices)

1*3*3 = 9 odd divisors.

[egybs]

The number is 450, not 420...

Prime factors of 450 are:

2,3,3,5,5

Remember that it's asking for all the ODD factors... So we know that 1 will be a factor... but 450 will not.

Now 2 won't be able to play any role, since we're looking for odds... so we can only multiply the odd numbers together.

so we have 3,3,5,5
How many different ways can we combine these?
4C1 = 4
4C2 = 6
4C3 = 4
4C4 = 1

In principle, we'd just add these all up together, but we need to remove a few possiblities:

For example, in 4C1, we could have 3,3,5,5... but there are two 3s and 2 5s... So we should subtract 2 possibilities.

In 4C2, we can get 15, 3 different ways... So subtract 3.
In 4C3, we can get 3*5*5 two different ways, and 5*5*3 two different ways, so subtract two.


Now let's sum them up:

4C1 -2 = 2
4C2 -3 = 3
4C3 -2 = 2
4C4 = 1


2+3+2+1 = 8

BUT, don't forget the factor 1... So we have 9 total.

D.

ok so i think i messed up on my math

420
/ \
20 21
/ \ / \
4 5 7 3
/ \
2 2

(2^2)(5^1)(7^1)(3^1)

2+1 + 1+1 + 1+1 + 1+1 = 9

[chidcguy]

The problem counts on us to forget 1. I left it out.