Extremely confused about this quant problem

[sayem619]

We see that the odd factors are 3^4, 5^2 and 7^1. I have two issues here:

1. Why is he multiplying 5X3X2 when the question asks “how MANY”? Shouldn’t be 5+3+2?
2. He says 3^0, 5^0, 7^0 should still be counted as odd factors. Why? He gives the example of “if you have 3 jeans, 2 pairs of socks, 3 shirts how many outfits can you make”? Well…why arent you making the 3 jeans into 3^0 jeans then?

[Leaderboard]

@sayem619

• Why is he multiplying 5X3X2 when the question asks “how MANY”? Shouldn’t be 5+3+2?

Nope - see the “Multiplication: The “Choice” Method” video on PrepSwift. The intuition is similar.

@sayem619

1. Well…why arent you making the 3 jeans into 3^0 jeans then?

I don’t understand what you are trying to say there.

[user8522]

@sayem619

1. He says 3^0, 5^0, 7^0 should still be counted as odd factors. Why? He gives the example of “if you have 3 jeans, 2 pairs of socks, 3 shirts how many outfits can you make”? Well…why arent you making the 3 jeans into 3^0 jeans then?

Let’s create an easier example. If you have 6! the odd factors are 5^1, 3^2. If you add them together, the result will be (2+3) - 5 such odd factors. However if you multiply them, the result would be 2*3 = 6.

Now, let’s test it. 6! = 720.

Here are all the factors.
1 | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 12 | 15 | 16 | 18 | 20
24 | 30 | 36 | 40 | 45 | 48 | 60 | 72 | 80 | 90 | 120 | 144
180 | 240 | 360 | 720

altogether 6 odd factors (1, 3, 5, 9, 15, 45)

If you want a more math explanation. For 3^2 you have 3 choices (zero 3s, one 3, two 3s), same goes for 5 (zero 5s, one 5). So like Leaderboard says, it comes down to the choice method.

so for 10!, the logic is the same. 5x3x2 = 30.

Hope that helps!