how many factors of 1000 can be divided by 20 without a remainder.
all u do is 20 goes into 1000 50 times so ur answers 50 rite? and another questions im doing a couple of different questions do i need to put them all on the same post xD
how many factors of 1000 can be divided by 20 without a remainder.
all u do is 20 goes into 1000 50 times so ur answers 50 rite? and another questions im doing a couple of different questions do i need to put them all on the same post xD
i agree.
taking mathstud's lead, i want to flesh this out a little more. i also found the prime factorization of 1000, then i factored 20 out of it. we want all possible multiples of 20 from this factorization and of course, 20 itself.
so $\displaystyle 1000 = 2 \cdot 5^2 \cdot 20$
now how many different combinations can you have in front of the 20? and of course, $\displaystyle 1 \cdot 20$ is an additional one
great minds think alike
huh? i don't get what you are saying
the point is, to be divisible by 20, you must be an integer times 20, thus, we only care about the factors that are a multiples of 20
now that i think about it, should we not consider negative factors as well?
xsriel, if you don't understand that, then look at it this way. You should know how to list the factors of 1000. They are 1 2 4 5 8 10 20 25 40 50 100 125 200 250 500 1000 . Which ones are divisible by 20 with no remainder? 20, 40, 100, 200, 500, 1000. That makes six. Maybe that will make it more clear to you.