To clarify, are you saying that the have to be composed of three single digit numbers?
Also, if one number has more than one acceptable composition, is it counted once of multiple times towards the 50? For example, 24 could be 2*3*4 or it could be 8*3*1; does that count as 1 or 2 examples?
i have to make a conjecture on if there is more or less than 50 numbers that can be written as product of 3 digit numbers w/o using the number 1 more than once as a factor. Example: 10 can be 1*2*5 and 24 can be 2*3*4
4 answers
yes i think they all have to be single digit numbers. and the second thing is that it counts as two examples
I'm assuming you are not using zero or negative numbers as components.
So then you have to think of all the combinations of 3 of the 9 single number digits, where 1 can only be used once.
For example:
1*2*2
1*2*3
1*2*4
1*2*5
1*2*6
1*2*7
1*2*8
1*2*9
There's 8 possibilities already, and as you can see there are eight more for 1*3*X.
Therefore, it's safe to assume that there are many more than 50 possibilities.
So then you have to think of all the combinations of 3 of the 9 single number digits, where 1 can only be used once.
For example:
1*2*2
1*2*3
1*2*4
1*2*5
1*2*6
1*2*7
1*2*8
1*2*9
There's 8 possibilities already, and as you can see there are eight more for 1*3*X.
Therefore, it's safe to assume that there are many more than 50 possibilities.
It is way more.
Reasoning is as such:
a*b*c
a can be any number from 1-9
b can be any number from 2-9
c can be any number from 2-9
(because 1 can only be used once)
therefore, you are just multiplying the total possibilities available.
9 possible in a
8 possible in b
8 possible in c
9*8*8 = 576
i hope that helps you.
Reasoning is as such:
a*b*c
a can be any number from 1-9
b can be any number from 2-9
c can be any number from 2-9
(because 1 can only be used once)
therefore, you are just multiplying the total possibilities available.
9 possible in a
8 possible in b
8 possible in c
9*8*8 = 576
i hope that helps you.