How Many 5-Digit Numbers from 4, 5, 6, 8, 9 are Divisible by 8?

[Matejxx1]

Homework Statement

From the numbers 4,5,6,8,9 we make 5 digits numbers (each number can be used only once).
h)How many of these numbers are divisible by 8?
The correct answer is 20

Homework Equations

a number is divisible by 8 if the last 3 digits are divisible by 8

If the hundreds digit is even, examine the number formed by the last two digits.
If the hundreds digit is odd, examine the number obtained by the last two digits plus 4

The Attempt at a Solution

Ok, so I started by trying to figure out how many number we could make that are divisible by 8
I divided the problem into 2 parts. First I calculated the possibilities if the hundredth digit is a odd number and then I did the same thing for the even number
a) since we have 3 even number there are 3 possibilities for the hundredth digit
and we also know that the other 2 number have to be divisible by 8 so that left me with 4 options (48,56,64,96)
so when I combined the hundredth number and these for options we get 3*4=12 options in total
b) we have 2 odd numbers so there are 2 possibilities for the hundredth digit
and we also know that the other 2 number + 4 have to be divisible by 8 so I got 2 options (68,84)
so when I combined the whole thing I got 2*2= 4 options in total

and this is where it kinda gets confusing for me I know that we have 16 numbers that can be divisible by 8 I got that from adding the even and odd parts together. Later on I figured that the first and second number can also be switched between themselves and that would not effect the first tree digits at all so I tried doing this
2!+12+2!+4=16+2*2!=16+4=20
I'm kinda wondering if this is an acceptable way of solving this because we never solved anything like this in school
and I would also like to see if any of you guys could give a more straightforward way of solving this because if this is right it seems way to complicated of a way of solving
Thanks for any feedback

 

[TheMathNoob]

Homework Statement

From the numbers 4,5,6,8,9 we make 5 digits numbers (each number can be used only once).
h)How many of these numbers are divisible by 8?
The correct answer is 20

Homework Equations

a number is divisible by 8 if the last 3 digits are divisible by 8

If the hundreds digit is even, examine the number formed by the last two digits.
If the hundreds digit is odd, examine the number obtained by the last two digits plus 4

The Attempt at a Solution

Ok, so I started by trying to figure out how many number we could make that are divisible by 8
I divided the problem into 2 parts. First I calculated the possibilities if the hundredth digit is a odd number and then I did the same thing for the even number
a) since we have 3 even number there are 3 possibilities for the hundredth digit
and we also know that the other 2 number have to be divisible by 8 so that left me with 4 options (48,56,64,96)
so when I combined the hundredth number and these for options we get 3*4=12 options in total
b) we have 2 odd numbers so there are 2 possibilities for the hundredth digit
and we also know that the other 2 number + 4 have to be divisible by 8 so I got 2 options (68,84)
so when I combined the whole thing I got 2*2= 4 options in total

and this is where it kinda gets confusing for me I know that we have 16 numbers that can be divisible by 8 I got that from adding the even and odd parts together. Later on I figured that the first and second number can also be switched between themselves and that would not effect the first tree digits at all so I tried doing this
2!+12+2!+4=16+2*2!=16+4=20
I'm kinda wondering if this is an acceptable way of solving this because we never solved anything like this in school
and I would also like to see if any of you guys could give a more straightforward way of solving this because if this is right it seems way to complicated of a way of solving
Thanks for any feedback

I am not an expert at this , but why did you multiply 3 * 4?. If you do that you can get repetitions and remember that each number can be used only once. I am still working on your problem and think the only way to do this is by counting.

 

[TheMathNoob]

I am not an expert at this , but why did you multiply 3 * 4?. If you do that you can get repetitions and remember that each number can be used only once. I am still working on your problem and think the only way to do this is by counting.

This is what I got. As you said, 48,56,64 and 96 are divisible by 8, so we have to find a way to match them with 4,6 and 8. We can't just do something like 3*4 because you may get something like 448 and this doesn't work because there is a repetition, so we have to do this by hand.

6->48
6,8->56
8->64
4,8->96

Those are the possible ways how we can match those numbers so the resultant outcomes would be

648,456,856,864,496,896.

The odd part is the same thing

As you said 68 and 84 each plus 4 are divisible 8, so we have to find a way to match them with 5 and 9

9,5-> 84
9,5->68

so the resultant outcomes are 984,584,568,968

so with even and odds, we got ten possible ways how the last three digits are divisible by 8. If we fill out the remaining spots with the two remaining numbers then it would be just be 10 ways, but as you said, we can switch the numbers, so for example if we pick 984, we can make 2 numbers from it, 65984 and 56984, so if we do the same thing with the other numbers, we get in total 20 possible ways.

 

[Matejxx1]

Thanks for the reply!
i see where i made a mistake and why 3*4 can't be used here.
So if i understood you correctly the correct way of solving this kind of problems is to count the number of possible ways (in this case 10) *spaces left factorial (in this case 2!)?
So let's say we added a sixth number let's assume that the number is still divisible by 10 ways.
So the correct answer then would be 3!*10=60?

 

[TheMathNoob]

Thanks for the reply!
i see where i made a mistake and why 3*4 can't be used here.
So if i understood you correctly the correct way of solving this kind of problems is to count the number of possible ways (in this case 10) *spaces left factorial (in this case 2!)?
So let's say we added a sixth number let's assume that the number is still divisible by 10 ways.
So the correct answer then would be 3!*10=60?

Yes, that's right.