Permutations

L

Loki123

Full Member
Joined
Sep 22, 2021
Messages
790
The answer is supposed to include 9! Times some number. I don't know how to do that. Can anyone please tell me. I am kinda running out of time, I don't ask for the whole problem to be done, but I would prefer some description of the problem or instruction rather than more questions.IMG_20220506_134331.jpg
 
You are correct in saying that the 9-digit number must end in 25 or 50.

You are also almost correct in saying that the 0 digits must be 0-8 OR 1-9 (not 0-9 as you have).

The third digit from the left (not the right as you have it, although that won't make a difference) is a 7.

So there are 4 cases to consider:

_ _ 7 _ _ _ _ 2 5 and filling in the spaces with 0, 1, 3, 4, 6, 8

_ _ 7 _ _ _ _ 2 5 and filling in the spaces with 1, 3, 4, 6, 8, 9

_ _ 7 _ _ _ _ 5 0 and filling in the spaces with 1, 2, 3, 4, 6, 8 (there already is a 0, so 1-9 in the fourth option won't work)

That should help.
 
Harry_the_cat said:
I'm not sure what you mean by "The answer is supposed to include 9! Times some number" ??
Click to expand...
For the result I live a line to write a number. It is followed by *9!. Basically what number times 9! Is the answer. If I calculate it like this I don't think I'll get 9! As I don't see it including permutations. Tho maybe I am supposed to get that part myself.
 
Harry_the_cat said:
I don't believe there could possibly be more than 9! nine-digit numbers that fulfill those conditions.
Click to expand...
I agree: in each of the 4 cases (i.e., *00, *25, *50, *75) we have at most 8 digits left for 6 positions, which is [imath]\frac{8!}{2!}[/imath] combinations. But [imath]4\times \frac{8!}{2!} = 2\times 8! < 9![/imath]
 
Loki123 said:
The answer is supposed to include 9! Times some number. I don't know how to do that. Can anyone please tell me. I am kinda running out of time, I don't ask for the whole problem to be done, but I would prefer some description of the problem or instruction rather than more questions.View attachment 32607
Click to expand...
You say that you are count all nine digit numbers that have a [imath]7[/imath] in the third place from the left and are divisible by [imath]225[/imath].
To do that you also say that the number should end in [imath]25\text{ or }50[/imath].
However, the number [imath]827693150[/imath] has the necessary conditions but is not divisible by [imath]225[/imath] !
SEE HERE Also this [imath]807693125[/imath]
For a number to be divisible by [imath]225=3^2\cdot 5^2[/imath] it must be divisible by both [imath]\bf{9\underline\text{ and }25}[/imath].
 
pka said:
You say that you are count all nine digit numbers that have a [imath]7[/imath] in the third place from the left and are divisible by [imath]225[/imath].
To do that you also say that the number should end in [imath]25\text{ or }50[/imath].
However, the number [imath]827693150[/imath] has the necessary conditions but is not divisible by [imath]225[/imath] !
SEE HERE Also this [imath]807693125[/imath]
For a number to be divisible by [imath]225=3^2\cdot 5^2[/imath] it must be divisible by both [imath]\bf{9\underline\text{ and }25}[/imath].
Click to expand...
Yes that's right, BUT the digits have to be either 0 to 8 (which add to 36, meaning the number is divisible by 9) OR 1 to 9 (which adds to 45 ...). There is no other set of 9 digits that will work. Loki123 has alluded to that.
The number you have given is missing the digit 4, so does not add to a multiple of 9 and so isn't a multiple of 225.
 
Harry_the_cat said:
Yes that's right, BUT the digits have to be either 0 to 8 (which add to 36, meaning the number is divisible by 9) OR 1 to 9 (which adds to 45 ...). There is no other set of 9 digits that will work. Loki123 has alluded to that.
The number you have given is missing the digit 4, so does not add to a multiple of 9 and so isn't a multiple of 225.
Click to expand...
Please tell me where in the actual question it is stated that only the first nine digits are used.
It is not in the actual question. It may be in Loki's work but not in the statement.
I would have Loki banned for this sort thing that happens again and again with Loki.
 
It doesn't actually state that. But because the number is divisible by 225 then it must be divisible by 9 as well as 25.
If the number is divisible by 9, then the digits must add to a multiple of 9. The only way this can happen with 9 different digits is if they are 0-8 inclusive (which add to 36) OR 1-9 inclusive (which add to 45). Any other combo of 9 different digits will not add to a multiple of 9 and hence the resulting 9-digit number will not be divisible by 9.
 
pka said:
Please tell me where in the actual question it is stated that only the first nine digits are used.
It is not in the actual question. It may be in Loki's work but not in the statement.
I would have Loki banned for this sort thing that happens again and again with Loki.
Click to expand...
I didn't make a mistake now.
 
Harry_the_cat said:
It doesn't actually state that. But because the number is divisible by 225 then it must be divisible by 9 as well as 25.
If the number is divisible by 9, then the digits must add to a multiple of 9. The only way this can happen with 9 different digits is if they are 0-8 inclusive (which add to 36) OR 1-9 inclusive (which add to 45). Any other combo of 9 different digits will not add to a multiple of 9 and hence the resulting 9-digit number will not be divisible by 9.
Click to expand...
Yes that was my thought process, I wrote for 45 what numbers I can have and what numbers I can have for 36.
 
You must log in or register to reply here.
Top