MHB Solving 6-Digit Number Quandaries: 1, 2, 3, 4, 5, 6

[Ciaran]

Hi there, I'm looking for some confirmation for the following problem:
The digits 1, 2, 3, 4, 5, 6 are written down in some order to form a six-digit number.
(1) How many such six-digit numbers are there altogether?
(2) How many such numbers are even?
(3) How many are divisible by 4?
(4) How many are divisible by 8?

For (1) I got 720, simply the permutations of 6 digits taking 6 at a time and for (2) I got 360. I'm not sure exactly how to proceed with (3) or (4). I know that (3) will help with (4) because if a number is divisible by 8, it must be divisible by 4.

 

[Opalg]

Hi there, I'm looking for some confirmation for the following problem:
The digits 1, 2, 3, 4, 5, 6 are written down in some order to form a six-digit number.
(1) How many such six-digit numbers are there altogether?
(2) How many such numbers are even?
(3) How many are divisible by 4?
(4) How many are divisible by 8?

For (1) I got 720, simply the permutations of 6 digits taking 6 at a time and for (2) I got 360. I'm not sure exactly how to proceed with (3) or (4). I know that (3) will help with (4) because if a number is divisible by 8, it must be divisible by 4.

The condition for an even number to be a multiple of 4 is

either the last digit is a multiple of 4 and the next-to-last digit is even
or the last digit is not a multiple of 4 and the next-to-last digit is odd.​

There is a similar condition for a multiple of 4 to be a multiple of 8, using the last three digits of the number. If the hundreds digit is even then the number formed by the last two digits must be a multiple of 8, and if the hundreds digit is odd then the number formed by the last two digits must not be a multiple of 8.

 

[Ciaran]

Thanks for your reply! How would I go about using this and would it have to be a case by case job? Do you agree with my answers for (1) and (2)?

 

[Opalg]

Thanks for your reply! How would I go about using this and would it have to be a case by case job? Do you agree with my answers for (1) and (2)?

Yes I do agree with your answers to (1) and (2).

For (3), as far as I can see you do need to work on a case by case basis, at least to some extent. There are not too many cases to consider. In fact, the last two digits of an even number must be
12, 14, 16, 24, 26, 32, 34, 36, 42, 46, 52, 54, 56, 62 or 64. Some of them are multiples of 4 and some are not. But in each case you have the same situation of having four digits to fill the remaining four places.

 

[Ciaran]

Is the answer for (3) 192?

 

[Opalg]

Is the answer for (3) 192?

Yes. (Sun)

 

[Ciaran]

Excellent! Is (4) 18?

 

[soroban]

Hello, Ciaran!

The digits 1, 2, 3, 4, 5, 6 are written down in some order to form a 6-digit number.
(1) How many such six-digit numbers are there altogether?
(2) How many such numbers are even?

For (1) I got 720, simply the permutations of 6 digits taking 6 at a time
and for (2) I got 360. . Right!.

(3) How many are divisible by 4?

A number is divisible by 4 if its last two-digit number is divisibe by 4.
There are 8 such endings: $\{12, 16, 24,32, 36,52,56,64\}$
The other 4 digits can be arranged in $4!$ ways.

Therefore: $\:8\cdot4! \:=\:192$ ways.

(4) How many are divisible by 8?

A number if divisible by 8 if its last 3-digt number is divisble by 8.
There are 14 such endings: $\:\begin{Bmatrix}136,152,216,256,264,312,352 \\ 416, 432,456,512,536,624,632 \end{Bmatrix}.$
The other 3 digits can be arranged in $3!$ ways.

Therefre: $\:14\cdot 3!\:=\:84$ ways.

 

[Ciaran]

Thanks a lot, guys! Perhaps I could have used the fact that the remainder after dividing a six digit number by 8 is 4d+2e+f- just a thought!