How many 8 digit Natural numbers are there whose digits are from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} which are divisible by 275?

Question

How many 8-digit Natural numbers are there whose digits are from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} which are divisible by 275? 275 = 5^2 * 11 There will be two cases - One in which 25 is at the end and one in which 75 is at the end and I thought that there would be two cases of each because

a b c d e f 2 5

(b+d+f+5)-(a+c+e+2) = 0 or 11 and the same with the case of 75

Now I tried making cases but couldn't find a way to solve the question properly. Can somebody help me with this

Answer 1

The smallest integer of eight digits does not containing zeros is $11\space111\space111$ and the largest is $99\space999\space999$. Since $11\space111\space111\equiv{11}\pmod{275}$ the first of our integers multiples of $275$ is $11\space111\space375$; similarly $99\space999\space999\equiv99\pmod{275}$ gives for the last one $99\space999\space625$. We need now to calculate $h$ in$$11\space111\space375+550h=99\space999\space625\Rightarrow h=161\space615$$ so there are in total $161\space617$ integers to consider from which we must discarded those that would contain zeros.

Trying to get a pattern to localized the integers containing zeros in the Arithmetic progressions $$11\space111\space375+550h\hspace2cm(1)\\99\space999\space625-550h\hspace2cm(2)$$ for the the values $h=1,2,3,\cdots,100$ we find $17$ integers to be excluded in $(1)$ and $18$ in $(2)$. It follows there are $165$ of the required integers plus all the analogues without zeros in $$11\space166\space375+550h\le 99\space944\space625; \space h\ge0$$ It is complicated to get a pattern if there is one and I think there is not possible a direct calculation other than brute force. I am maybe wrong about this.

Answer 2

You can use Mathematica to verify your result.

DO NOT admit repeat digits

(*Generate the list of 8-digit numbers divisible by 275*)
numbers = Range[Ceiling[10000000/275]*275, Floor[99999999/275]*275, 275];

(*Filter out numbers with 0 or repeated digits*)
validNumbers = 
  Select[numbers, 
   FreeQ[IntegerDigits[#], 0] && DuplicateFreeQ[IntegerDigits[#]] &];

(*Get the count*)
Length[validNumbers]

$$ \color{red}{972} $$

admit repeat digits

(*Generate the list of 8-digit numbers divisible by 275*)
numbers = Range[Ceiling[10000000/275]*275, Floor[99999999/275]*275, 275];

(*Filter out numbers with 0 or repeated digits*)
validNumbers = Select[numbers, FreeQ[IntegerDigits[#], 0] &];

(*Get the count*)
Length[validNumbers]

$$ \color{red}{96621} $$