Understand the problem

The three-digit number 999 has a special property: It is divisible by 27, and its digit sum is also divisible by 27. The four-digit number 5778 also has this property, as it is divisible by 27 and its digit sum is also divisible by 27. How many four-digit numbers have this property?
Source of the problem
Israel 2014, Problem 4
Topic
Combinatorics, Number Theory
Difficulty Level
6/10
Suggested Book
Excursion in Mathematics by Bhaskarcharya Prathistan

Start with hints

Do you really need a hint? Try it first!

Let’s write the problem mathematically, i.e. in terms of the equations. Let’s write the condition mathematically. Let $abcd$ be a four-digit number, with $1\le a\le9$ and $0\le b,c,d\le 9$, and $a,b,c,d$ positive integers. Then we need to have $1000a+100b+10c+d=27n$ and $a+b+c+d=27m$, where $n,m$ are positive integers. Now, we have to count the number of such solutions. Observe that the sum of the digits can be at most 36. So, m = 1. Hence, a+b+c+d = 27. This leads to $111a+11b+c=3(n-1)$ . This implies $2b+c$ being a multiple of $3$. Now, this implies we have quite a number of cases to investigate. Let’s do them one by one patiently.
b c a+d
0 3,6,9 27,24,21,18
Out of this b = 0, c = 9, a+d = 18, a = 9, c =9 is the only possibility considering the maximum possibble value of a+d being 18.  
b c a+d
1 1,4,7 25,22,19
  None of the cases is fine as the maximum possible value of a+d is 18.
b c a+d
2 2,5,8 23,20,17
This gives rise to b = 2, c = 8, a+d = 17. Hence two solutions 8289 and 9288.
b c a+d
3 0,3,6,9 24,21,18,15
Thus you can understand this gives rise to 4 solutions: ($6399,7398,8397$ and $9396$)
b c a+d
4 1,4,7 22,19,16
This gives rise to three solutions: ($7479,8478,$ and $9477$).
b c a+d
5 2,5,8 20,17,14
Only the last two choices are acceptable; the former gives us two solutions, and the latter 5 (for a total of seven solutions).  
b c a+d
6 0,3,6,9 21,18,15,12
The last three choices are acceptable and give us $1+4+7=12$ solutions.
b c a+d
7 1,4,7 19,16,13
The last two choices are acceptable and give us $3+6=9$ solutions.
b c a+d
8 2,5,8 17,14,11
All choices are acceptable and give us $2+5+8=15$ solutions.
b c a+d
9 0,3,6,9 18,15,15,9
All choices are acceptable and give us $1+4+7+9=21$ solutions. Therefore, there are $1+2+5+3+7+12+9+15+21=75$ four-digit numbers with this property.

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