Let $$N$$ be the number of ways of distributing $$52$$ identical balls into $$4$$ distinguishable boxes such that no box is empty and the difference between the number of balls in any two of the boxes is not a multiple of $$6$$. If $$N=100a+b$$, where $$a, b$$ are positive integers less than $$100$$, find $$a+b$$.

Let the four distinguishable boxes contain $$x_1,x_2,x_3,x_4$$ balls respectively.

Conditions:
  1. $$x_i \ge 1$$ for every $$i$$ (no box is empty).
  2. For every pair $$i\neq j$$, the difference $$x_i-x_j$$ is not a multiple of $$6$$.
     ⇔ $$x_i \not\equiv x_j \pmod{6}$$, i.e. the residues $$x_i \pmod{6}$$ are all distinct.

1. Remove the “at-least-1” restriction by writing $$x_i = y_i+1$$ with $$y_i\ge 0$$.
    Then $$y_1+y_2+y_3+y_4 = 52-4 = 48$$.

2. Work with residues modulo $$6$$.
  Put $$x_i = 6k_i + r_i$$ where $$0\le r_i \le 5$$ and the four residues $$r_1,r_2,r_3,r_4$$ are distinct.
  The sum condition becomes

$$6(k_1+k_2+k_3+k_4)\;+\;(r_1+r_2+r_3+r_4)=52$$

⇒ $$k_1+k_2+k_3+k_4=\dfrac{52-(r_1+r_2+r_3+r_4)}{6}\,.$$

Because $$52\equiv 4 \pmod{6}$$, we need $$r_1+r_2+r_3+r_4\equiv 4 \pmod{6}$$.

3. Find all sets of four distinct residues whose sum is $$\equiv 4 \pmod{6}$$.
  Let the two omitted residues be $$s,t$$. The six residues sum to $$0+1+2+3+4+5=15\equiv 3 \pmod{6}$$, so

$$r_1+r_2+r_3+r_4 = 15-(s+t)\equiv 4 \pmod{6}\; \Longrightarrow\; s+t\equiv 5 \pmod{6}\,.$$

The only unordered pairs satisfying this are $$\{0,5\},\{1,4\},\{2,3\}$$. Hence the three admissible sets of residues are

A : $$\{1,2,3,4\}$$     B : $$\{0,2,3,5\}$$     C : $$\{0,1,4,5\}$$.

4. For each set, compute $$k_1+k_2+k_3+k_4$$.

The sum of residues in every set is $$10$$, so

$$k_1+k_2+k_3+k_4 = \dfrac{52-10}{6}=7$$

for all three cases.

5. Count the $$k_i$$ solutions for a fixed assignment of residues to boxes.

Case A: No residue is $$0$$. All $$k_i\ge 0$$.
  Number of non-negative solutions of $$k_1+k_2+k_3+k_4=7$$ is $$\binom{7+4-1}{4-1} = \binom{10}{3}=120$$.

Cases B and C: Exactly one residue is $$0$$. For that box we need $$x_i\ge 6$$, i.e. $$k_i\ge 1$$. Put $$k_i = 1 + k_i'$$ with $$k_i'\ge 0$$.
  Then $$k_1'+k_2'+k_3'+k_4'=6$$, giving $$\binom{6+4-1}{4-1} = \binom{9}{3}=84$$ solutions.

6. Multiply by the number of ways to assign residues to the distinguishable boxes.

• Set A provides $$4! = 24$$ assignments, each contributing $$120$$ solutions:
  Total $$=24 \times 120 = 2880$$.

• Sets B and C each give $$4! = 24$$ assignments with $$84$$ solutions each:
  Total $$=2 \times 24 \times 84 = 4032$$.

7. Total number of admissible distributions

$$N = 2880 + 4032 = 6912$$

8. Write $$N = 100a + b$$ with $$a,b \lt 100$$.
  Here $$N = 69\cdot100 + 12$$, so $$a = 69,\; b = 12$$ and $$a+b = 81$$.

Final answer: 81