# Determine the number of six-digit integers (no leading zeros) in which no digit may be repeated and divisible by 4?

Determine the number of six-digit integers (no leading zeros) in which no digit may be repeated and divisible by $4$?

I've tried solving this problem, but the result is different with the solution provided in the book... This is my way :

in order the number to be divisible by $4$, the last two-digit must be divisible by $4$, so the possibilities are (I've group them) $(24,64,84,28,48,68), (20,40,60,80), (12,32,52,72,92,16,36,56,76,96)$ so I'll have three cases

the first case : the number of six-digits integer that divisible by $4: 7 \cdot 7 \cdot 6 \cdot 5 \cdot 3 \cdot 2 = 8820$

the second case: the number of six-digits integer that divisible by $4: 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 1 = 6720$

the third case: the number of six-digits integer that divisible by $4: 7 \cdot 7 \cdot 6 \cdot 5 \cdot 5 \cdot2 = 14700$

so, the total is $30240$ but the answer provided in the book is $33,600$

What am I'm missing? Is this the correct approach?