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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?

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Hint: The way you approach is correct. However, you are missing some digits. For instance 04, 08 can be in the last digit so the number divides by 4. If you complete your list, the answer would be correct.

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  • $\begingroup$ ahh..., correct.., I missed the two possibilities of the last two-digit... :D Ok.., I get it..., thank you... :) $\endgroup$ – chihiroasleaf Feb 9 '13 at 11:02

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