How to Construct DFA and NFA for Languages Divisible by 6 or 10?

[logicalman]

L = {a^k / k is divisible by 6 or 10 (or both)}

Give a DFA and NFA for the above.

Can anyone do it?

My 2 cents:
This is my DFA for the above language.

(q_x) = accepting state where x is any integer.

>q_0-->q_1--> ... (q_6)...q_28--->(q_30)

 

[logicalman]

c'mon time is running out!

 

[wais]

-->q_31--> ... (q_36)...q_58--->(q_60)-->q_61--> ... (q_66)...q_88--->(q_90)-->q_91--> ... (q_96)...q_118--->(q_120)-->q_121--> ... (q_126)...q_148--->(q_150)-->q_151--> ... (q_156)...q_178--->(q_180)-->q_181--> ... (q_186)...q_208--->(q_210)-->q_211--> ... (q_216)...q_238--->(q_240)-->q_241--> ... (q_246)...q_268--->(q_270)-->q_271--> ... (q_276)...q_298--->(q_300)-->q_301--> ... (q_306)...q_328--->(q_330)-->q_331--> ... (q_336)...q_358--->(q_360)-->q_361--> ... (q_366)...q_388--->(q_390)-->q_391--> ... (q_396)...q_418--->(q_420)-->q_421--> ... (q_426)...q_448--->(q_450)-->q_451--> ... (q_456)...q_478--->(q_480)-->q_481--> ... (q_486)...q_508--->(q_510)-->q_511--> ... (q_516)...q_538--->(q_540)-->q_541--> ... (q_546)...q_568--->(q_570)-->q_571--> ... (q_576)...q_598--->(q_600)-->q_601--> ... (q_606)...q_628--->(q_630)-->q_631--> ... (q_636)...q_658--->(q_660)-->q_661--> ... (q_666)...q_688--->(q_690)-->q_691--> ... (q_696)...q_718--->(q_720)-->q_721--> ... (q_726)...q_748--->(