#19
Jan405, 04:45 PM

P: n/a




#20
Jan505, 07:00 PM

P: n/a

matt,
I have one more question about your proof. S = 33 (mod 10), but does S really need to be divisible by 3? 43 = 33 (mod 10), but 43 isn't divisible by 3. 



#21
Jan605, 04:11 AM

Sci Advisor
HW Helper
P: 9,398

what proof? I didn't say that I'd proved it, I offered two things for you to think about that might prove it. One of them does more easily than the other. I've no idea if the sum of the first r!'s is not divisible by 9 or not, but that seems a reasonable thing to consider. Divisbility by 3 and the modulo 10 bit are completely unrelated. The mod 10 bit proves it straightforwardly.
And I've no idea what's puzzling you in your question. Of course there's no reason for two numbers in the equivalence class of 3 mod 10 to both be divisible by 3, there's no reason for either of them to be divisible by 10. IF they are both divisible by 3, so is their difference, which, by definition, is also divislbe by 10, and hence by 30. And? 


#22
Jan605, 05:58 AM

P: n/a

Oh, sorry. I misread what you posted. I know that the mod 10 bit proves (since 33 isn't a square mod 10), but I thought you'd approached it from a different angle.




#23
Jan605, 06:18 AM

Sci Advisor
HW Helper
P: 9,398

I had. A little check, which I intended you to do demonstrates the div by 9 thing doesn't actually help.



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