moneydude242

I believe this is a typo, because after a bit of thinking, wouldn't the next number be 191?

I could try to help you with this question, but could you please clarify what the blanks are in this problem? What I am referring to is " both of which are greater than  " and "The resulting product is less than  ". Thank you! 

Ohh, ok I understand. The 14 each time carries onto the next digit, which causes you to add 14. Thank you so much for the help, and your patience! 

So I'm not the best at math, but I'll give this a shot. We have the set {1,2,2,4,4,4}. First off, we have 6! = 720, not counting any duplicates. Because there are 3 copies for each time we rearrange the set, (because some of them are us just rearranging the fours), we divide 720 by 3, to get 240. We repeat this process with the two 2s, and get 120. Now, I'm not 100% sure about this answer, so Melody (or anybody else), if you see this and it's wrong, could you please correct me? Thank you!

Ok, I get that, but how did you get 14 as a remainder? I didn't see any division... (sorry, but I really don't understand)

Thank you for telling me how this works, but could you please explain why you multiplied 20 by a 4 and got 140?

Ok, so right now you've got me confused. Do you mean that \(7+77+777+7777 = 8500?\) I believe the sum of those four numbers is 8638... Also, what do you mean that "The sum of each group is 85000000, which has a 7 in the hundreds place"?

Thank you!

Uh, but when you were multiplying the (n-2)(2n-3), wouldn't it multiply/add to -7n?