a quick math challenge

12272013, 05:03 AM
Post: #1




a quick math challenge
Of the 362,880 9digit numbers containing each of the digits 19, how many of those 362,880 9digit numbers are prime?


12272013, 05:12 AM
Post: #2




RE: a quick math challenge
None. They're all divisible by 3.
Graph 3D  QPI  SolveSys 

12272013, 12:47 PM
Post: #3




RE: a quick math challenge
Correct. I teach the divisibilityby3 rule to my middle school math students, but I know that it is promptly forgotten by the time they get to high school.


12272013, 04:16 PM
Post: #4




RE: a quick math challenge
Quote: I teach the divisibilityby3 rule to my middle school math students, but I know that it is promptly forgotten by the time they get to high school. Really? I thought that this rule is known to everyone and it's as nearly common knowledge as divisibility by 2. Even people that I would call innumerate know this when I've asked them. katie 

12272013, 04:43 PM
Post: #5




RE: a quick math challenge
The kids forget it almost immediately. I might teach it to them in 6th grade, and in 7th grade they've forgotten it. They do generally know, however, that all even numbers are divisible by 2.
Regarding adults, I suspect that if you took a manonthestreet poll, you'd probably find fewer than 5% would know the divisibilityby3 rule. 

12272013, 10:53 PM
Post: #6




RE: a quick math challenge
All these numbers are divisible by 9 too.
 Pauli 

12282013, 11:57 AM
Post: #7




RE: a quick math challenge
(12272013 04:43 PM)Don Shepherd Wrote: Regarding adults, I suspect that if you took a manonthestreet poll, you'd probably find fewer than 5% would know the divisibilityby3 rule. Somewhere back in my head I might have found a 40 year old memory of that rule. If I would have searched for it, because it probably wouldn't have crossed my mind to apply it to this challenge in the first place... 

12282013, 10:06 PM
Post: #8




RE: a quick math challenge
I teach 9th grade Alg 1, and am happy to report approx 80% of my students rember the 2, 3, and 5 rules. Cannot figure out why they can't remember the by 6 rule!
I'm a math teacher. Of course I have problems. 

12292013, 02:19 PM
(This post was last modified: 12292013 02:20 PM by Waon Shinyoe.)
Post: #9




RE: a quick math challenge
(12282013 10:06 PM)Les_Koller Wrote: I teach 9th grade Alg 1, and am happy to report approx 80% of my students rember the 2, 3, and 5 rules. Cannot figure out why they can't remember the by 6 rule! Excuse me but what's the 2,3, and 5 rules? I think a 9digit number contains 1~9 can be divided by 3 because (abcdefghi) mod 3 = (1*9 + (0+1+2)*3 ) mod 3 = 18 mod 3 = 0 (I'm a high school student in P.R.China.) 

12292013, 03:37 PM
Post: #10




RE: a quick math challenge  
12292013, 06:09 PM
(This post was last modified: 12292013 06:10 PM by Katie Wasserman.)
Post: #11




RE: a quick math challenge
Quote:I teach 9th grade Alg 1, and am happy to report approx 80% of my students rember the 2, 3, and 5 rules. Cannot figure out why they can't remember the by 6 rule! Very interesting that we seem to have a real split in people's experience with this. I wonder how we could conduct an unbiased poll on the general population's knowledge of divisibility rules. No doubt that the US will be low on the list of industrialized countries as far as percentage of population that know this, but within the US I suspect it might be very different region by region. katie 

12292013, 06:31 PM
Post: #12




RE: a quick math challenge
(12282013 10:06 PM)Les_Koller Wrote: I teach 9th grade Alg 1, and am happy to report approx 80% of my students rember the 2, 3, and 5 rules. Cannot figure out why they can't remember the by 6 rule!For me was simple to remember the 3, 2 and 6, a bit more difficult the divisibility by 7, I'd right now to check again on wiki the last one. Rules by mind are made to be forgotten if non applied, I don't mean daily, but often........ 

12312013, 03:03 PM
Post: #13




RE: a quick math challenge
(12292013 06:31 PM)aurelio Wrote: a bit more difficult the divisibility by 7Similar to divisibility by 11, but you pair the digits off into threes before checking them out. Is 123,456,789 divisible by 7? Well, 123  456 + 789 = 456 which is not a multiple of 7, so neither is the original number. Can anybody explain why this works for 7 (and 13)? 

12312013, 03:05 PM
Post: #14




RE: a quick math challenge  
12312013, 05:38 PM
Post: #15




RE: a quick math challenge
(12312013 03:03 PM)Curlytop Wrote: Can anybody explain why this works for 7 (and 13)?\[ \begin{align} 1000&\equiv&6 (7) \\ &\equiv&1 (7) \\ \end{align} \] \[ \begin{align} 1000&\equiv&12 (13)\\ &\equiv&1(13) \\ \end{align} \] \[ \begin{align} 123,456,789&=&123\times1000^2+456\times1000+789 \\ &\equiv&123\times(1)^2+456\times(1)+789 (7) \\ &\equiv&123456+789 (7) \\ \end{align} \] Cheers Thomas 

12312013, 07:25 PM
Post: #16




RE: a quick math challenge
Yes that's it. It all revolves around 1001 being a multiple of 7, of 11 and of 13. (In fact it is 7 x 11 x 13).
Did anybody catch the comment about pairing the digits off into threes? One of mine as a kid, which they never let me forget. 

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