a quick math challenge

[Don Shepherd]

Of the 362,880 9-digit numbers containing each of the digits 1-9, how many of those 362,880 9-digit numbers are prime?

[Han]

None. They're all divisible by 3.

[Don Shepherd]

Correct. I teach the divisibility-by-3 rule to my middle school math students, but I know that it is promptly forgotten by the time they get to high school.

[Katie Wasserman]

Quote: I teach the divisibility-by-3 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.

[Don Shepherd]

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 man-on-the-street poll, you'd probably find fewer than 5% would know the divisibility-by-3 rule.

[Paul Dale]

All these numbers are divisible by 9 too.

- Pauli

[Maximilian Hohmann]

(12-27-2013 04:43 PM)Don Shepherd Wrote:  Regarding adults, I suspect that if you took a man-on-the-street poll, you'd probably find fewer than 5% would know the divisibility-by-3 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...

[Les_Koller]

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!

[Waon Shinyoe]

(12-28-2013 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 9-digit 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.)

[Thomas Klemm]

cf. Divisibility rule

[Katie Wasserman]

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.

[aurelio]

(12-28-2013 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........

[Curlytop]

(12-29-2013 06:31 PM)aurelio Wrote:  a bit more difficult the divisibility by 7

Similar 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)?

[Curlytop]

(12-28-2013 10:06 PM)Les_Koller Wrote:  Cannot figure out why they can't remember the by 6 rule!

Nor can I! Is the number divisible by 3? Is it even?

[Thomas Klemm]

(12-31-2013 03:03 PM)Curlytop Wrote:  Can anybody explain why this works for 7 (and 13)?

\[ \begin{align}
1000&\equiv&6 (7) \\
&\equiv&-1 (7) \\
\end{align} \]
\[ \begin{align}
1000&\equiv&12 (13)\\
&\equiv&-1(13) \\
\end{align} \]
\[ \begin{align}
123,456,789&=&123\times1000^2+456\times1000+789 \\
&\equiv&123\times(-1)^2+456\times(-1)+789 (7) \\
&\equiv&123-456+789 (7) \\
\end{align} \]

Cheers
Thomas

[Curlytop]

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.