Brain Teaser #17: Autoraptor's ID Number

147258369

Edit: I think Im one digit off...but for the life of me I cant figure it out. Is this even possible?

 

Clues for #9 and #1 are no use ... the rest are tough to come up with!
*back to scratching my head*

 

The fifth digit has to be 5.
The fouth digit has to be either 2 or 6.

And divisible by 9 is a moot point.

 

well you also know the 5th number has to be 5. My number is all right except for the 8th digit. I cant find anything to switch it around with either.

 

Switch your 2 and 6.

I think that does it. Checking now...

 

Nope, my bad. I was way off.

 

damn you peter I am so damn close...I HATE these things.

 

ANSWER DELETED BY TILLMAN

See http://216.239.41.104/search?q=cache:Unjad3h8FYIJ:www.brianhetrick.com/casio/p112899.html+%22nine-digit+number+%22+%22divisible+by+nine%22&hl=en

If you want the answer. I decided that just answering it would be unfair, as my math skills are horrible....

 

well I saw Tillman's answer and it checks...Just cant figure out how they did it.

 

Its summer for me, i refuse to do math problems until after labor day

 

I was actually going to write a program to do it, too...

 

We know the answer, it's on my link....

 

Yea! I got my own brain teaser! Thanks!

 

A program? Awwww, that is so *ucking lame! This is dumb. Next time get one with a clever answer, not just some stupid programming that will try millions of solutions until it nails it!!!

 

This is actually an interesting problem to tackle mathematically (i.e. without the aid of a computer program). Does take a little trial and error though. This is roughly how I did it :

I indexed the digits from 1 to 9, as d1, d2, ..., d9

The condition of divisibility by 9 has no significance since the sum of digits from one to nine (= 0.5*9*10 = 45) is divisible by 9 anyway. It is trivial to see that the divide by one for the last digit has no meaning either. So only the other conditions are important.

Now d5 has to be 5 since 0 is not included. Also, every even indexed digit has to be even and every odd indexed digit therefore has to be odd.

We've already assigned 5 -> d5. So we have the set {1,3,7,9} to play with for the remaining odd digits {d1,d3,d7,d9}. We can't really refine this at the present, or at least I don't know how.

But we can do something with the evens. We have {2,4,6,8} to play with. Observe that d1d2d3d4 has to be divisible by 4 and d1d2d3d4d5d6d7d8 has to be divisible by 8 (and hence also by 4). The rule for div by 4 is that the last two digits have to be div by 4. We know that d3 and d7 are both odd. The only two digit numbers where the tens place is odd that are divisible by 4 have 2 or 6 in the units place (e.g. 12 or 36). They definitely cannot have 4 or 8 in the units place (e.g. 38 or 54 are not divisible by 4).

So we know that d4 = {2 or 6} and d8 = {6 or 2} (the converse). By elimination, d2 is either {4 or 8} and hence d6 = {8 or 4} (the converse).

We've narrowed things down but not quite enough for comfort. There's nothing easy we can do with the divide by 7 rule, cause it's complicated but we can try to leverage with the divide by 3 stipulations. Let's see :

d5 = 5 and d6 = {4 or 8}. I have to admit I got lucky here. I chose d6 = 4 and that led me quickly to the right answer. Here's how :

d5 = 5 and d6 = 4. The sum of d5 and d6 is divisible by 3. Since d1d2d3d4d5d6 has to be divisibly by 6 (and hence by 3), the sum of digits from d1 to d4 has to be divisible by 3 as well.

Now d4 = {2 or 6}. If d4 = 6, then the sum of d1 to d3 has also got to be divisible by 3. d2 = {4 or 8}. I considered the case of d2 = 8 first. I looked at it in terms of triples {d1,d2,d3} that summed to a multiple of 3.

Namely, {1,8,3}, {3,8,1}, {1,8,9} and {9,8,1}. For each triple I constructed the number formed (considering the prior assumptions as well) up to d7 (letting this vary between the only odd numbers remaining, which are 7 and either 9 or 3 depending on which triple was being considered) and tested division by 7. The only triple that satisfied was {3,8,1} with d7 = 7. Since d4 = 6 (by initial assumption), d8 has to be 2.

We're almost done. We now have up to d8. All that remains is to set the remaning odd number (9) into d9.

So we have the solution : 381654729.

This did require a fair bit of trial and error but it isn't too tedious once you narrow the cases down. Oh, and a calculator is very handy to test division by 7.

 

Calculator? Now you disapoint me!!! Tssss ... LMAO!! Good stuff. I like this much better than the program!

 

Lol, thanks ! There's unfortunately no "elementary" rule for divide by 7. There is an unwieldy rule that says "take the last digit, double it, subtract that from the rest of the number, and if the result is divisible by 7, so is the original number", but I just couldn't figure out how to work that into the scheme of my attack on the problem.

Fun stuff.

 

And of course, if the number you get after the hoopla isn't easily divisible 7, you could keep doubling and subtracting until you get one that is/isn't!

FYI, I didn't immediately start coding. I got all the way up to Turb0's 7th paragraph on my own before someone posted the solution. I know I could do it if I wanted, I was just going for speed!!!