Comments on: CZ. A Parlour Trick

By: mathsoldier

mathsoldier — Mon, 29 Oct 2007 03:09:17 +0000

by adding in a new type of step to the process, namely adding the original number at various points, you are allowed to “confuse” the audience more and are able to keep the divisions much more simplified. Here is my example for a 9 digit number

Start with a person, lets call him Bob, picking any 9 digit positive integer. Tell Bob to “double” the number creating a 18 digit integer that is his 9 digit number written twice and write it down on a piece of paper, of course at no point are you allowed to see this paper. Now follow these steps
1) tell Bob to divide this 18 digit number by 7 and give the result to the person next to him
2) have this new person divide the number by 11 and hand it to someone else
3) now have the number divided by 13 and again the result is passed on
4) now divide by 19 and this time have the paper passed back to Bob
5) now Bob adds the original 9 digit number to the number he was given and passes it on
6) divide by 20 and pass on
7) divide by 11 and pass back to Bob again
8) again Bob adds the original 9 digit number and passes it on
9) divide by 10 and pass it on
10) divide by 12 and pass on
11) divide by 2 and give it back to Bob

now miraculously the paper that Bob finally recieves contains his original number :-D
now I found this out by noticing that after dividing the “doubled” number by 7,11,13, and 19 you end up with 52579 times the original number. Now that number is prime, but you add one to it by adding the original number and you have 52580 which breaks down nicely to 239 with additonal divisions by 20 and 11. Now again 239 is prime but adding one by adding the original number gives us 240 which again breaks down nicely with divisons by 12 and 2. Thus by throwing in a couple of simple additions we are able to perform the trick with a 9 digit number without having to do very complicated divisions. I’m sure similar techniques could be applied to the other digit sizes.

By: strauss

strauss — Sat, 27 Oct 2007 16:28:36 +0000

That’s an interesting idea, to look for numbers of this general form that have some reasonable factors. Looks like 1001 is going to be the champion, but

111111 = 3 * 7 * 11 * 13 * 37

100010001 = 3 * 7 * 13 * 37 * 9091

Kyle’s idea, given in the next segment, is to pretend that the factors are chosen randomly! This way you can have fewer steps, and stranger divisions.

(“Double a four-digit number, then divide by, oh, I don’t know, how about 73? …”)

By: rschranz

rschranz — Sat, 27 Oct 2007 02:02:46 +0000

There’s no good one if you “double” a 2-digit number, 101 being prime, but “tripling” a 2-digit number is OK, since 10101 is 3x7x13x37. Tripling a 3-digit number also works but 1001001 is 3×333667, so it doesn’t make for a very good trick: “OK, now divide it by 333667…”

By: John Dalbec

John Dalbec — Wed, 24 Oct 2007 01:11:04 +0000

“doubling” a 3-digit number multiplies it by 1001=7*11*13. Dividing by these prime factors in succession reverses the “doubling” operation to produce the original number.
“doubling” a 2-digit number multiplies it by 101 which is prime. Dividing by 101 produces the original number, but that’s not surprising.
“doubling” a 4-digit number multiplies it by 10001=73*137. This is not as good because there are only two prime factors.
“doubling” a 9-digit number multiplies it by 10^9+1=(10^3+1)(10^6-10^3+1)=1001*999001=7*11*13*19*52579.
“doubling” a 15-digit number multiplies it by 10^15+1=(10^3+1)(10^12-10^9+10^6-10^3+1)=1001*999000999001=
=7*11*13*211*241*2161*9091.