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slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
I am looking for a mathematical method that I can research to accomplish the following. Given an eight digit number or so, I need to determine what combination of three different, three digit numbers can be multiplied together in sequence that would yield the eight digit number. In some cases the eight digit number selected may require a variable number of digits to equal the number. An example (I believe) would be y = a x b x c, with y being known and a, b, and c to be determined. I'm led to believe it involves factorials but the procedure eludes me. I can provide more details if necessary. I could go to Secret Sessions to let you know the specific use if requested or PM me.
Thanks in advance for any assistance you may provide. John
John
"A poor workman always blames his tools" |
slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
More detail about the effect and not the secret.....
1) Someone copies the numeric serial # from a dollar bill on a piece of paper. 2) Each of three other participants write down a three digit number (on occasion numbers can be less than three) on a piece of paper. 3) Another person multiplies the three numbers together. 4) The answer is compared to the dollar's serial number and they match.
John
"A poor workman always blames his tools" |
landmark![]() Inner circle within a triangle 5197 Posts ![]() |
John, the problem you pose is not trivial.
Essentially you are asking to find the prime factors of a large number. In fact, that is such a difficult problem to answer in a general way with a specific formula, that many forms of professional encryption are based on that difficulty. So the short answer is: 1) there may well be more than one way of factoring the number into three factors; and 2) forget it. On the other hand, if you can get a hold of a dollar bill with a specific eight digit serial number, then it is possible. Take any three three-digit prime numbers and multiply them together. The resulting answer will have only those three numbers as three-digit factors. For example, 233 x 251 x 293 = 17135519. So if you can find a serial number of 17135519 you can be sure that only the three numbers mentioned will give you that result. But I don't think that will be very helpful to you unfortunately.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
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slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
Landmark, thank you for your input. I hate to abandon it so let me give you some more detail.
Among other things, I subscribe to the Tarbell Lectures offered by Penguin Magic and Dan Harlan. The last one was this effect demonstrated. He said only certain dollar bills have a serial number that can be broken into factors and that by searching the internet you can find such a site that you could input the serial number and get the three factors to perform this effect. Again only certain serial numbers allow you to do this effect and you do it by trial and error going thru each dollar bill and saving the ones that work. I'll check my Tarbell books to see if I can get more info or possibly contact Dan because it's a great routine.
John
"A poor workman always blames his tools" |
slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
Got it! Problem solved with landmark's help. Routine is good to go....
John
"A poor workman always blames his tools" |
landmark![]() Inner circle within a triangle 5197 Posts ![]() |
John, just be careful that the serial number is not a number that can be factored in more than one way. Not necessarily likely, but not impossible.
For example, 60503040 = 234 x 320 x 808. Okay, but that same number 60503040 also equals 936 x 640 x 101. If that's not a problem for your effect, fine. But depending on the nature of the effect, it could screw you up.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
Landmark, that actually helps the routine. I could use the same bill with different source numbers or I could get input from two different "teams" with the same end result. You've given me much more to think about.
Thanks again for your valued input.
John
"A poor workman always blames his tools" |
landmark![]() Inner circle within a triangle 5197 Posts ![]() |
Oh, glad it's a bonus then!
Let us know how then how the routine goes down with an audience.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
jstreiff![]() Special user 701 Posts ![]() |
The original question asked about the number of 3 digit numbers required to produce an 8 digit number. Empirically it appears that three randomly selected numbers in the rough range of 220 through 460 will always produce an eight digit number.
John
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slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
Thanks for your response jstreiff, I may not have stated it clearly enough but the original question was to find a method to be able to determine what the three, three digit numbers were that generated a known (given) eight digit number. So I had the end result but was looking for the unknown source numbers.
John
"A poor workman always blames his tools" |
DanHarlan![]() V.I.P. 999 Posts ![]() |
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slowkneenuh![]() Regular user After 5,278+ posts, only credited with 133 Posts ![]() |
I'm using a different site but thanks for an alternative one.
By the way Dan, I really enjoyed that effect, your teaching and your enthusiasm for magic. Particularly impressive was that significant upgrade to a classic effect. I am looking forward to your upcoming lectures and am a big fan of your products also!
John
"A poor workman always blames his tools" |
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