I'm stumped

Aug 18, 2019
6
0
Hey gang. I used to do a card trick which had 5 piled of cards all equal amounts. The hard part comes when you add an ace into the mix and request someone to move whichever cards they want but the piles have to go back to the original values. Not removing any cards. Adding aces is repeated one at a time Once the equationn is complete u till all 4 aces are in the mix. Now from memory the number in the piles was 5x piles of five. So four corners with a 2 and 3 in each corner 2+3=5... And just a number 5 card in the middle... Hence 5 piles of 5. Add an ace onto the 5 in the middle and request someone rearrange them to give you 5 piles of five again. Is this correct and if so how does it work? I definitely remember doing this trick but can't remember if it was piles of 5 and if it was. I don't remember what cards to move to which locations... Thanks for your help..
 

Josh Burch

Elite Member
Aug 11, 2011
2,966
1,101
Utah
I'm not sure of the name of the routine but I think I understand what you are getting at. To clarify can you answer these questions?

Was the routine based on adding up the cards? Or adding up the values of the cards?

How many piles of cards were there altogether?

What did the cards look like on the table? Were they arranged in a line? Or a square?
 
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Aug 18, 2019
6
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IMG_20190816_122240.jpg
[/URL][/IMG]
I'm not sure of the name of the routine but I think I understand what you are getting at. To clarify can you answer these questions?

Was the routine based on adding up the cards? Or adding up the values of the cards?

How many piles of cards were there altogether?

What did the cards look like on the table? Were they arranged in a line? Or a square?
Hey bud. Thanks for your reply. You add the value of the cards. It is 5 piles in total and it is layed out in a square shape with 1 card in the middle. Like Pic. Cheers. I'm super excited and hopeful that you have the answers..
 
Aug 18, 2019
6
0
I'm not sure of the name of the routine but I think I understand what you are getting at. To clarify can you answer these questions?

Was the routine based on adding up the cards? Or adding up the values of the cards?

How many piles of cards were there altogether?

What did the cards look like on the table? Were they arranged in a line? Or a square?
Does that help Josh? Cheers
 
Jan 26, 2017
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Virginia
IMG_20190816_122240.jpg
[/URL][/IMG]
Hey bud. Thanks for your reply. You add the value of the cards. It is 5 piles in total and it is layed out in a square shape with 1 card in the middle. Like Pic. Cheers. I'm super excited and hopeful that you have the answers..
I'm not sure if it's just me, but I don't see the image here
 

Josh Burch

Elite Member
Aug 11, 2011
2,966
1,101
Utah
Aug 18, 2019
6
0
It was mathematical. No tricks. Just some how it was done and seemed impossible but was indeed easy once you know the combo. One ace goes in. Cards get moved around. 5 piles of 5 again with All cards being used. Then another ace. Repeat this for all 4 aces
 

RealityOne

Elite Member
Nov 1, 2009
3,744
4,076
New Jersey
Were the cards picked up and dealt into piles again?

Did the middle value increase each time due to the addition of the ace (e.g. 5, 6, 7, 8, 9)? Because it is impossible to have the 11 cards (including the ace) add up to 25 because 25+1 is 26. Also, could there be more or less than two cards in each pile? That would seem necessary.

So, for example, was the second deal something like side piles: A+2+2, 5, 3+2,3+2 and the middle pile: 3+3?

Next deal could be: 3+A+A, 3+2, 3+2, 3+2 with the middle 5+2

If that is the case, you can probably stack the cards to come up with the result and the different number of cards can be handled through a the dealing sequence. If not, I'm stumped.
 
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Aug 18, 2019
6
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All cards can be picked up and re dealt if necessary but from memory it wasn't necessary. The 1st addition was only a couple of cards moved. But yes the values all become the same number again. And I know mathematically it seems impossible. But the only way I can make sense of it is. It's a division. So 26 divided by 5 isn't a whole nimber. It's a decimal 5.2 then it goes into 5.4 5.6 and 5.8.. Never into 6 as a whole number. Number 5 may not have been the magic number in the equation. But I'm almost certain it was. I'll have to have a play with other numbers but it won't make any difference it will always give a .2 decimal
 

Josh Burch

Elite Member
Aug 11, 2011
2,966
1,101
Utah
Were the cards picked up and dealt into piles again?

Did the middle value increase each time due to the addition of the ace (e.g. 5, 6, 7, 8, 9)? Because it is impossible to have the 11 cards (including the ace) add up to 25 because 25+1 is 26. Also, could there be more or less than two cards in each pile? That would seem necessary.

So, for example, was the second deal something like side piles: A+2+2, 5, 3+2,3+2 and the middle pile: 3+3?

Next deal could be: 3+A+A, 3+2, 3+2, 3+2 with the middle 5+2

If that is the case, you can probably stack the cards to come up with the result and the different number of cards can be handled through a the dealing sequence. If not, I'm stumped.

This is about as close to a solution as I can imagine. (Personally, I thought it was based on the same "math" as Steinmeyers Bermuda triangle/indian coin scam)

@Mrwoody1985 I'm curious, where did you learn this? Was it a book? The internet? A friend?

And one more question. You said that you used to do this trick? Or did you see someone else do it?

I'm really interested because I love crazy, old, mathematical tricks like this. These are the kinds of tricks that many magicians overlook.
 
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