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Here's a _much_ better party/bar trick:
1. Point to the drink glass in front of you 2. Ask friend if they think the circumference or height of the glass is larger 3. Friend will invariably think that height is larger 4. Wager for a drink that that's not the case 5. Use a straw to measure the circumference and then height 6. Enjoy your free drink
Unless you've picked an a Champaigne flute you'll always win. To make it even more fun, for (4) wager that the circumference will be _two_ times the height to make it sound more incredulous. For typical glasses you may go up to three!
I learned this trick from _Things to Make and Do in the Fourth Dimension_ (
https://www.amazon.com/Things-Make-Fourth-Dimension-Mathemat...
), overall good book.
How this goes in real life:
"Was your number... 3741?"
"No."
"Heh, I'm afraid that's impossible based on number theory. You must have made a mistake in the step where I asked you to choose a second 4-digit number by scrambling the four digits of the first number and subtracting the larger number from the smaller. Try it again, I'll wait."
"No."
A good magician would ask them to have a calculator out... or present only to an audience of geniuses.
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A different issue by picking 2222 (anything that ends up with all 0s or 9s ofc)
Not something I would bring up at a party because people are drinking and not really interested in doing addition/subtraction and will DEFINITELY pick simple numbers. By the time you get into "it has to be non-zero digits" they have already become bored.
It feels like my little brother running up to me and screaming "What's 729 squared?" and without a beat "531,441" , "I'm smarter than you!"
345 also doesn't work if you pick 435 as the shuffle, for example.
Your example does work for this.
Corrected.
Yeah, I donno. Even if I were able to convince someone at a party to go along with this "trick", I don't think they'd be especially impressed when after scrambling and subtracting 4 digit numbers, I'm able to wow them with a single digit.
Actual Math Party tricks are standard affairs: just jokes about Math.
-- Answer: A natural log.
-- Answer: Because Dec(imal) 25 == Oct(al) 31.
The Sheriff and Derivative meet up, and a showdown was about to occur. Knowing that it will all be over soon, they exchanged pleasantries before the final showdown.
The Sheriff introduces himself: "I'm e^x, the Sheriff of this town. You can't dare to derive me!"
Upon hearing the name: the evil derivative gives a toothy grin and says "My name is d/dy".
Etc. etc. Assuming you're comfortable telling these kinds of jokes of course. As usual, you need to read the room and see if the jokes would fly in the company you're in.
Q: What do you get when you cross an hippo and an aardvark?
A: HippoAardvarkSinTheta
Q: What do you get if you cross a mosquito and a mountain climber?
A: Nothing, you can't cross a vector and a scalar!
Q: What's yellow and equivalent to the axiom of choice?
A: Zorn's lemon!
What's purple and commutes?
An Abelian grape.
I'm being honest here. Reading along I randomly picked 987 from my brain. I scrambled that to 897. Subtracting the smaller from the larger I got 90, per the instructions. So I am unable to tell you the remaining digits starting with a number that isn't 9 or 0.... Maybe I'm a "you must be fun to talk to at parties" kind of guy.
Although not mentioned in the article, the trick can handle this. The rule should be "remove a digit that isn't 0", and if the sum of remaining digits is divisible by 9 then the removed digit must be 9 as well. That will cover everything except the case when the original number and the scrambled number are the same.
yeah for first 10000 digits there is about a 2%-5% chance (around 20%-5% for first 2k, then falls below 2% for remaining 2/3s) of selecting one that doesn't work. obviously picking something like 1111, 2222, ... won't work either since you can't rearrange it to anything but 0s.
I did the same thing, starting from 100. I guess maybe "no digits" is a sufficient response to end up with 90?
lol
I would bet a decent chunk of money the author has never successfully tried this at a party.
Asking people to subtract two 4 digit numbers and picking digits that are not 0 and 9 is absolutely not a โparty trickโ.
Maybe title would have to be The Math Party Math Trick.
Back in grade school when playing with a calculator, I came across this weird pattern when adding various combinations of 3 digit numbers on the number pad.
I wonder if there is something similar at foot as with this trick.
As follows, taking a calculator and the number grid, so a series of additions of a row added to the reverse of that row added to the second row, etc etc.
So following a horizontal pattern:
123 + 321 + 456 + 654 + 789 + 987 = 3,330
Now, let's see the sum when we use a vertical pattern:
147 + 741 + 258 + 852 + 369 + 963 = 3,330
Then, you can also do the diagonals:
Diagonally NW to SE:
748 + 847 + 159 + 951 + 263 + 362 = 3,330
Second Diagonal NW to SE:
784 + 487 + 159 + 951 + 623 + 326 = 3,330
Diagonally SW to NE:
142 + 241 + 753 + 357 + 869 + 968 = 3,330
Second Diagonally SW to NE:
421 + 124 + 753 + 357 + 689 + 986 = 3,330
Just a silly occurrence that they all sum to the same value 3,330.
Notice that in this arrangement
1 2 3 4 5 6 7 8 9
opposite values add up to 10: 1 + 9, 2 + 8, 3 + 7, 4 + 6.
Why these numbers do this is simply because the sequence is laid down in left to right, top-to-bottom order in a symmetric way, and these values pairs end up equidistant elements from the opposite ends of the sequence. If you put your index fingers on 1 and 9, and then count upwards with the left one and down with the right, you can see how this plays out.
And this summing to 10 to do with why the last digit of the sums you are seeing is always 0: 3330.
For instance in 123 + 321 + 456 + 654 + 789 + 987, the last digit is (3 + 1) + (6 + 4) + (9 + 7). We can rearrange these six numbers into (1 + 9) + (4 + 6) + (7 + 3) = 10 + 10 + 30 = 30.
Ok, so now we have a 0, and a carry of 3.
Next, note that since the diagonally opposite elements add to 10, all the three-element traces that pass through the center 5 necessarily add up to 15: (1 + 5 + 9) = (2 + 5 + 8) = (3 + 5 + 7) = (4 + 5 + 6) = 15.
In calculating the second digit of the sum you have 2 5 and 8, which occur twice: (2 + 2 + 5 + 5 + 8 + 8) = 30. Combine that with the carried 3 and you get 33. Put down the 3 and carry the 3.
Then again, the 100's digit is just mirror image of the ones: it adds up to 30, which combines with the carried 3 to make 33.
With 784 + 487 + 159 + 951 + 623 + 326, though you have rearranged the digits to form corner triangles, that is just a red herring. If you look at the ones digits inside this sum, you have 4 7 9 1 3 6. These is just the set made up of the left and right columns of the square, which we know can be put into 3 pairs adding to 10, making 30. Again we get our 0 to put down and 3 to carry.
The middle digits, the tens, are once again 8 5 and 2, doubled up again: another 15 x 2 = 30: put down 3, carry 3.
And so it goes.
It stays true for digits 1,3,7,9 i.e. 13+31+79+97 = 17+71+39+93
also for 1,2,4,5 with 12+21+45+54 = 14+41+25+52
Maybe there is some pattern.
Something similar we used to do in school to find answers for multiplication table of 9. We used to write all tables from 1 to 20, recite & them read them back from memory.
We would go by writing 9, ten times in ten total rows, like
9
9
....
9
Then from tenth row come up by writing x (sign of multiplication)
9x
Now go down again writing 1,2,3 in each row like
9x1
9x2
....
9x10
Now come up writing = in every row.
Now go down writing digits 0 to 9 like
9x1=0
9x2=1
....
9x10=9
Now write 0 to 9 again going up, ending up with table.
9x1=09
9x2=18
9x3=27
.....
9x10=90
Hereโs one of my favorites - super easy problem that almost everybody gets wrong.
On a piece of paper, write the following numbers as a vertical column:
4000
40 20 30 10
For best results, 4000 should be first and 10 last. Now take a second piece of paper and cover the number list. Ask someone to add the numbers up out loud as you move down the list, uncovering one number at a time starting with 4000. (First you expose the 4000, next the 40, and so on.) In nearly every case the person will call out the following: 4000, 4040, 4060, 4090, 5000. And typically they wonโt believe you when you tell them theyโre wrong, sometimes even after giving them the paper to redo their math.
Edit: formatting
I guess it's a good party trick if you've never been to a good party.
some python to do this trick and find counterexamples that won't work
import random as r
def fnc(b):
bstr = list(str(b))
if bstr.count(bstr[0]) == len(bstr):
return "can't shuffle"
max_ct = len(bstr)*10
# do while
r.shuffle(bstr)
b_shuf = int(''.join(bstr))
diff = abs(b-b_shuf)
bstr = list(str(diff))
i = r.randint(0,len(bstr)-1)
while bstr[i] == '0' or bstr[i] == '9':
i = r.randint(0,len(bstr)-1)
bstr = list(str(b))
r.shuffle(bstr)
b_shuf = int(''.join(bstr))
diff = abs(b-b_shuf)
bstr = list(str(diff))
max_ct -=1
if max_ct == 0:
return "bollocks!"
del(bstr[i])
s = sum(map(int,bstr))
i =0
while (i+s) % 9 != 0:
i+=1
return bstr,i,diff
When I was a kid, I loved the Mathemagic book [1] IIRC this was one of the tricks described in the book.
[1]
https://www.amazon.com/Mathemagic-Raymond-Blum/dp/0806983558
This guy must go to parties with Good Will Hunting.
When he said "now subtract the two 4 digit numbers" I felt a whole room full of eyes glaze over.
That sort of trick is a terrible opener. You want something simple for an opener. A 'tweener' trick yeah something like this is OK, but read your audience. I may have binge watched all of scam school.
On the most recent episode of Penn & Teller's "Fool Us" a magician fooled them with a simple mathematical trick that could probably be done as a party trick.
This episode was filmed under COVID restrictions so the magician was appearing via video but earlier he had a box delivered to the studio. He asked Penn to open the box and remove the top item. It was a big foam rock.
Then he had Teller look in and remove another item, which turned out to be a big pair of scissors.
Finally, he had Alyson (the host) remove the big wad of paper that was left.
He then wrote a prediction on a big piece of paper, and told one of them (Penn I think) to pick any two of the items and have the people holding those items to swap them. This was done, and then the prediction was revealed which was something like "Penn loses to Teller" which was correct.
He then wrote another prediction down, and this time had them do two swaps. Again the prediction was right.
Then he had them do more swaps, revealed that he had not written a prediction down instead saying he made his final prediction before starting the trick. He asked them to un-wad the paper and sure enough, it contained a prediction and it was right.
Penn and Teller thought the trick worked using multiple outs. E.g., if the final configuration did not work with what was written on the wadded up paper (which until he told them to un-wad it they had no idea contained a prediction) he had some other prediction hidden in the box that he would have revealed instead.
In reality, given three people, P, T, and A, holding in some order rock, paper, scissors, there are only two possible configurations as far as who beats whom goes: P > T > A > P and P < T < A < P.
Any swap toggles between these two configurations. The magician controlled packing the box and told them how to unpack it, so he knew it would start in P>T>A>P. So he writes down "Teller beats Penn", gets them to do a swap, and the prediction is right. It doesn't matter which two do the swap.
Same for the rest of the rounds. He knows how many swaps he is going to ask them to do, so he knows which of the two states they will end up in.
The only contingency he has to deal with is if when he asks Teller to remove an item from box Teller had grabbed the wad of paper instead of the scissors. That's probably not a big risk because Teller would probably think the paper was just padding for shipping.
If that does happen, the magician just needs to get in an odd number of extra swaps before going to the prediction on the wad of paper.
It was a bit surprising that this fooled Penn and Teller because they have had in their act tricks that used the same general idea of something having only a couple of states so that while it seems like the person they are doing the trick to is making choices leading to a combinatorial explosion in possible outcomes it really is only flipping a bit.
I can't wait to trick my drunk friends with this one
Watch "Last year at Marienbad" for the ultimate math party game :-)
Oh thanks for getting me spending about an hour on Wikipedia reading about this movie, the relation to Godard movies, and so much else :-)
Will definitely watch this one soon!