From reader Noble Abraham comes this question:

Is it possible to create a pool of distinct four digit numbers, so that the sum of 8 randomly chosen numbers from this pool is 35712? If yes, how?

I posted this in Quora also.

The basis of this question is actually a performance by a magician. He first wrote down his prediction on a sheet of paper. Then he picks someone from the audience to randomly select 8 four digit numbers from about 1000 (according to him) distinct four digit numbers kept in a bowl. The magician then adds these 8 numbers and finds the sum as 35712, which happens to be his prediction as well. I feel that the pool of numbers has a high probability that sum of any 8 numbers is 35712. Also, think 35712 can be replaced by any 5 digit number, and re-create the pool accordingly.

Am I wrong in asking the question. Or should I think that the fellow has got magical powers. He has chosen at least 11/12 numbers from the bowl that were distinct; 8 for performing the magic and 3/4 to show that they are distinct, before actually going in to the act.

Thank you so much for your valuable time.

You came to the right place because I have studied “Mentalism” i.e. mental magic, i.e. tricks which make it appear that one has paranormal powers, for some time. The top lesson gleaned from years of reading is that the audience never remembers what happened.

I have confirmed this wisdom time and again, both in my own “performances” and in the work of professional magicians. So I hope you won’t take it badly if I suggest you might not have perfectly recalled the exact sequence of events.

Now, there are three broad ways this trick can be accomplished. I won’t tell you exactly how the first two work, but we’ll try to figure out the third. It will be obvious that the methods in all three can be mixed.

The first is substitution. That is, the picking is genuine, as it appears. The magician really does have a bowl of many 4-digit numbers, all different. Eight are pulled out and somebody—probably not him—does the addition. Usually, the magician will have it done on a large chalkboard so that the audience can see the numbers, or he’ll have one or two people use calculators so that the sum can be verified. It helps (but it not necessary) the magician to be able to add rapidly. This is a skill that can be mastered easily.

The sum is announced, perhaps double-checked, then the magician reveals his “prediction”, which is found to match. Off the top of my head, I can name about a dozen ways that the magician “swaps” the prepared prediction with another “prediction” he made after learning the true sum.

The second is a force. Here, the picking is not genuine, despite appearances. There are literally hundreds of ways, with new ones invented monthly, whereby a magician can make it seem that you have a free choice where in reality the outcome is predetermined. (Sort of like how neurologists view all of human behavior.) Of course, he needn’t force every number of the eight picked. He would only have to force enough of them to cause the final solution to belong to a small set (a dozen or less) of possible sums. He could then use any number of methods to reveal the “prediction.” A cheesy, but effective, way to do this is to have, say, four different predictions in your suit pockets, two in the breast and two in the outside pockets: you pull out the one that was the final force. A miracle!

The third is mathematical. That is, the 4-digit numbers on the slips are designed such that picking eight of them force a predetermined single sum of 35712. There cannot be 10,000 slips with all the 4-digit numbers 1000, 1001, 1002,…, 9998, 9999 because, of course, by chance you could end up with the sum 8028 (the lowest possible) or 79964 (the highest possible). So the bowl must be loaded with slips such that the sum is fixed. If such a set of numbers exists, such a trick would be called “self working.” If this method is used, I’d lay my money on a set of sums, not just 35712, married to a substitution.

I saw the solution posted on Quora—the writer there suggests labeling all slips “4464”. This works mathematically, but it makes for a poor performance. You can get away with it only by walking to 8 different, widely separated audience members, have them silently pick a number, then have them add it to the sum shown on, say, a calculator. If you’re blustery enough you might get away with this, but chances are you’ll get caught.

So I leave it for a homework problem for everybody. Does a set of N different 4-digit numbers exist such that pulling 8 out of N leads to a sum of 35712 every time?

Categories: Fun

Only if N = 8.

I’d agree with Mr. Sears.

Think about the state of affairs after drawing 7 numbers: there is only one number that can be added to the sum of the 7 to total 35712. Therefore, the last draw must depend on the previous 7. There would have to be some mechanism for representing the state of the system and constraining the choice of the final number.

If N = 9, then, after 7 draws, there is still only 1 number that can bring our total to the required 35712, yet there is a choice of 2. Unless those 2 numbers are the same, you would expect the wrong sum 50% of the time, were you to repeat the experiment a large number of times.

Therefore, the 2 numbers must be the same, which contradicts our requirement that the 9 numbers be distinct.

That’s settled then … it’s magic!

Speaking of magic numerical feats: I’ve heard of individuals who, by employing a Conjuring Interval in their Mind Magic, are able to produce a number with amazing properties. But I’ve also heard in other circles that isn’t magic all; simply Slight of Mind.

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Watched a program on paranormal debunking where the host had 8 (I think) different people give him a number. He predicted the sum in advance. Each had a card in their possession which when they turned over had the number they had given. Said magic wasn’t explained.

I’m certain there are sets of numbers adding to N and maybe the result could be forced by switching bowls each containing slips of a single value. OTOH I once saw Penn & Teller stuff a fortune cookie so calculating the answer in real time then stuffing the prediction isn’t out of the question.

I’m sure that I have seen Derren Brown doing similar tricks but don’t recall him saying how he did it.

I had a feeling that 35712 would divide by a large power of 2 and factorised it to 128*9*31. If the individual numbers were also divisible by high powers of 2 then this might be a clue as t how the trick was done if it was only last number that would be “forced” to take a particular value to make the total add up.

Derren Brown famously did a live performance where throughout the show he had been subliminally imprinting various terms on the audience’s subconcious, including some numbers and a choice of daily newspaper, so later on when he asked audience members to choose a newspaper, article, line, and word, they chose exactly what he wanted. It was interesting because after the fact he showed flashbacks of the points where he had said these things, and in retrospect, in isolation they were blindingly obvious. In context, though, viewers just forgive the occasional slip of the tongue, not realizing it’s deliberate.

I think that the best example of Derren leading people in their choices of what to say or do was with the blind athlete asked to describe a past event. Derren had written down fake details of the event and got the guy to reproduce these details as he recounted the event. When presented with Derren’s “predictions” in braille, the guy was left wondering how Derren had known all the details beforehand. You can rule out what the participant saw in this case so you have the full trick in audio.

It is worth noting that Derren usually filters his participants down to the most easily led bad liars to get the best results and you only see the successes on TV. He is still quite successful in live performances though; far better than “real” TV psychics when doing their exact specialism.

So I leave it for a homework problem for everybody. Does a set of N different 4-digit numbers exist such that pulling 8 out of N leads to a sum of 35712 every time?No (unless N=8). In every set of 8 numbers, replacing one of the 8 with a number not in the set gives a sum different from that of the original set.

George, your mentioning of that famous show by Derren Brown reminds me of the other occasion when he planted the fiendish suggestion in the audience’s minds that they would forget they had been to his show immediately they left. Of course they did. People who were interviewed in the foyer immediately afterwards had no idea what they had just been to.