This came in a magic trick box that our daughter picked for her birthday. Exhausted from all the Bar Mitzvah preparations for her older brother and her own demands for a Harry Potter birthday party, we didn't have any time to buy her a present or even think of anything original to serve as a present. As soon as she realized that no wrapped packages should be expected, she pragmatically proposed that on our next trip to a toy store she will pick something that she likes, I will wrap it and we all pretend it was a surprise.

She chose two sale boxes: one with a hundreds of hip-hop tattoos and another with magic tools. Among dice, fake nails and blood in the magic box there were these cards.

Now, pay attention, because it starts to be complex.

Our daughter, the magician, would ask you to think of a number between 1 and 63. Do not reveal this number!

Then, she will show you these cards, one-by-one.

After seeing each card you will say whether the number you picked is on this card or not.

And then, after these 6 yes-no answers, she will guess your number!

Let's try.

I pick 23.

A-yes

B-yes

C-yes

D-no

E-yes

F-no

She magically guesses: 23!

How does she know?

Quite simple. The instructions on this magic set tell her to add up all the left corner numbers from the "yes" cards.

A+B+C+E

1+2+4+16=23

Unbelievable!

Try it yourself and think how this could possibly work. Two puzzle points for any lead here.

Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.

## 9 comments:

The top left number of each card is a power of 2. By saying yes or no, you tell if that power of 2 is in your number.

23=2^0+2^1+2^2+2^4.

Notice that here 2^3 is missing, because you said "no" to the D card.

Any number can be written as a sum of powers of 2, those cards will work for any number from 0 (yes, 0) to 63.

This is using binary decomposition. Each card represents all the numbers in the 1-63 range that have a "1" in a specific position of a 6-digit binary number. Each card's first number represents the corresponding "base" binary, i.e. 1, 10, 100, 1000, etc. When all such base binaries are summed up, we basically get the full value. For example, 23 is 010111, which means it's a sum of these binaries:

1 (A)

10 (B)

100 (C)

10000 (E)

Neat trick, indeed!

Welcome back! We all missed you, I'm sure.

This trick works on binary, a mathematical language whose entire vocabulary consists of only two words -- in this case yes or no.

Yes means the two to the power of ? is present.

No means that it is not. If you have a binary number like

100101 what you are being told is that

2^0 is present so you have a 1

2^1 is not present you have a 0

2^2 is present so you have a 4

2^3 is not present

2^4 is not present

2^5 is present so you have a 32

The total is 37. That is the unique representation of 37 in binary.

The cards will record 37 on the first card (A) and the 3rd card (C) and on the sixth card (F) 32. All other cards do not contain the "magic" binary number.

All numbers from 1 to 63 are uniquely represented on the 6 cards.

Notice 63 is on all 6 cards. That's because 63 is 111111 in binary.

If we add the first number found on each card B,C,D,E and F, the result is an even number so to guess an odd number, we have to add one figuring on the card A. Every uneven number figures on card A.

Now, an even number could be written 2 n with n, natural number.

When n is 1, 2 n is 2 and 2 is also the only even prime number. Some prime numbers such as 3, 5 or 11 are going to be on this card. 7 for example is 1+2+8 so 7 is also found on card D where n is 4.

6 is 2+4, there is no need to cover n=3 because the numbers divisible by 3 have been covered.

This game uses the property of numbers, divisibility as well as addition and multiplication.

I really like that! Thanks for sharing.

I think my son has these same magic number cards.

The left top number on each of the cards is a power of 2.

2^0 = 1; 2^1 = 2; 2^2 = 4; 2^3 = 8; 2^4 = 16; 2^5 = 32

The numbers in this problem range from 1 to 63. Each of the numbers 1 to 63 can be expressed as a sum of some of the powers of 2 from 2^0 (=1) to 2^5 (=32). 63 is sum of all the these powers of 2 (1+2+4+8+16+32).

Each number = d0*(2^0) + d1*(2^1) + d2*(2^2)+d3*(2^3) + d4*(2^4) + d5*(2^5), where each di = 0 or 1; i=0....5

Each card i starts with a power of 2 (2^i).

The sequence of numbers on each card is in sequential order by 1 up to 63, except for that after every (2^i)th number, the numbers skip by 2^i + 1.

TracyZ

oops ....

I meant to write that for each card j (j=1,2,3,4,5,6),

the card numbers start at 2^(j-1), and go sequentially with increases of 1, except for that after every 2^(j-1)th number, the numbers skip by 2^(j-1)+1.

For card 1(A): the numbers are all the odd numbers from 1 to 63.

For card 2(B): the numbers start at 2,3, then skip by 3 to 6,7, then skip by 3 to 10, ....

For card 3(C), the numbers start at 4,5,6,7, then skip by 5 ((2^2)+1) to 12,13,14,15, then skip to 20...

and so on.

For card 6(F), the numbers start at 32, and go sequentially by 1 to 63, and never skip because 2^5+1=33, and the sequence ends before it gets there.

TracyZ

Wow, you are smart!

You explained it better than I could.

Anne-Marie - I am not sure I got your answer but I think you come very close to discovering the power of 2 trick through addition and multiplication.

A clean example of math doing magic here. Two puzzle point for each of you!

Wow Its really incredible . I think this is not a magic tricks . I think it's a mathematical tricks.

Thanks for this great tricks. many of my friend know this . but i don't . Now i learned it . Thanks to share this post .

easy magic tricks

Thanks for sharing. It's really awesome trick. Magic tricks revealed from quite long time. Everyone these days all the time wonders how magicians perform all those fascinating, unimaginable magic tips and illusions?

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