Divisible by Seven (worst card trick ever?) - Numberphile

Featuring Professor Tony Padilla.

NUMBERPHILE

Videos by Brady Haran

[AgentM]

We had Parker Squares... Is this a Tony Trick?

[martinepstein]

People like to point out that he misses the occasional obvious multiple of 7 when working his algorithm but I don't blame him. If he took an extra second to look for tricks at every step he could lose time in the long run, and if he's here to show us an algorithm he might as well use it.

[ds]

It's either this or the Parker's magic square. Penn & Teller level.

[TrimutiusToo]

He is bad at this. 351428 is of course divisible by 7 because 35, 14 and 28 are divisible by 7...

[handsome_man]

did this trick in some bars last night. I was pulling chicks like there was no tomorrow

[hallfiry]

This video might need an edit. At 3:58, you can read his password from his fingers...

[esotericVideos]

That's a Parker Square of a magic trick

[liesalllies]

Numberphile cringe. I can sense the regret from behind the camera. Love it 😂

[EllipticGeometry]

I had a slightly different algorithm in mind. We know 351429≡1(mod 7). You can obviously subtract the remainder to get a multiple of 7. The result is 351428.

Moving on to the tens digit. Take the remainder from the ones digit (1) and multiply by 5, yielding 5. You can use −2 or anything else congruent mod 7 if you prefer. Remember this number. Subtract it from the tens digit. That yields 351379 which also changes another digit. Fix that by adding 7 to the tens digit: 351449. You may be free to add or subtract another 7: what matters is that you don’t carry or borrow.

Moving on to the hundreds digit. Take the 5 you remembered and multiply by 5 again, yielding 25≡4(mod 7). Now remember this 4. Subtract it from the hundreds digit: 351029.

Moving on to the thousands digit. Take the 4 you remembered and multiply by 5 again, yielding 20≡6(mod 7). Now remember this 6. Subtract it from the thousands digit and add 7 to compensate for the borrow: 352429. And so on.

What I like about Tony’s version is that you’re more in touch with the numbers. Mine has the advantage that you can efficiently skip to any digit, even if the number is googolplex-sized. Observe that 5⁶≡1(mod 7): to find by what to multiply the remainder so you can subtract it from a given digit, all you need to know is the remainder of that digit position divided by 6. And instead of a googolplex, you’re dealing with a mere log(googolplex)=googol. You can find the remainder of that in log(googol)=100 time. Long division won’t take that long, but you can combine the digit sum and evenness using the Chinese remainder theorem to save more time. Now enjoy your walk over to the digit you want to modify. That will still take a googol-type eternity. Just a much shorter eternity than when you have to work on all digits along the way.

[Gairhoth]

It’s probably easier to just divide by 7 by long division

[florencefortyseven]

*That was a little bit unclear, so I went through the solution myself, and I just want to share the notes I made.*

The last two digits of the first number are 28. 28 is 0 mod 7. So you need to find a number, ending with 9, that is 0 mod 7. That is, of course, 49. Put a 4 down.

Shift one position. Look at the second-to-last digit and the third-to-last digit. That makes 44. That's 2 mod 7. So you need to find a number, ending with 2, that's 2 mod 7. That is, of course, 72. Put a 7 down.

Shift one position. 17. That is 3 mod 7. So you need to find a number, ending with 4, that is 3 mod 7. That is, of course, 24. Follow the same pattern. Put a 2 down.

Shift one position. 52 is 3 mod 7. That gives you 31. Put a 3 down.

Shift one position. 33 is 5 mod 7. That gives you 75. Put a 7 down and there you are.

*If you practice it, you could totally do it on command.*

[brodaclop]

What makes this trick slightly less impressive when you do it modulo 9 is that the answer is the exact same card in every single row.

This is genius.

[jnthegreat]

you can't do this for ANY number. to be divisible by 2, for example, or 5. if a large number ends in anything but a 0 or 5 there's nothing you can do to turn it into a number divisible by 5 by changing a middle digit.

[filipsperl]

So what happens after you put the cards in the array? Where do you see wheter the original number is divisible or not?

[tomd]

Instead of this card trick 🙄 to see if a number is a multiple of 7, you could convert it to octal and cross add that. For example, 102 is 146 in base 8. 1+4+6=11. 11 is not a multiple of 7, so neither is 102. Conversely, 84 in base 10 is 124 in base 8. 1+2+4 = 7, which is of course a multiple of 7; therefore, so is 84.

It works by the same principle and for the same reasons as the divisibility by 9. In fact, you can do the cross-sum divisibility check for any number b-1, where b is the base > 3. So, to check for divisibility by 11, you can convert the number to base 12, and if the resultant cross-sum is a multiple of 11, then so is the original number.

[rangerocket]

0:59 [INAUDIBLE]you just flip one of the digits? ya I will.

[hymnz]

What’s the so big deal about divisibility by 7.. Separate the last digit, double it and subtract it from the remaining part. If the answer is 0 or divisible by 7, the full number is divisible by 7.. Example: 4046 -> 404-12 -> 392 -> 39-4 -> 35 (divisible by 7) => 4046 is divisible by 7 (17*2*17*7) ; another 214, 301, 528 -> 21, 430, 136 -> 2, 143, 001 -> 214, 298 -> 21, 413 -> 2, 135 -> 203 -> 14 => divisible by 7

[martinepstein]

I'm rehashing my challenge from the other video.

You start with a number that has more than 3 digits, and you take the alternating sum of groups of 3 digits starting from the right. For example, 4, 647 -> 647 - 4 = 643. There are three prime numbers p such that the new number is divisible by p if and only if the original number was divisible by p. Tony explained that one of those primes is 7. What are the other two?

[rgqwerty]

Im a bit confused. Where do the numbers in the array with the gaps come from? Just random?

[zepatrik]

You could speed up the process by having a deck of cards with every card attached to 5 other cards of the same type to it (e.g. with a rubber band) so then you can just spread them out on the table to create the 2d array