Analyzing Rüdiger Gamms performance (division by 109)

Mental Calculation, Mathematics, and Puzzles
1

While watching YouTube videos’of Rüdiger Gamms performance in order to find out how he does the higher powers I realized something.
In at least three video’s I saw him do a division by 109. There is a reason for this.

One example is this video:

The calculation starts at 0:50.
Watch the video first before reading how this is done.

So first I will show how to do this calculation mentally.
If it looks difficult, dividing by 109, consider dividing by 110 first.
For example, let’s say we need to divide 550 apples evenly over 110 buckets.
You would immediately see that this will lead to 5 apples per bucket.
If we now take 1 bucket away, and keep the 5 apples inside that bucket as remainder, we actually have divided 550 apples over 109 buckets, leading to 5 apples per bucket and a remainder of 5.

This is what we will be doing.
We will divide by 110 and make a correction to any remainder for dividing by 109 instead of 110.

In the video, Rüdiger divides 93 by 109. Let’s do this.
93 is less than 109, so we call out ‘0.’.

In your mind add a zero: 930.
Closest factor of 110 is 880 or mathematically => 880 < 930 < 990. So call out the 8.
The remainder of dividing 930 by 110 is 50, because 930 minus 880 = 50.
Now make the correction for division by 109 instead of 110. Think about the 110 buckets with apples again. We take away 1 bucket of the 110 buckets with each 8 apples, so 880/109 = 8 with a remainder of 8, making a total remainder of 50+8=58.

550/109 = 5, remainder 5
880/109 = 8, remainder 8. See the pattern?

Putting it all together. We do the division in 2 steps. First divide by 110, then correct the remainder and to keep things simple, the correction is exactly the same as the number we call out!

So 930 - 880 = 50, add the extra remainder 8, leading to a total remainder of 58. add 0: 580

580 / 110 = 550/110=5. Call out the 5, remainder = 30, add 5, total remainder is 35. Add zero.
350 / 110 = 330/110=3. Call out the 3, remainder = 20, add 3, total remainder is 23. Add zero.
230 / 110 = 220/110=2. Call out the 2, remainder = 10, add 2, total remainder is 12. Add zero.
120 / 110 = 110/110=1. Call out the 1, remainder = 10, add 1, total remainder is 11. Add zero.
110 / 110 = 110/110=1. Call out the 1, remainder = 0, add 1, total remainder is 1. Add zero.
10 / 110 = 0/110=0. Call out the 0, remainder = 10, add 0, total remainder is 10. Add zero.
100 / 110 = 0/110=0. Call out the 0, remainder = 100, add 0, total remainder is 100. Add zero.
1000 / 110 = 990/110=9. Call out the 9, remainder = 10, add 9, total remainder is 19. Add zero.
190 / 110 = 110/110=1. Call out the 1, remainder = 80, add 1, total remainder is 81. Add zero.
810 / 110 = 770/110=7. Call out the 7, remainder = 40, add 7, total remainder is 47. Add zero. etc.

Is this interesting?
If it is, I can show you 2 ways of speeding this up, making the mental calculation easier an a lot speedier!
Let me know, please.

2 Likes
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2

Thanks for a good way to explain it. For some reason, he usually divides by a prime number, which is interesting because audiences tend to give him something like 99 or another “Schnapszahl”.

3

My skeptical mind would call this ‘misdirection’.
For by calling it a prime number, it gives the number mystical properties, which in this case do not apply.

The first way to speed this up is instead of dividing by 110, divide by 11. Then, Instead of adding 5 to 80, just concatenate 5 to 8.
So in the example, think about 93.
See 88 as the nearest multiple of 11.
Subtract 93 and 88 in your mind, giving 5.
Now concatenate 8 and 5: 85.

  1. Nearest multiple of 11 is 77. Subtract: 8 concatenate the 7: 87
    87-77=10 &7 : 107
    107-99=8 & 9 :89
    89-88=1 & 8:18 etc.

See how easy this is? I bet with a bit of practice anybody on this site can do this quickly!
But there is an even quicker way…

1 Like
4

This is a good example where memory work can help to find a solution.

5

There is! 109 has a nice property. I will demonstrate this property with a much smaller number, with 7.
The decimal representation of 1/7 is
1/7 = 0.1428571428571428571428…
Now look at the decimal representation of 2/7, 3/7, 4/7, …
2/7 = 0.2857142857142857142857…
3/7 = 0.4285714285714285714285…
4/7 = 0.5714285714285714285142…
5/7 = 0.7142857142857142857142…
6/7 = 0.8571428571428571428571…
Notice something. We get the exact same digits as 1/7, just with a different starting point. The results of 2/7 or 3/7 are just cyclic permutations of 1/7. You only have to memorize the 6 digits (142857) and need a simple multiplication to calculate x/7.

I’ll give you an example: We want to calculate 5/7.

  1. We know that 1/7 is a bit bigger than 0.14.
  2. We multiply 0.14 with 5, we get 0.7. So we know that our result is a bit bigger than 0.7
  3. Now we look into our memorized digits 142857. We notice, that we have a 7 in there, and when we cycle the 7 to the beginning we get 714285. 0.71 is a bit bigger than 0.7, so we found the correct cycle.
    Now we can say, that 5/7 = 0.714285714285714285714…

Another example: 3/7

  1. 1/7 = 0.14 We know that.
  2. We multiply 0.14 with 3, this gives us 0.42. So 3/7 is a bit bigger than 0.42.
  3. We look into 142857 and see, that we have a 42 in it. So we cycle it and say:
    3/7 = 0.428571428571428571428…

I think you’ll get the point. The same thing goes with 109.
The decimal representation of 109 has 108 reapeting digits.
1/109 = 0.
00917431192660550458715596330275229357798165137614
67889908256880733944954128440366972477064220183486
23853211009174311926605504587155963302752293577981
65137614678899082568807339449541284403669724770642
20183486238532110091743119266055045871559633027522
93577981651376146788990825688073394495412844036697
2477064220183486238532110091…
The 00917431192660550458715596330275229357798165137614
67889908256880733944954128440366972477064220183486
23853211 repeats over and over. A multiple of 1/109 has the same digits, just with a different beginning.

First you have to memorize the 108 digits. You can do that with simple mnemotechnics.
Now we want to calculate 93/109, like Rüdiger Gramm.

  1. We know, that 1/109 is a bit bigger than 0.00917.
  2. We multiply 0.00917 with 93, we get ~0.852.
  3. Now we look in 00917431192660550458715596330275229357798165137614
    67889908256880733944954128440366972477064220183486
    23853211 for a 852. We find it: 00917431192660550458715596330275229357798165137614
    67889908256880733944954128440366972477064220183486
    23853211. (Notice: There is no 852 in the repeating decimals, but we have 853, which is a bit bigger than 852.)
    We cycle and get as result:
    93/109 = 0.85321100917431192660550458715596330275229357798165137614
    67889908256880733944954128440366972477064220183486
    23    85321100917…

It’s really simple.

Notice: This trick works only with numbers x, which have x-1 repeating digits.
It works with 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, …
You may wanna look at Cyclic number - Wikipedia

1 Like
6

Fantastic Kinma. I remember Tammet doing similar calculations and you calling it out here. You must be some sort of mathematical genius yourself! :wink:

7

Thanks Kinma, excellent post.

There is a good scientific paper which gives some insight into how Gamm achieves his calculation feats here:
http://psychiatry.wustl.edu/Resources/LiteratureList/2001/January/Pesenti.pdf

This confirms that he knows the digits to divisions by many periodic (cyclic) prime numbers, such as 109. The full quote is:

“He has extensive knowledge about mathematical properties of numbers. For example, he knows many periodic prime numbers, that is, prime numbers whose inverses have as many recurrent decimal positions as the prime number itself minus 1, and he knows their corresponding period. (For example, dividing 1 by 113 results in a number with 112 decimals, constituting a period that is repeated ad infinitum.)”

The paper also confirms that he knows many power multiplications (“To raise two-digit numbers to the second up to
the fifth power, R. Gamm retrieves the answers directly from memory”).

So it has been no secret that Rudiger Gamm’s calculations rely heavily on memory, since this paper was written in 2001. You wouldn’t get this impression from the media, which typically presents his most impressive looking feats, which are precisely those which rely most on the pre-memorised answers.

Calculations in mental calculation competitions, in which Rudiger Gamm has done extremely well, are a much better measure of true mental calculation ability, but the media doesn’t report on them because they don’t give such a good show. Instead, we see feats like this one which are 90% memory and 10% calculation.

8

Kinma,

The method in your original post sounds like Craig Aitken’s method for dividing by numbers ending in 9, as described at the following link:

(Under “Mental Arithmetic”, click on “Aitken’s lightning divison”.)

9

Pesenti did a good job. I read his other study as well. The possibility to build higher powers out of lower ones is either a magical trick or he speaks about his way to memorize. Is there any advantage you have if you know the first five or six powers in raising to even higher ones? It IS certainly much easier to raise numbers with more digits but Rüdiger is doing the other thing :-).

Anyone ever tried the fifth?

10

Jakube, that’s exactly it. By memorizing the sequence, just calculate the first couple of digits, tune in to the sequence and rattle digits!

Do you know this about the sequence?
If you cannot memorize the 108 digits, just memorize the first 54.
The second half can be calculated by subtracting from 9!

009174311926605504587155963302752293577981651376146788
990825688073394495412844036697247706422018348623853211
_______________________________________________________+
999999999999999999999999999999999999999999999999999999

This is also true for the 1/7 sequence:
142
857
___+
999

Greymatters, my example about the 110 buckets with apples sounds indeed very similar to Craigs Aitken’s method.
This method can be done with any division that ends in a 9. Whether this is 29, 109, or 69.
Or 28, or 108. Just double the extra remainder in these cases.

Yan, thanks for the compliment!

11

While working on my blogpost for today ( Grey Matters: Blog: Leapfrog Division II ) about “leapfrog division” for fractions with denominators ending in 1, I ran across this video of Rudiger Gamm again: https://www.youtube.com/watch?v=x5OOMmxI9bs

It starts with him mentally working out 62/167, and figured he must be using the same, or at least a very similar technique. 62/167 is equal to 186/501. Two modifications as mentioned in the original article, and we get:

186/501 becomes 186/5
186/5 becomes 185/5

Now, we work out the calculation of the decimal (subtracting the quotient from 99, instead of 9 as in my blog post):

185 ÷ 5 = 37 (remainder 0 )
062 ÷ 5 = 12 (remainder 2)
287 ÷ 5 = 57 (remainder 2)
242 ÷ 5 = 48 (remainder 2)
251 ÷ 5 = 50 (remainder 1)
149 ÷ 5 = 29 (remainder 4)
470 ÷ 5 = 94 (remainder 0 )

…and on and on. Verify these digits with Wolfram|Alpha: 62/167 - Wolfram|Alpha

I’d guess Rudiger Gamm says something like, “Make it hard…choose a prime denominator” (Probably less than 200 or so, too) That way, he eliminates all even denominators, as well as denominators ending in 5. That would give him an excellent chance of working with a denominator ending in 9 or 1, or at least getting a denominator ending in 3 or 7, which can be scaled by 3 to get denominators ending in 1 or 9.

12

I agree with the thought that he changed 167 into 501.
Or 1002. The reason being that it is common to change what looks difficult to what is easy.

I started this threat with the observation that he uses 109 in a lot of videos. This cannot be a coincidence.
Off camera he probably steers the interviewer into choosing certain numbers.

That way we cannot see why the number 109 gets chosen so often.