# How do you calculate higher powers of numbers?

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#1 31 December, 2012 - 08:15
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#### How do you calculate higher powers of numbers?

This is a somewhat old discussion (from Brainboard) which I just want to give to another group of interested persons. It was started by a numbers friend (me) who had seen Rüdiger Gamm and was astonished about his skills. For some reason, higher powers can not be solved by shorter ways and are a great challenge. Just for fun, how would you calculate higher powers in your head?

31 December, 2012 - 13:49
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Do you know the URL of the thread on Brainboard? Maybe we could read it with Google Translate.

31 December, 2012 - 19:10
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The mental multiplication is explained here:
http://superconscious1.blogspot.nl/2008/01/mental-multiplication.html

Now if you want to do this from left to right, do the steps in the order 5, 4, 3, 2, 1 instead of 1, 2, 3, 4, 5.

The best part is the criss-cross method.

1 January, 2013 - 08:41
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In this video he shows to calculated 83 to the n-th power, starting with 1, then 2 etc.

Here is how to do this. You will see that memory is extremely important in calculating this.

83 ^1 = 83.
83^2 = 83X83.
In your mind put these 2 numbers under each other:
83
83
__ X

Now we split the calculation up into part, starting from the parts that give the numbers tot he left, which are the numbers that give us the highest amount.

83 X 83 =
80 X 80 +
3 X 80+
80 X 3 +
3 X 3.

80 X 80 = 6400
3 X 80 = 240
80 X 3 = 240
3 X 3 = 9

Now do the addition from left to right: 6889

Again without the zero's:

83 X
83

First only look at the leftmost numbers
8 X 8 = 64
There might be a carry, so just remember '64'.

second step. Move your attention one digit to the right. First do this for the multiplicand:
.3 X
8.
In your mind see the 3 and the 8, do the multiplication and remember 24.

then for the multiplier:
8. X
.3
In your mind see the 3 and the 8, do the multiplication and remember 24.

Now add the number from the previous 2 steps: 24 +24 = 48. Now add the 48 to the 64, but remember that since we shifted 1 digit the 48 is ten times lower than the 64. In your mind, see this:
64....
.48...+
688

The last step is 3 X 3 = 9.
688.
...9+=

6889.

1 January, 2013 - 11:39
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Next step. 83^3.

Remember from the previous post that 83^2 = 6889.
83^3 = 6889 X 83.

6889
83 X

In order to understand the algorithm, let's break apart the calculation.
6889 =
6000 +
800 +
80 +
9

and 83 =
80 +
3

So 6889 X 83 =

6000 X 80 +
800 X 80 +
80 X 80 +
9 X 80
+
6000 X 3 +
800 X 3 +
80 X 3 +
9 X 3

This is the exact same as:

6 X 8 X 10.000 +
8 X 8 X 1.000 +
8 X 8 X 100 +
9 X 8 X 10 +
6 X 3 X 1.000 +
8 X 3 X 100 +
8 X 3 X 10 +
9 X 3

Now regroup so that the 10.000's are grouped together, the 1000's are grouped together, the 100's are grouped together etc.:
6 X 8 X 10.000 +

6 X 3 X 1.000 +
8 X 8 X 1.000 +

8 X 8 X 100 +
8 X 3 X 100 +

9 X 8 X 10 +
8 X 3 X 10 +

9 X 3

Now go back to the calculation:
6889
83 X

See if you can spot the 10.000's, the 1.000, the 100's, the 10's and the singles directly from the numbers. If you can, you now understand why it is called criss-cross multiplication.
If you cannot spot these, i'll try to show you like this:

10.000's:
6xxx
8x X

1.000's:
6xxx
x3 X

and:
x8xx
8x X (see the cross?)

100's:

x8xx
x3 X

xx8x
8x X (see the cross?)

10's :
xx8x
x3 X

xxx9
8x X (3rd cross)

and, last, the singles:

xxx9
x3 X

Now, can you do the calculation in your head from left to right?

1 January, 2013 - 14:01
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Thats an interesting way to split it in groups. However, doing a fifth becomes challenging and going higher starts to be too though, at least for me.

The discussions on brainboard were:

Rüdiger says that he puts numbers together but this can not be proven by maths but looks like a mnemotechnical aid he uses. This way he finds the ending of 75^6 in 45^50. If anyone can proof that it would be great.

1 January, 2013 - 23:24
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RetoCH wrote:
For some reason, higher powers can not be solved by shorter ways and are a great challenge. Just for fun, how would you calculate higher powers in your head?

If you want 100% accuracy, this is a major challenge.
On Brainboard I saw somebody ask about 17^200 and I have no idea how to do this with 100% accuracy.
The answer is 247 numbers long!
Using a full, precision calculator we see that 17^200 = 1229657818953615363634064177522701398683653711601707664057125506857630498777292504373533929451303187906364621481616464391925225392475534663863941904565705353494294088812262655119980838406538250506869340806656182146496426397201584783786258459216001

A normal scientific calculator would give you this: 1.22965781895361e+246
And this number is derived from logaritms and polynomals.

Using logarithms you can calculate the answer to maybe 3-5 numbers precision, which is very impressive already in my book.

1 January, 2013 - 23:28
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How would you like this thread to continue? I could talk about how to do these numbers using logarithms or do you want to stay with 100% accuracy?

2 January, 2013 - 02:35
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As you already pointed out, 100 percent accuracy is not the way to go for really high ones. I wonder if you can show a way to reach a 5.th to 10.th power in the head with 2 digits. That seems to be doable but IS a major achievement.

Even remembering all this powers is not easy :-o.

3 January, 2013 - 13:08
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Let's do some mental magic.
But first a bit of theory about logarithms.

Logarithms change multiplication into addition, division into subtraction, powers into multiplication and roots into division.

If you can mentally multiply, you can calculate higher powers using logarithms!

Ok, let's give some properties. All logarithms are base 10.

log(aXb) = log(a) + log(b) => Multiplication become addition.

log(a/b) = log(a) - log(b) => Division becomes subtraction.

log(a^b) = log(a) X b => Taking powers become multiplication.

An example:

If we want to calculate 18^5, we take the log(18), multiply this with 5 and then take the anti-log or calculate 10^answer.

This sounds more difficult than it is! I will first show you how to do this on paper, after that we will see how many numbers we need to memorize in order to calculate this mentally.

so, 18^5:

We take the log(18) = 1,2553....

multiply this by 5:

1,2553 X 5 = 6,2765

The 6 from the answer means million and the 0,2756 gives you the numbers. 10^0,2765 = 1,89, so the answer is 1,890,000.

Full accuracy gives: 18^5 = 1,889,568.

Now let's do this mentally.

First we memorize all numbers 1-10:

log 1 = 0

log 2 = 0.30103

log 3= 0,4771

log 4= log 2 + log 2 = 0.6021

log 5= log(10/2) = 1 - log 2 = 0.6990

log 6= log 2 + log 3 = 0.7882

log 7= 0.8451

log 8= 3 X log 2 = 0.9031

log 9 = 2 X log 3 = 0.9542

log 10 = 1

So if we memorize 2, 3, and 7, we can calculate the rest. Start doing this daily to get used to quickly going from the number to the logarithm and back.

2 more important numbers:

log 0.9 = log 9/10 = log 9 - log 10 = 0.9542 - 1 = -.0458.

log 1.1 = 0.0414. Remember this a 'plus 10%'

log 1.01 = 0.0043. Remember this a 'plus 1%

These 2 are used to calculate/interpolate numbers between the ones you memorized.

Let's do 18^5 mentally.

18 = 20 X 0.9.

The log is log 20 + log 0.9

log 2 = 0.30103, so log 20 = 1.3010.

log 18 = 1.3010 - 0.0458 = 1.2552.

Alternatively you could have added log 2 and log 9.

1.2552 X 5 = 6.2760 (divide 1.2552 by 2 and move the decimal point)

6 means million, 0.2760 gives us the numbers. Let's focus on the 0.2760

log 2 = 0.3010, so it is lower than 2.

0.301 - 0.276 - 0.025.

0.025 / 0.0043 = almost 6, so it is less than 6% lower than 2. Remember that 0.0043 was plus 1%.

2 minus 6% = 1,88. It was a little lower than 6%, so let's round this up to 1.89 to account for the difference.

Full accuracy gives: 18^5 = 1,889,568.

Another one. Let's do 33^5.

Log 33 = log 30 + log 1.1 (remember that log 1.1 was 'plus 10%').

log 30 = 1 + log 3 = 1.4771.
log 1.1 = 0.0414.
add those 2: 1.4771 + .0414 = 1.5185

5 X 1.5185 = 7.5925 (divide by 2)
7 means 10 million and focus on 0.5925

log 4 = 0.6021, so it is almost 4
0.6021 - 0.5925 = 0.0096.
.0096 / .0043 = 2 1/4, so 2 1/4 percent.

4 minus 2 1/4 % = 3.9.

3.9 x 10 million = 39 million.

Full accuracy gives 33^5 = 39,135,393.

4 January, 2013 - 13:56
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Let's do a couple more:
Recall that:
log 1 = 0
log 2 = 0.30103
log 3= 0,4771
log 4= log 2 + log 2 = 0.6021
log 5= log(10/2) = 1 - log 2 = 0.6990
log 6= log 2 + log 3 = 0.7882
log 7= 0.8451
log 8= 3 X log 2 = 0.9031
log 9 = 2 X log 3 = 0.9542
log 10 = 1

log 0.9 = log 9/10 = log 9 - log 10 = 0.9542 - 1 = -.0458.
log 1.1 = 0.0414. Remember this a 'plus 10%'
log 1.01 = 0.0043. Remember this a 'plus 1%

For example:
22^5.

22 = 10*2*1.1. So log 22 = log10 + log2 +log1.1 = 1 + 0.30103 + 0.0414 = 1.3424.
1.3424 X 5 = 6.712.
6 means million and 0.712 is close to 0.699. 10^0.699 is 5, so it is a little over 5. How much?
.712 - .699 = .013. .013 / .0043 = 3. So it is 3% over 5 = 5.15.

22^5 = 5,153,632. See how close we are?

Here is a tip if you want to get proficient with this stuff.
Memorize them without. So log2 = 3010, 1% = 43, 10% = 414
Learn how to quickly interpolate:

1.2 = 1 +10% +10% - 1% = 414 +414 - 43 = 785.
log 1.2 = 0,0792

If you can quickly guess log1.2, you can calculate log24 when you know log20.

Let's do another one.
37^5
37 = 40 - 7.5%
40 = 16020
If -10% = 458, then -7.5% = 458/4X3 = 114X3 = 342
16020 - 342 = 15678
15678 X 5 = 78390
7 means 10 million, so 8390 gives us the numbers
8451 = log7.
8451 - 8390 = 61
61/43 = 1.4
So we look for 1.4% less than 7 = 6.9.
The number is 69 million.
37^5=69,343,957. See how close we are?

Does this all make sense? It is a lot of addition and multiplication and a bit of linear interpolation.
Nothing that you cannot do in your head with a little bit of practice and a good memory.

First learn all logs below 10, then by using the 10% and 1% numbers, guess 11, 12, 13 etc.
If you want to really impress people, memorize all logs below 100.
This will give you extreme speed and a lot less need for interpolation.

6 January, 2013 - 13:16
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Thanks for a doable way to calculate higher ones without the need to memorize a lot.

7 March, 2014 - 14:46
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Hi Reto,

I saw Rudiger do his show again.
I am convinced now that the way he does this is by memorizing the numbers.
There is no calculation going on when he recites the higher powers.

8 March, 2014 - 16:37
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I just saw this video yesterday:

19 March, 2014 - 01:51
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Kinma wrote:
I am convinced now that the way he does this is by memorizing the numbers.
There is no calculation going on when he recites the higher powers.

To do it by memorization, how many numbers would one have to memorize? In the video above, he probably would have needed to prepare for up to at least 8520. If it's memorized, the recall speed is mindblowing. This is from the video:

7312 = 22,902,048,046,490,258,711,521
5417 = 282,288,975,128,239,507,545,882,230,784
7717 = 117,582,402,033,097,174,749,136,828,787,597
8320 = 240,747,534,123,068,430,155,382,383,568,465,195,601
7216 = 521,578,814,501,447,328,359,509,917,696

(I didn't double check for typos...)

8320 would require 13 three-digit images memorized with names.

Regardless of whether it's calculated or memorized, it's one of the greatest mental feats I've ever seen. I've watched the video at least 20 times already. :)

19 March, 2014 - 04:57
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Josh Cohen wrote:

To do it by memorization, how many numbers would one have to memorize?

Assuming he knows all numbers from 0-100 to the maximum power of 20, that would be 94,851 digits!
Staggering indeed.

19 March, 2014 - 05:14
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He has a special brain indeed. It is not known how exactly he does this.
I have had discussions in the past with Reto how he does this and we both don't exactly know.

We know how to make this calculations, but not with 100% accuracy.
He himself does not say either.

21 March, 2014 - 20:53
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At 11:00 in the video below, he talks a little bit about technique:

21 March, 2014 - 21:23
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Josh, Can you write out his saying about the technique? I'm quite bad at listening english :<

22 March, 2014 - 01:29
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He explains that he uses memorization and calculation.
He uses memorization foremost, then calculation

He explains that he memorized the powers of 300,000 numbers.

He uses memorized numbers as a tool for calculating. And gives an example for square roots.
If you memorize the root of 45 and you want to calculate the root of 11.25, you take the memorized number and divide it by 2.

He does not explain how he memorizes the 300,000 numbers btw. Just that he does.

22 March, 2014 - 20:03
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wow. It's great task to master powers of 300,000 numbers. Like impossible mission.

25 March, 2014 - 02:15
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Keep in mind, he can do the first 20 powers really FAST: That doesnt mean he can do more, you`ll find much longer recitations (up to the power of 200!). If you look how much digits that is, you get near a million of digits- the question about HOW to store such a number still is a miracle.

Great to see the discussion coming up again!

7 April, 2014 - 04:22
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To make things more human, in a somewhat weird japanese championships, Rüdiger made a mistake in "putting together" the 17.th power of a number. As far as my knowledge goes, this was the first reported error ever (!) in higher powers.

Learning powers is just plain hard and -possibly- boring. Still working up to the fifth...

1 June, 2015 - 11:59
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While I'm waiting for 3x3 multiplication to get easier I thought I might start to play with logarithms.

The long string addition and subtraction seems like a good memory practice for the major system and a good stretch for holding longer numbers in my head.

Has anyone created practice tools software/spreadsheet for calculating logs mentally. I'm sure I can make something in google docs or excel but if it's already out there it would be nice to have. I prefer practicing the skill to building the tools for this hobby :)

thanks,
Robert

1 June, 2015 - 13:13
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With regard to some of the exponent calculations above it would seem to me a lot easier to use

a^2 + b^2 + 2ab + b^2 ... a^2 + b^2 + c^2 + 2ab + 2ac + 2ac .. etc for

44^2 444^2...
lots of easy steps this way and you can just keep expanding upwards by 2's

similarly
for sum of cubes
a^3 + b^3 = (a + b) (a^2 - ab + b^2)

i.e. 84^3

I'm pretty sure Euler has a formula for sums of fourth and fifth powers in his book on elements of algebra. I'll have to check this evening.

Is there a general algebraic solution to sums of powers for decomposing....

xyz ^ n such that
x^n f(j) + y^n f(j) + z^n f(j) + f(xyz) ...

i.e. 123^12

3 June, 2015 - 11:26
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RobertFontaine wrote:
Has anyone created practice tools software/spreadsheet for calculating logs mentally.

4 June, 2015 - 00:46
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"I'm pretty sure Euler has a formula for sums of fourth and fifth powers"

Robert, from the Binomial theorem we can derive formulas
for all kinds of powers of (a+b)^n, for n integer

The expansions get complicated after 4th power,but all those formulas definitely exist.

e.g. we should find that cube of 594245 = 84.072736. (8 correct, rounded). One such correct result was marked with 8 points. For all 10 tasks the maximum was 80 points. The gold medal winner Jan van Koningsveld, got 50 points, that's an average of 5 correct cube digits. I got a 7th place with 33 points. The task was only 10minutes, that's why in order to save time for calculating decimals,I memorised the first 100 cubes. But I in case I forgot any cube, I also had in mind the formulas in order to calculate, in case needed
a^3 + b^3 = (a + b) (a^2 - ab + b^2)
and (a+b)^3 = a^3 + 3*a^2*b + 3*a*b^2 + b^3
and (a-b)^3 = a^3 - 3*a^2*b + 3*a*b^2 - b^3

I talked with Andreas Berger who came 2nd then and he also told me he had memorised the 100 cubes. I know that's not calculation, but sometimes it's useful to have a few more stuff read for recall out of your mental inventory. Besides, it's only 90 new numbers (assuming that mostmental calculators know the first 10 cubes (1,8,27,64,125, 216, 343, 512, 729
And actually for all non-integer roots the fun starts after the decimals, that's where the real algorithm starts to unfold.

About the memorised cubes, since the task contain 6-digit roots I payed more attention to the cubes from 46-99, which yield a six-digit cube. It's not like we memorised 10,000+ powers like Gamm did. The task is simpler. I think with a few mnemonics those cubes can also be memorised, they seem bit like telephone numbers.

46 97336
47 103823
48 110592
49 117649

50 125000
132651
140608
148877
157464
166375

175616
185193
195112
205379
216000

226981
238328
250047
262144
274625

287496
300763
314432
328509
343000

357911
373248
389017
405224
421875

438976
456533
474552
493039
512000

531441
551368
571787
592704
614125

636056
658503
681472
704969
729000

753571
778688
804357
830584
857375
884736
912763
941192
970299

All the above have integer cube root (46-99)

Nodas

20 June, 2016 - 10:10
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To add a newer clip with some interesting answers, I finally come back to this place :-).

This is Rüdiger interviewed by Nelson Dellis.

26 June, 2016 - 11:42
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