Zum gleichen Thema ein völlig anderer Ansatz, bin da per Zufall drüber gestolpert:
Re: [Mental Calculation] How to raise numbers to the power of n
Hi Mario.
Ok, here's an example - how to raise 38 to the power 5.
Firstly, that triangle of numbers:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
As we are seeking the fifth power of a number, we need this level as it has 6
numbers (we will need to find 6 terms, with powers of numbers ranging from 0
{not 1} to 5). These numbers are the coefficients of the terms to be created.
Total = 1x(X^5)x(Y^0) + 5x(X^4)x(Y^1) + 10x(X^3)x(Y^2) + 10x(X^2)x(Y^3) +
5x(X^1)x(Y^4) + 1x(X^0)x(Y^5). Remember, of course, X^0 (or indeed Y^0) = 1.
As we are looking for the value of 38^5, we can say X=30 & Y=8, and we can treat
X as X=3, provided we remember to multiply by the correct power of 10 at some
point.
Total = (3^5)x1x10^5 + 5x(3^4)x8x(10^4) + 10x(3^3)x(8^2)x10^3 +
10x(3^2)x(8^3)x10^2 + 5x3x(8^4)x10 + 8^5
Total = 24300000 + 32400000 + 17280000 + 4608000 + 614400 + 32768 = 79235168.
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There is another technique, involving 'markers' (the use of modular
mathematics).
By using modular maths, it is possible to obtain an answer to a problem without
actually performing the task itself as such. An example of this is as follows:
Problem: Find the fourth power of 47.
47 leaves a remainder of 5 after subtraction of the maximum multiple of 7, this
is shown as 47 = 5(7). In a similar manner, 47 = 3(11) and 47 = 8(13).
These markers can all be treated in the same way as the number(s) they
represent.
Since 47=5(7), 5^4=625, and 625=2(7), the answer to the problem is of the form
X=2(7).
Since 47=3(11), 3^4=81, and 81=4(11), the answer to the problem is of the form
X=4(11).
Since 47=8(13), 8^4=4096, and 4096=1(13), the answer is of the form X=1(13).
As we see the forms of 2(7) and 4(11), we can take the number 2 (which is, of
course, 2(7)) and add multiples of 7 until we find a number that is also 4(11).
So; 2, 9, 16, 23, 30, 37... and there we are. The number 37 is both 2(7) and
4(11). As 7x11=77, we can now say the answer is of the form X=37(77).
Taking this process oen further step forward, we can now add multiples of 77 to
37 until we find a number which is also of the form X=1(13). This leads us via
10x77 = 770 to add to the 37 to give 807. We can thus now say that the correct
answer to the problem is of the form X=807(1001), since 77x13=1001.
Another method of analysing problems & potential solutions is via the use of
modulo 9, what I call the 'marker' of each number. This is very easy to
calculate; simply add the values of the digits of a number, and keep adding the
digits of the answer until you arrive at a single-digit number. In this case,
the marker of the original number is 2, since 4+7=11 and 1+1=2. Now we see
2^4=16, and 1+6=7, so the marker of the answer must be 7. Or to put it another
way, the answer is of the form X=7(9).
Moving the numbers forward, we find that we need to add 5x1001=5005 to 807
(giving us 5812) in order to find a match. The answer now appears as being of
the form X=5812(9009).
Now to determine the last two digits of the answer:
47^4 = (47^2)^2, and 47^2 = 2209, thus ending in __09. Taking the square of this
two-digit number, and recognising the last two digits (indeed in this case the
only two digits) of the answer, we find the last two digits of the answer must
be __81.
The partial answer we have so far is 5812, ending with __12, and this requires
the addition of 41x09 to create a number ending with __12. This means we must
add 41x9009 = 369369 to 5812, giving us 375181 as our new partial answer. Since
we must not now change the final two digits of this answer, this is now regarded
as being modulo 900900 (9009 x 100). Therefore the answer is now of the form
X=375181(900900).
As we have already seen, 47^2=2209, and the final answer is the square of this
number. Without actually working out the whole number, we can see at a glance
that it will be between 4.5 & 5 million. With multiples of 900900 to be added to
375181, it soon becomes obvious that we need to add 5x900900 = 4504500 to 375181
to give us our final answer - which thus becomes 4879681. This is indeed the
correct answer to the problem of 47^4.
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A somewhat more impressive result is taken from the problem of finding the value
of 29^7 using much the same process. Since we have now already seen this process
in action in detail, I shall move this one on somewhat more quickly.
29 = 1(7), also 7(11) & 3(13). Taking the secenth power in each case, the answer
is seen as being of the form 1(7), 6(11) and 3(13). With the modulo workings for
7 & 11, we see the number 50 emerging, and the use of modulo 13 gives us a
partial answer of X=666(1001).
Looking at the marker of 9 (or 0, if you prefer), and seeing the required marker
is 2, we add 1x1001 = 1001 to reach 1667 and thus the partial answer of
X=1667(9009).
Seeking the last two digits, we use the fact that 29^7 = (29^2)^3 x 29.
29^2 = 841 (ending with __41), so we have 41x41x41x29 to show us the last two
digits.
As 41x41 = 1681 (ending __81), also 41x29 = 1189 (ending __89), we now see 81x89
as the 'signpost' to the last two digits of the answer. 81x89 = 7209, which ends
with __09, and so we now see that the last two digits of the answer to the
problem are __09.
Working as before, this means that we need tro ad 38x9009 = 342342 to 1667 to
reach a sub-total (so far) of 344009; this indicating that the answer is of the
form X=344009(900900).
Since we are looking for the 7th power of 29, the answer must obviously be a
multiple of 29, i.e. it must be of the form X=0(29). As 344009 = 11(29) and
900900 = 15(29), we need to add 7x900900 = 6306300 to 344009 in order to reach a
number which matches all the required conditions so far; this number is
therefore 6650309.
At the moment, we see that the answer is of the form X=6650309(26126100).
Now we can take the rather unusual step of using a modulus which is greater in
value than the original number. 29= -2(31), and (-2)^7 = -128. Against modulo
31, this leaves us with -4; and since 31-4=27, we can say that the final answer
to the problem must be of the form X=27(31). Our present partial answer is
6650309, which is 3(31), and the present modulus is 26126100, itself being
13(31). In order to create a number of the form X=27(31), we need to add
9x26126100 = 235134900 to 6650309, thus giving us 241785209. The modulus is now
31x26126100 = 8099909100, so the answer now appears as being of the form
X=241785209(809909100).
Now we try to estimate the approximate value of the final answer.
To estimate 29^7, try working with 30^7 and then taking away 1/30th of the
result seven times over. 30^7 = 3^7 x 10^7, i.e. 2187 x 10^7, or 21870 million.
Taking away 1/30 of the available value seven times, we go through 21141, 20436,
19756, 19097, 18461, and 18146 to reach 17831. Therefore the approximate value
of thefinal answer will lie in the region of 17831 million.
To get there, we must thus add 21x809909100 (our present modulus) to 241785209,
this giving us a final answer of 17008091100+241785209 = 17249876309. This is
indeed the value of 29^7.
Please let me know if you'd like me to explain this in more detail or with
further examples of the method involved.
Best regards,
George Lane
To be tested, vielleicht inspiriert es jemanden für einen ähnlichen Lösungsweg?