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# welp

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i need to ask this again because someone marked it as answered before i understood it. WAIT UNTIL I UNDERSTAND NEXT TIME

why do the powers of 11 have the same digits as the pascal triangle?

11^0=1

11^1=11

11^2=121

and so on.

Sort:

#1
+79846
+1

OK...OBT....see if this helps

Row 0  =  1 * 10^0      =  1  =   11^0

Row 1 =   1 * 10^1  + 1*10^0   =  11     =   11^1

Row 2 =   1*10^2  + 2*10^1 +  1 *10^0   =  121    =  11^2

Row 3 =   1*10^3  + 3*10^2 + 3*10 + 1*10^0  =  1331    =  11^3

Look  at row 5  which  is      1   5    10    10    5     1

So we have

1*10^5   +  5*10^4  + 10*10^3  + 10*10^2  + 5*10^1  + 1 * 10^0   = 161051  =  11^5

Do you see the pattern  ???

CPhill  Dec 1, 2017
edited by CPhill  Dec 1, 2017
edited by CPhill  Dec 1, 2017
#2
+91237
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Hi Chris,

I saw you last answer too and I have only just worked out what the question even meant :)

This question is making me think about tthings that I do not think I have considered before.

I might put my 2cents worth in later.

OBT do you undertand the link between Pascal's triangle and combination mathematics?

(CPhill used the combination maths notation in his previous answer )

also OBT,

you should have referenced your last question thread so that people could jump between the two.

https://web2.0calc.com/questions/yaaaaay-i-need-help-again

Melody  Dec 1, 2017
edited by Melody  Dec 1, 2017
#3
+91237
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Also OBT,

We usually mark questions as answered as soon as we, or another member, answer them.

If you are not happy with the answer then please just politely say so (along with a thank you for the answerer's efforts) and ask us, a moderator, to remove the tick. We will be quite happy to do this for you.

Melody  Dec 1, 2017
#4
+91237
+1

This will probably go over your head OBT but this is how expansions work for powers.

$$(a+b)^n=\binom{n}{0}*a^nb^0+\binom{n}{1}*a^{n-1}b+\binom{n}{2}*a^{n-2}b^2+\dots \binom{n}{n}*a^{0}b^{n}\\~\\ now\;\;\binom{n}{0},\;\binom{n}{1},\binom{n}{3},\;\dots \binom{n}{n}\\ \qquad \qquad \text{Is the nth row of pascal's triangle (if the first row is taken as 1,1)} \\\\~\\ \text{Lets see what hapens if a=10 and b=1}\\ (10+1)^n=\binom{n}{0}*10^n*1^0+\binom{n}{1}*10^{n-1}*1+\binom{n}{2}*10^{n-2}*1^2+\dots \binom{n}{n}*10^{0}*1^{n}\\ (10+1)^n=\binom{n}{0}*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots \binom{n}{n}*10^{0}\\ (10+1)^n=\binom{n}{0}*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots \binom{n}{n}*10^{0}\\ (10+1)^n=1*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots 1*10^{0}\\$$

Now I can see that the digits of this number will give pascals triangle BUT only while every nCr term is a single digit.

In other words it will only work up to  11^4 = 14641

I'll look at 11^4 and try and explain:

$$11^4=(10+1)^n\\ =1*10^4+\binom{4}{1}*10^{3}+\binom{4}{2}*10^{2}*+\binom{4}{3}*10^{1}+\binom{4}{4}*10^{0}\\ =1*10^4+\binom{4}{1}*10^{3}+\binom{4}{2}*10^{2}*+\binom{4}{3}*10+\binom{4}{4}\\ =1*10000+\binom{4}{1}*1000+\binom{4}{2}*100+\binom{4}{3}*10+\binom{4}{4}\\ =1*10000+4*1000+6*100+4*10+1\\ =14641$$

It will not work for 11^5 becasue there will be carry, try it and you will find out. :)

Melody  Dec 2, 2017
#5
+354
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jeez i am going nuts with these odd answers...sorry if im being a bother

#6
+91237
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You are not a bother, you just have not learned some of this stuff yet :)

Melody  Dec 4, 2017

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