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So, I stumbled upon the following question.

Using binomial theorem compute $102^6$.

Now, I broke the number into 100+2.

Then, applying binomial theorem

$\binom {6} {0}$$100^6(1)$+$\binom {6} {1}$$100^5(2)$+....

enter image description here

I stumbled upon this step. How did they add the humongous numbers? I am really confused.

Kindly help me clear my query.

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  • $\begingroup$ They're powers of ten.... And you don't mean "embezzled." Maybe you mean "confused." $\endgroup$ Mar 4 '17 at 4:39
  • $\begingroup$ Correct! I need to work on my English skills, man! Thanks for the help tho! $\endgroup$ Mar 4 '17 at 4:46
  • $\begingroup$ @symplectomorphic - actually, they probably meant "bamboozled", as it means something similar to "confused", but can be easily mixed up with "embezzled". $\endgroup$
    – Glen O
    Mar 4 '17 at 7:42
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The long way:

$$ \begin{align} & 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\ =\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\ =\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10} + \color{red}6 \cdot 10^9 + \color{red}1 \cdot 10^8 + \color{red}6 \cdot 10^7+\color{red}2 \cdot 10^6 + \color{red}4 \cdot 10^5+\color{red}1 \cdot 10^4+\color{red}9\cdot 10^3 + \color{red}2 \cdot 10^2 + \color{red}6 \cdot 10^1 + \color{red}4 \cdot 10^0 \end{align} $$

The latter is precisely the representation in base $\,10\,$ of $\;\color{red}{1126162419264}\,$.

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    $\begingroup$ Thanks for the help! Perhaps the most convincing and noteworthy answer. $\endgroup$ Mar 4 '17 at 10:52
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That number is $1\; 12 \; 6\; 16 \; 24\;192\; 64$ (space added for emphasis). Notice a relationship with the coefficients of the powers of $10$?

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$10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$ $= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$

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  • $\begingroup$ Thank you very much, sir! I really appreciate it. $\endgroup$ Mar 4 '17 at 10:51
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Since it's all powers of ten you could add those quite easily. If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.

$\text{1 12 6 16 24 192 64}$ Those powers just end up just putting those digits in the right order because we use a number system with ten digits.

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  • $\begingroup$ Ok! Why not 10112...? $\endgroup$ Mar 4 '17 at 4:50
  • $\begingroup$ You see say you're adding something like a $12 \times 10^{10}$. You can see that those powers are all spaced evenly. By that I mean to say whenever you have say n digits to be added, over there they had decreased the exponent of 10 by n $\endgroup$
    – SBM
    Mar 4 '17 at 4:53
  • $\begingroup$ Moreover, saying a number equals something like $123.xyz$ is the same thing as $1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + x \times 10^{-1} \ldots$ and so on. As long as you keep those evenly spaced, it remains same. By x, y and z I meant arbitrary digits not variables $\endgroup$
    – SBM
    Mar 4 '17 at 4:56
  • $\begingroup$ I got your point. $\endgroup$ Mar 4 '17 at 5:04
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Start adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term. $$ ...2400000\quad +\\ .....19200\quad +\\ .......64\quad $$ ...and so on till the first.

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