Moving decimal on multiplying or dividing by power of 10 How does "moving decimal" works on multiplying or dividing by any number which is power of 10? For example: 1.23×100 = 123 in this example only value is changed while digits remains same

The question, as I understand it, is due to the fact that someone said that to multiply $1.23$ by $100,$ all you have to do is erase the decimal point between the $1$ and the $2$ and write a new one to the right side of the $3$ without changing anything else about the way the number is written. (I think you know that when a base-ten numeral is written without a decimal point, there is an "implied decimal point" on the right-hand side of the rightmost digit; that is why you said the decimal point is "moved" rather than "removed".)
Now you rightfully think we should only believe this if there is a good reason to believe that it really works. And this reason should have to do with what it really means when we write numbers in these ways.
To expand on one answer already posted, it all comes down to how a place-value system like base ten works. The key point is that each place where a digit appears in a number has a place value determined by its position relative to the decimal point. So the $3$ in $1.23$ is in a place with place value $\frac{1}{100}.$ To find out what number a string of decimal digits (with a decimal point) actually represents, we have to multiply each digit by its place value and add the results:
$1.23=1\times <!--\; \times \; -->1+2\times <!--\; \times \; -->\frac{1}{10}+3\times <!--\; \times \; -->\frac{1}{100}.$
Looking at the number this way, one relatively simple answer to the question of what happens when we multiply by $100$ is that the multiplication distributes over the addition, and then we use this to work out new place values of all the digits:
$\begin{array}{rl}1.23\times <!--\; \times \; -->100& =(1\times <!--\; \times \; -->1+2\times <!--\; \times \; -->\frac{1}{10}+3\times <!--\; \times \; -->\frac{1}{100})\times <!--\; \times \; -->100\\ & =1\times <!--\; \times \; -->1\times <!--\; \times \; -->100+2\times <!--\; \times \; -->\frac{1}{10}\times <!--\; \times \; -->100+3\times <!--\; \times \; -->\frac{1}{100}\times <!--\; \times \; -->100\\ & =1\times <!--\; \times \; -->100+2\times <!--\; \times \; -->10+3\times <!--\; \times \; -->1\\ & =123.\end{array}$
The way I think about this that really makes sense to me is not that the decimal point moves two places to the right, but rather that every digit of the number moved two places to the left (to a place worth $100$ times what its old place was worth). But since I am lazy, I notice that instead of erasing every digit and moving it two places over, I can just erase one little dot and move it two places, and I get exactly the same resulting arrangement of symbols. It's much like the scene in a film where a group of soldiers are standing at attention in a line, and their commander asks one volunteer to step forward. Everyone except one soldier steps backward, and therefore that hapless soldier (who is now one step in front of everyone else) becomes the volunteer.
Another way to look at this (if you consider just multiplication by $10$ at first) is as a carry operation. We a carry operation occurs if we multiply $1.23$ by $5.$ Looking first at the $\frac{1}{100}$ place, we find a $3.$ Since $3\times <!--\; \times \; -->5=15=10+5,$ we have $5$ in $\frac{1}{100}$ place, but the $10$ is "carried" by adding $1$ to the $\frac{1}{10}$ place. Since $1.23$ has $2$ in the $\frac{1}{10}$ place, and $2\times <!--\; \times \; -->5=10,$ there is nothing else to add in the $\frac{1}{10}$ place (we end up with just the carried $1$ there) and we carry a $1$ into the $1$'s place. In the $1$'s place, $1.23$ has a $1,$ and since $1\times <!--\; \times \; -->5=5$ we take the $5,$ add the carried $1,$ and put the sum $5+1=6$ in the $1$'s place:
$1.23\times <!--\; \times \; -->5=\mathrm{6.15.}$
Multiplication by $10$ can be treated similarly, but it is easier, because in each place we always get something like $1\times <!--\; \times \; -->10=10,$$2\times <!--\; \times \; -->10=20,$ or $3\times <!--\; \times \; -->10=30,$ that is, whatever number we found in each place is the same number that gets carried to the next place to the left, and when it gets there we find that the number in that place had a product that ended in zero, so we add zero to the carried digit and get the same digit we started with one place to the right.
So after multiplication by $10,$ every single digit ends up one place to the left of where it started.
To multiply by $100$ or $1000$ you can just multiply by $10$ two or three times. As a result, each digit gets moved left two or three times.

For example: 1.23×100 = 123
What's happening here is you're considering $1.23={\displaystyle \frac{123}{100}},$ so multiplying by $100$ cancels the denominator,
$\begin{array}{rl}1.23\times <!--\; \times \; -->100& ={\displaystyle \frac{123}{100}}\cdot <!--\; \cdot \; -->100\\ & =123\cdot <!--\; \cdot \; -->{\displaystyle \frac{100}{100}}\\ & =123\end{array}$
What's happening whenever you move any decimal point is similar, since a decimal looks like this in general:
$123.56={10}^2+2\cdot <!--\; \cdot \; -->{10}^1+3\cdot <!--\; \cdot \; -->1({10}^0)+{\displaystyle \frac{5}{{10}^1}}+{\displaystyle \frac{6}{{10}^2}}$
So moving a decimal point is really shifting the powers of $10$ which are present.