Question

The digit 5 appears twice in the number 255,120. How does the total value of the 5 on the right compare to the total value of the 5 on the left?

Answer 1

Answer:

The value of the first "$5$" in the number $255,\!120$ is ten times that of the second "$5\!$" in this number.

Step-by-step explanation:

What gives the number "$255,\!120$" its value? Of course, each of its six digits has contributed. However, their significance are not exactly the same. For example, changing the first $\verb!5!$ to $\verb!6!$ would give $2\mathbf{6}5,\!120$ and increase the value of this number by $10,\!000$. On the other hand, changing the second $\verb!5!\!$ to $\verb!6!\!$ would give $25\mathbf{6},\!120$, which is an increase of only $1,\!000$ compared to the original number.

The order of these two digits matter because the number "$255,\!120$" is written using positional notation. In this notation, the position of each digits gives the digit a unique weight. For example, in $255,\!120\!$:

$\begin{array}{|r||c|c|c|c|c|c|}\cline{1-7}\verb!Digit!& \verb!2! & \verb!5! & \verb!5! & \verb!1! & \verb!2! & \verb!0!\\\cline{1-7}\textsf{Index} & 5 & 4 & 3 & 2 & 1& 0 \\ \cline{1-7} \textsf{Weight} & 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0}\\\cline{1-7}\end{array}$.

(Note that the index starts at $0$ from the right-hand side.)

Using these weights, the value $255,\!120$ can be written as the sum:

$\begin{aligned}& 255,\!120\\ &= 2 \times 10^{5} + 5 \times 10^{4} + 5 \times 10^{3} + 1 \times 10^{2} + 2 \times 10^{1} + 0 \times 10^{0} \\&=200,\!000 + 50,\!000 + 5,\!000 + 100 + 20 + 0 \end{aligned}$.

As seen in this sum, the first "$5$" contributed $50,\!000$ to the total value, while the second "$5\!$" contributed only $5,\!000$.

Hence: The value of the first "$5$" in the number $255,\!120$ is ten times that of the second "$5\!$" in this number.  

Answer 2

What this person above said