Questions & Answers

Question

Answers

Answer

Verified

148.5k+ views

Hint: Apply the method where you subtract smaller roman notations from bigger standard ones. For example, $4=5-1=V-I=IV$.

Complete step-by-step answer:

So, first let’s remember what the list of standard notations in Roman numerals looks like.

So, one by one the list goes like this :

$\begin{align}

& 1=I \\

& 5=V \\

& 10=X \\

& 50=L \\

& 100=C \\

& 500=D \\

& 1000=M \\

\end{align}$

This list shows the notation for standard numbers upto $1000$, but other numbers can be formed by the addition or subtraction of these standard notations only.

Thus, for example, whenever we write a Roman numeral signifying a bigger number in front of a Roman numeral signifying a smaller number, then the smaller number is considered to be subtracted from the bigger number.

For example, if we want to write $4$, we use the Roman numerals of $5$ and $1$. We know that, $4=5-1$. Converting this statement into Roman numerals, we’ll get that $4=V-I=-I+V$. Now, we can write this subtraction by writing $I$ first, and follow it up with the $V$. So, ultimately, $4=5-1=V-I=-I+V=IV$, which indeed we know is correct, from our common knowledge.

Similarly, even $9$ is written as $IX$ in Roman numerals, because $X$ written in front of $I$ essentially means writing $10$ in front of $1$. This in turn means, that because we wrote a bigger number in front of a smaller number, it signifies subtraction of the smaller number from the bigger number. So, $IX=X-I=10-1=9$.

Now, the number we have to convert here is $92$.

First, let’s break $92$ up into the sum of a multiple of $10$, and another whole number lying between $1$ and $9.$

Doing so, we can say that, $92=90+2=90+1+1.$

Now, let’s tackle the multiple of $10$ that we have here, or $90$. $90$ can be written as $100-10=-10+100$. We know that we have standard notations in Roman numerals, that exist for both, $100$ and $10$. So, $-10+100$ written in Roman numerals, will be $-X+C$. Now, since $C$ is bigger than $X$, the subtraction can be written as $-X+C=XC$. Therefore, we come to the fact that $90=100-10=-10+100=-X+C=XC$.

Now, to tackle the whole number between $1$ and $9$ part, we know that $2=1+1$, and we already have standard notation for $1$ in Roman numerals. So, $2=1+1=I+I=II$.

Now, combining the Roman numerals obtained by tackling the two parts separately, we get that $92=90+2=XC+II=XCII$.

Hence, $92$ in Roman numerals is written as $XCII$.

Note: Addition in Roman numerals is simply written as the Roman numeral for the smaller number being added in front of the Roman numeral for the bigger number being added. But, the moment you write the bigger number in front of the smaller number, it means you intend to subtract the smaller number from the bigger number.

Complete step-by-step answer:

So, first let’s remember what the list of standard notations in Roman numerals looks like.

So, one by one the list goes like this :

$\begin{align}

& 1=I \\

& 5=V \\

& 10=X \\

& 50=L \\

& 100=C \\

& 500=D \\

& 1000=M \\

\end{align}$

This list shows the notation for standard numbers upto $1000$, but other numbers can be formed by the addition or subtraction of these standard notations only.

Thus, for example, whenever we write a Roman numeral signifying a bigger number in front of a Roman numeral signifying a smaller number, then the smaller number is considered to be subtracted from the bigger number.

For example, if we want to write $4$, we use the Roman numerals of $5$ and $1$. We know that, $4=5-1$. Converting this statement into Roman numerals, we’ll get that $4=V-I=-I+V$. Now, we can write this subtraction by writing $I$ first, and follow it up with the $V$. So, ultimately, $4=5-1=V-I=-I+V=IV$, which indeed we know is correct, from our common knowledge.

Similarly, even $9$ is written as $IX$ in Roman numerals, because $X$ written in front of $I$ essentially means writing $10$ in front of $1$. This in turn means, that because we wrote a bigger number in front of a smaller number, it signifies subtraction of the smaller number from the bigger number. So, $IX=X-I=10-1=9$.

Now, the number we have to convert here is $92$.

First, let’s break $92$ up into the sum of a multiple of $10$, and another whole number lying between $1$ and $9.$

Doing so, we can say that, $92=90+2=90+1+1.$

Now, let’s tackle the multiple of $10$ that we have here, or $90$. $90$ can be written as $100-10=-10+100$. We know that we have standard notations in Roman numerals, that exist for both, $100$ and $10$. So, $-10+100$ written in Roman numerals, will be $-X+C$. Now, since $C$ is bigger than $X$, the subtraction can be written as $-X+C=XC$. Therefore, we come to the fact that $90=100-10=-10+100=-X+C=XC$.

Now, to tackle the whole number between $1$ and $9$ part, we know that $2=1+1$, and we already have standard notation for $1$ in Roman numerals. So, $2=1+1=I+I=II$.

Now, combining the Roman numerals obtained by tackling the two parts separately, we get that $92=90+2=XC+II=XCII$.

Hence, $92$ in Roman numerals is written as $XCII$.

Note: Addition in Roman numerals is simply written as the Roman numeral for the smaller number being added in front of the Roman numeral for the bigger number being added. But, the moment you write the bigger number in front of the smaller number, it means you intend to subtract the smaller number from the bigger number.

Students Also Read