A Story of Numbers | IT

Question 12

How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals? Try to find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX.

Solution

We can multiply Roman numerals by using repeated addition and the distributive property, carefully combining and simplifying symbols at each step.

Step 1 — Multiplying V × L

To multiply $V \times L$ , we will add $L$ to itself $V$ times. The Roman numeral $V$ means five. So, we need to add $L$ five times.

Let us write out $L$ added five times. $L + L + L + L + L$ Now, we combine all the $L$ symbols together. $LLLLL$ We know that two $L$ 's make a $C$ ( $LL = C$ ). So, we can group the $L$ 's and simplify. $LLLLL = (LL) + (LL) + L$ $= C + C + L$ $= CC + L$ $= CCL$

$\boxed{CCL}$

Step 2 — Multiplying L × D

To multiply $L \times D$ , we will add $D$ to itself $L$ times. The Roman numeral $L$ means fifty. Adding $D$ fifty times directly would be very long. Instead, we can break $L$ into smaller units. We know that $L$ is equal to $X$ added five times ( $L = X+X+X+X+X$ ). So, $L \times D$ is the same as $(X+X+X+X+X) \times D$ . This means we first calculate $X \times D$ . Then we add that result five times.

Let us calculate $X \times D$ . To do this, we add $D$ to itself $X$ times. The Roman numeral $X$ means ten. So, we need to add $D$ ten times. $D + D + D + D + D + D + D + D + D + D$ Now, we combine all the $D$ symbols. $DDDDDDDDDD$ We know that two $D$ 's make an $M$ ( $DD = M$ ). So, we can group the $D$ 's and simplify. $DDDDDDDDDD = (DD) + (DD) + (DD) + (DD) + (DD)$ $= M + M + M + M + M$ $= MMMMM$ So, $X \times D$ is $MMMMM$ .

Now, we use this result to find $L \times D$ . We need to add $MMMMM$ five times (since $L = X+X+X+X+X$ ). $MMMMM + MMMMM + MMMMM + MMMMM + MMMMM$ Combine all the $M$ symbols. $MMMMMMMMMMMMMMMMMMMMMMMMM$ There are $25$ $M$ 's in total.

$\boxed{MMMMMMMMMMMMMMMMMMMMMMMMM}$

Step 3 — Multiplying V × D

To multiply $V \times D$ , we will add $D$ to itself $V$ times. The Roman numeral $V$ means five. So, we need to add $D$ five times.

Let us write out $D$ added five times. $D + D + D + D + D$ Now, we combine all the $D$ symbols. $DDDDD$ We know that two $D$ 's make an $M$ ( $DD = M$ ). So, we can group the $D$ 's and simplify. $DDDDD = (DD) + (DD) + D$ $= M + M + D$ $= MM + D$ $= MMD$

$\boxed{MMD}$

Step 4 — Multiplying VII × IX

To multiply $VII \times IX$ , we will use the distributive property. First, let us break down each Roman numeral into its parts. $VII = V + I + I$ . $IX = X - I$ .

Now, we apply the distributive property. $(V+I+I) \times (X-I)$ This expands to: $(V \times X + I \times X + I \times X) - (V \times I + I \times I + I \times I)$ Next, we perform the basic symbol multiplications. We know that: $V \times X = L$ (because $X+X+X+X+X = L$ ) $I \times X = X$ $V \times I = V$ $I \times I = I$ Substitute these results back into the expression: $(L + X + X) - (V + I + I)$ Simplify the terms inside each parenthesis. $(LXX) - (VII)$ Now, we need to perform the subtraction $LXX - VII$ . To subtract Roman numerals, we can convert symbols to smaller units if needed. We need to subtract $V$ and $II$ from $LXX$ . $LXX$ has $L, X, X$ . To subtract $V$ , we convert one $X$ into $VV$ . $LXX = L X V V$ Subtract $V$ : $L X V V - V = L X V$ Now, we need to subtract $II$ from $LXV$ . $LXV$ has $L, X, V$ . To subtract $II$ , we convert $V$ into $IIIII$ . $LXV = L X IIIII$ Subtract $II$ : $L X IIIII - II = L X III$ So, the final product is $LXIII$ .

$\boxed{LXIII}$

Answer

(i) V × L = CCL (ii) L × D = MMMMMMMMMMMMMMMMMMMMMMMMM (iii) V × D = MMD (iv) VII × IX = LXIII

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Q12

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