The largest product

Suppose you are given the digits $2, 3, 5$, each to be used exactly once, and asked to fill them into $\square\square \times \square$ to get the largest possible product. There are only a few arrangements, but the largest isn't immediately obvious. The pattern that emerges has a beautiful algebraic explanation , and it generalises to bigger problems too.

Concept

The problem. Place each of the given digits exactly once into the boxes of a fixed pattern (like $\square\square \times \square$ or $\square\square \times \square\square$) so as to make the largest possible product.

Why the answer isn't obvious. Each digit's value depends on its position: in a tens place it counts ten times more than in a units place. So you want big digits in high-value positions , but exactly which positions are highest-value depends on the arrangement.

Small case: $\square\square \times \square$ with digits $\{2, 3, 5\}$.

There are $3! = 6$ arrangements. Let's try the ones with the largest product candidates:

The largest is $\mathbf{32 \times 5 = 160}$. The biggest digit ($5$) goes alone, and the next two ($3, 2$) sit in the two-digit factor , with the larger one in the tens place.

Why? With digits $a, b, c$ and $a > b > c$, the candidate $bc \times a$ equals $(10b + c) \cdot a = 10ab + ac$. Swapping the role to $ac \times b$ would give $(10a + c) \cdot b = 10ab + bc$. The difference is $ac - bc = c(a - b)$. Since $a > b$, this is positive , meaning the first form $(10ab + ac)$ is larger. So putting the largest digit alone wins.

Bigger case: $\square\square \times \square\square$. Among $4!$ orderings of $\{a > b > c > d\}$, the maximum is

$\text{(}\ ad\ \text{)} \times \text{(}\ bc\ \text{)},$

with the two largest digits in the tens places and the two smaller digits in the units places. Formal proof uses repeated swap arguments like above.

A general principle. When forming products by placing digits, rank positions by their place-value weight, and pair the largest digits to the largest weights. Often the placement also distributes large digits across the factors so that both factors grow together , products grow fastest when factors are close in size.

Connection to area. Visualise each two-digit number as the side of a rectangle and the product as its area. For a fixed perimeter (sum of digits), the area is maximised when the rectangle is closest to a square , which is exactly the pattern of distributing big digits across both factors.

Worked examples

Example 1. Largest product of $\square\square \times \square$ using $\{2, 3, 5\}$.

• Apply the rule: largest digit alone. Answer: $32 \times 5 = 160$.

Example 2. Largest product of $\square\square \times \square$ using $\{1, 4, 9\}$.

• Largest is $9$, alone. Other two: $41$. Product: $41 \times 9 = 369$.
• (Check: $14 \times 9 = 126$, $91 \times 4 = 364$, $49 \times 1 = 49$. So $369$ wins.)

Example 3. Largest product of $\square\square \times \square\square$ using $\{1, 2, 3, 4\}$.

• Pair big with small across factors. Try $41 \times 32 = 1312$ and $42 \times 31 = 1302$ and $43 \times 21 = 903$. Largest is $\mathbf{41 \times 32 = 1312}$.

Example 4. Smallest product (for contrast) of $\square\square \times \square$ using $\{2, 3, 5\}$.

• Mirror the rule: smallest digit alone. Smallest is $2$. Other two: $53$. But $53 \times 2 = 106$ vs $35 \times 2 = 70$ , the smallest two-digit factor is $35$. So smallest product is $35 \times 2 = 70$, not $53 \times 2$.
• The pattern "use the smallest digit alone with the smallest two-digit number" , equivalent to pushing small digits to high-value spots , gives the smallest product.

Try it yourself

1. Largest $\square\square \times \square$ using $\{1, 5, 8\}$.
2. Largest $\square\square \times \square$ using $\{0, 3, 7\}$ (no leading $0$).
3. Largest $\square\square \times \square\square$ using $\{2, 4, 6, 8\}$.
4. Smallest $\square\square \times \square$ using $\{2, 7, 9\}$.
5. With $\{3, 5, 7\}$, by how much is the largest product bigger than the smallest?
6. Argue (in words or with algebra) why the largest digit always goes in the highest place-value position.
7. Largest $\square\square\square \times \square$ using $\{1, 2, 3, 4\}$.
8. Two students are given the same three digits and asked to find the largest product. One gets $73 \times 5 = 365$ and the other $53 \times 7 = 371$. Who is correct? Why?

Activity

Digit duel. Roll three dice. The three numbers shown are your digits. With paper and pen, write down all possible arrangements of $\square\square \times \square$ and find the maximum product. Repeat the experiment $10$ times. Does the "largest digit alone, larger of the rest in tens place" rule always win? Record any roll where the rule seems to fail , and check your arithmetic.