# What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?

What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.

Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?

• I asked a question that used this result (and has a nice, clean answer!) several years ago: See MSE 226983 for more! Commented Jun 12, 2016 at 3:29
• (1+2+...+8+9)^2 Commented Jun 12, 2016 at 18:45

The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$

Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$

So the top two rows sum to $(1+2)\times 45 = 135$

Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$

The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$

We want to find the following: $$\sum_{i=1}^9 \sum_{j=1}^9 ij$$ Factor out the $i$ from the first summation: $$\sum_{i=1}^9 i\left(\sum_{j=1}^9 j\right)$$ Note that $\sum_{j=1}^9 j=45$. $$\sum_{i=1}^9 i\cdot 45$$ Factor out the $45$: $$45\left(\sum_{i=1}^9 i\right)$$ Note that $\sum_{i=1}^9 i=45$. $$45\cdot 45=2025$$

Use the distributive principle. The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)

Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$

Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:

First, show that this is true for $n=1$:

$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$

Second, assume that this is true for $n$:

$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$

Third, prove that this is true for $n+1$:

$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$

$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$

$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$

$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$

$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$

$\frac{n^4+2n^3+n^2}{4}+\frac{n^3+2n^2+n}{2}+\frac{n^3+2n^2+n}{2}+n^2+2n+1=$

$\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$

$\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$

$\frac{n^4+6n^3+13n^2+12n+4}{4}=$

$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$

Please note that the assumption is used only in the part marked red.

Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.

I will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.

Let us start with a small size version. If you take the top $3\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:

$$s= 1\times 1 +1\times 2 + 1\times 3 + 2\times 1 +2\times 2 + 2\times 3 + 3\times 1 +3\times 2 + 3\times 3 \,.$$ You see that you can rewrite this sum of nine products as: $$s = (1+2+3)\times (1+2+3)\,,$$ using the distribution of multiplication over additions.

This form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself.

It can be reassuring to verify it works: you get $36$, which you can check by hand ($6+12+18$). You can test it with the $2\times2$ or $4\times4$ matrix, to verify it is not a coincidence.

The hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\ldots,i,\ldots,n)\times(1,\ldots,j,\ldots,n)$.

Now you have to find the sum of integers. If you forget the generic expression for the sum of the $n$ first integer, have this figure in mind:

You pack together two triangles which give you a $n\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle.

So finally the answer is $\left(n(n+1)/2\right)^2$. With $n=9$, you get $2025$.

Once you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.

• The link to "I like visual proofs" is dead. Besides that, I think the name "visual proof" is not suitable here. Perception may be betraying, so visual remarks should be backed up by some other argument. This makes it not related to "visual"! Commented Jun 12, 2016 at 16:23
• @Jasper I have modified the answer. The visual aspect was just a reminder for a formula given in other answers Commented Jun 12, 2016 at 16:49