Hi Sylvia,

I am going to use base seven.

In base seven there are seven digits, 0, 1, 2, 3, 4, 5, and 6.Hence my multiplication table will look like

The first row is easy since $0 \times b = 0$ for any number $b$ in any base.

Also multiplication is commutative, $a \times b = b \times a$ so once you complete a row you can fill in the corresponding column.

The second row and second column are also easy since $1 \times b = b \times 1$ for any $b$ in any base.

$2 \times 2 = 4 \mbox{ and } 2 \times 3 = 6$ so the next two entries in the third row are $4$ and $6$ but $2 \times 4$ is eight and I don't have a digit for eight. But eight is seven plus 1, that is one seven and one more so in base seven, eight is $11_7.$ Similarly $2 \times 5$ is ten which is seven plus three so in base seven, ten is $13_7.$ Likewise $2 \times 6$ is twelve which is seven plus five so in base seven, twelve is $15_7.$ Thus I can complete the second row and column.

I will find one more entry, $5 \times 6.$ $5 \times 6$ is thirty. If you divide thirty by seven you get 4 with a remainder of 2. Thus thirty is four sevens and two more so in base seven, thirty is $42_7.$ Hence I get

You can complete the table.

Write back if you need more assistance,
Penny