Remembering multiplication of these two numbers: $7 \times 8 = 56$ and $9 \times 6 = 54$

Question

I have almost mastered multiplication table up to 9x9 however, I'm having problems with the following two.

7 x 8 = 56 and 9 x 6 = 54

For some reason my brain thinks that 56 and 54 are somewhat the same and sometimes confuses the two multiplication problems. I was wondering if there is a way I can remember this easier. If someone on here has some rhyming of some sort.

When in doubt I always resort to doing 49+7 or 45+9 to figure out the two, because 7x7 and 9x5 is easy for me.

Thanks

Answer 1

Multiples of $9$ in the multiplication table have the property that the sum of its digits is always $9$. So $9 \times 6$ cannot be $56$. Also, $9 \times n$ starts with digit $n-1$.

Answer 2

Try to see it in as many ways as you can (as when you use $49+7$ and $45+9$). For $9$ is nice that (when $n \in (0,1,\dots,9)$) $$9 \times (n+1)= n\cdot 10+(9-n)$$ so $9 \times (5+1)= 5\cdot 10+(9-5)$ (its ugly to see but nice to understand).

A nice exercise could be to write the multiplication table just with the number factorized, I mean $56=7 \times 2^3$ and $54=3^3\times2$.

However, if you're looking for "mnemotechnical stuffs", as to memorize $\pi$'s digits, the best in my opinion is the Phonetical Conversion. I'm surprised is still not being teached at school: with little effort you gain a lifetime powerfull instrument. With that, the multiplication table will be just a picture in your mind.

Answer 3

Getting the 9's table to memory used to be tough for me until I was told of a shortcut; the key is to note the symmetry exhibited by the 9's table:

$$\begin{matrix}09&90\\18&81\\27&72\\36&63\\45&54\end{matrix}$$

where the ones on the left going downward are from $9\times1$ to $9\times5$, and the ones on the right going upward are from $9\times6$ to $9\times10$. (This gives an insight into why.) When I was very young, I had difficulties remembering the multiplication tables; after being told of this property, I figured I could "fold over" the 9's table in half and get away with just remembering the first five, and that part of the multiplication tables subsequently became less difficult for me.

Take heed also of lhf's comments. Any multiple of 9 should have its digits sum to 9 or a multiple of 9. Even if you extend the table to, say, $9\times 13=117$, the digits of the answer should still sum to 9 or a multiple: $1+1+7=9$.