## Multiplication

The purpose of this guide is to tell you which tricks to use when faced with a mutiplication problem. I'd like to present things in two ways: First, I'll go through different "rules of thumb" about which tricks to use when (For example: "If the numbers you want to multiply are close together, then...") While I like heuristics, I think they can become overly normative; so I also want to give you a list of example problems, and talk about the tricks you might use for each. Ok; let's start!

# Rules   of   Thumb

You want to multiply the numbers $$n$$ and $$m$$ together...

• Are $$n$$ and $$m$$ both less than $$20$$?

If so: I think that base numbers are the best way to go. For example: $$17 \cdot 18$$ $$= 10 \cdot 25 + 56 = 306$$ can be done almost instantly, as can most multiplications under $$20$$ (if you use base numbers).

• Are $$n$$ and $$m$$ very close together, say $$\left| n - m \right| \leq 10$$, and of the same parity (i.e. both even or both odd)?

If so: I think that multiplying by squaring is the best option. For example: $$35 \cdot 45 = 40^2 - 25 = 1575$$ can be computed very quickly. Similarly for $$29 \cdot 31 = 900 - 1 - 899$$.

• Are $$n$$ and $$m$$ of the same parity and is $$(n+m)/2$$ easy to square?

Then: Again, I think that multiplying by squaring is a good bet.

• Are $$n$$ and $$m$$ close together? (Forget about parity.)

Then using base numbers is a good idea. I've only covered the rudements of this method at Ars Calcula; for more advanced versions of this trick, you should see Handley's Speed Mathematics.

• Is either $$n$$ or $$m$$ equal to $$25$$? More generally, does either $$n$$ or $$m$$ end in $$5$$?

We've seen tricks for multiplying by $$25$$ and for multiplying by something ending in $$5$$.

• Is either $$m$$ or $$n$$ easily broken into prime factors?

If so, try multiplying by factoring. For example: To compute $$14 \cdot 91$$, you might compute $$7 \cdot 91 = 637$$ and then double to get $$1274$$.

• Is either $$m$$ or $$n$$ equal to $$11$$?

We have a trick for that.

• Does either $$m$$ or $$n$$ have a lot of repeating digits or digits that are multiples of one another?

If so, you might want to re-use some calculations

• Did none of the above work?

Then you should try a brute-force method, such as direct multiplication or (much better, in my opinion) distributed multiplication. I think that distributed multiplication is the way to go. I've only talked about it a bit at Ars Calcula; if you want to master the method, I recommend Art Benjamin's Secrets of Mental Math.

# Learning   by   Example

Now let's look at some examples. I wrote a short python script (please feel free to use it) to generate multiplication problems $$m \cdot n$$ where both $$m$$ and $$n$$ are between $$0$$ and $$100$$. Here are $$10$$ problems produced by that script, along with the methods I would use to solve them.

• $$23 \cdot 74$$

$$25 \cdot 74$$ is $$1/4$$ of $$7400$$ so... $$1850$$. From that we subtract $$2 \cdot 74 = 148$$. So subtract $$200$$ to get $$1650$$. Then add $$52$$ to get $$1702$$, final answer. We multiplied by $$25$$, then corrected (i.e. subtracted $$148$$ so as to use distributed multiplicaton. The actual subtraction was done via subtracting by adding.

Alternatively: We could have used distributed multiplicaton directly: $$23 \cdot 74$$ is $$1480 + 210 + 12$$, so... $$1702$$.

• $$16 \cdot 30$$

This one is easy: $$3 \cdot 16 = 48$$, so $$480$$ is our final answer.

• $$99 \cdot 77$$

$$100 \cdot 77 = 7700$$, and we subtract $$77$$ (i.e. subtract $$100$$ and then add $$23$$) to get $$7623$$.

Alternatively, we can try multiplying by factoring: $$11 \cdot 77 = 847$$ using the $$11$$ trick. Now use a variation of that trick (see Handley's Book) to multiply by $$9$$. We get that $$9 \cdot 847 = 7623$$.

• $$37 \cdot 22$$

We'll use factors and the $$11$$ trick. $$37 \cdot 11 = 407$$, and doubling that gives $$814$$. Final answer.

• $$33 \cdot 5$$

Use the trick for numbers ending in $$5$$: Half of $$330$$ is $$165$$.

• $$77 \cdot 81$$

We'll multiply by squaring. $$79^2 = (80-1)^2 = 6400 -159 = 6241$$. Subtract $$4$$ to get the final answer: $$6237$$.

Alternatively: Use factors and the $$11$$ trick.

• $$31 \cdot 69$$

I vote for distributed multiplicaton. $$30 \cdot 69 = 30 \cdot 70 - 30 = 2070$$. Now add $$69$$ to get $$2139$$, final answer.

• $$46 \cdot 85$$

Let's do distributed multiplication: We'll compute $$50 \cdot 85$$ and then subtract $$4 \cdot 85$$. Both of these products are easy to compute $$50 \cdot 85$$ is half of $$8500$$, i.e. $$4250$$. And $$4 \cdot 85 = 2 \cdot 170 = 340$$. So subtract $$400$$ from $$4250$$ and then add $$60$$ to get $$3910$$.

• $$35 \cdot 57$$

Let's do multiplying by squaring. $$(35 + 57)/2 = 46$$, and $$46^2 = 2116$$. (Remember: There's a trick for squaring numbers near $$50$$.) Now we subtract $$11^2 = 121$$ to get the final answer of $$1995$$.

• $$97 \cdot 99$$

The best way to deal with this one is with base numbers. We have that $$100 \cdot 96 = 9600$$, then we add $$3$$ to get the final answer of $$9603$$.

As a final note: Remember that you can easily check your answer by casting out nines.