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.