Multiplication Made Easy
Real Easy
A nice shortcut to the product of two, two-digit numbers exists if the following
conditions are met:
1. The tens digits are the same, and
2. The ones digits sum to 10.
Increase one of the tens digits by one and multiply it by the other tens digit, multiply
the ones digits together and write the two products next to each other.
e.g. 43 × 47 = 2021 5 × 4 and 3 × 7
It can be seen by using a calculator that the method will work if these conditions are
met. There is the case where we will need to use a zero in the tens column; 31 × 39
for example. 1209
Why does this work?
If we write a two digit number – ab – then we recognise that the value of the a is 10a.
Our numbers can then be written as (10a + b) × (10a + (10 – b))
Multiplying this out we obtain 10a2 + 100a -10ab + 10ab + b(10 – b)
Now we can factorise the first two terms and note that the next terms are equal but
have opposite signs and so sum to zero. The last term could be multiplied out as well
but it is actually nicer to leave as it is.
100a(a +1) + b(10 – b)
Now a was our original tens digit and a + 1 is the number one more than that digit.
Multiplying by 100 places that product in the desired position. Leaving the b(10 – b)
as is means that our last task is multiplying the original ones digits.
Well what happens with other cases?
Pretty Easy
Cases like 73 × 78 Tens digits identical, ones digits do not sum to 10
(10a + b) × (10a + c)
100a2 + 10ac + 10ab + bc
100a2 + 10a(b + c) + bc
Multiply the tens digits and call the answer hundreds, add the ones digits, then
multiply by the tens digit and then by 10, then multiply the ones digits. Collect the
results.
This algorithm is relatively easy to remember.
7 × 7 = 49 3 + 8 = 11 11 × 7 = 77 3 × 8 = 24
4900 + 770 + 24 = 5694 Which (as expected) is also correct.
What the..?
Let us take cases like 54 × 76 for example. Tens digits different but ones digits sum
to 10.
We have (10a + b) × (10c + (10 – b))
This produces 100ac + 100a – 10ab +10bc + b(10 – b)
Giving 100a (c+1) – 10ab + 10bc + b(10 – b)
Now the first and last sections of this can be converted to rules;
Increase one of the tens digits by one, multiply and this product will be hundreds, the
last section is simply multiplying the ones digits together as earlier.
The middle section can be described as multiply the ones digit of the first number by
the tens digit of the second number then subtract the product when you multiply the
ones digit of the first number by the tens digit of the first number. Get this answer and
multiply it by ten. Combine all of your results.
But are these rules worth remembering!
54 × 76
5 × 8 = 40, call it 4000
4 × 6 = 24
4 × 7 = 28 4 × 5 = 20 28 – 20 = 8 8 × 10 = 80
4000 + 24 + 80 = 4104 which fortunately is the correct answer!
Haven’t We Met Before?
Where there is no link between the numbers
34 × 72 for example
we can express this as (10a + b) × (10c + d)
The expansion of this produces
100ac + 10ad + 10bc + bd
This is our standard algorithm, though written in a form that suggests computing the
largest – and therefore most important – section first.