The longhand way of multiplying any two-digit natural number of the form
is to find (10a+b)(10c+d) (= 100ac + 10ad + 10bc + bd). This is no great challenge--the method I was taught in grade school involves at most seven lines of paper:
d*10a + d*b
10(c*10a + c*b)
100ac + 10ad + 10bc + bd
This method saves a modicum of space by abbreviating the addition to a matter of taking each digit as it comes out of the multiplication ('carry'ing digits over ten to the next column), and it accounts for the multiplication by 10 as above. However, although it is a wonderful system for multiplying numbers, in the special case of two-digit numbers the multiplication can often be done in your head (zero lines of paper). Save the rainforest and learn these handy tricks.
Multiplying By Eleven
14 99 32 87
x11 x11 x11 x11
---- ---- ---- ----
154 1089 352 957
If you're multiplying a two-digit number (call it ab) by eleven, you can find the product with enormous celerity and efficacy, and you may thereby greatly impress your peers. Here's how:
1) Put b on the right. In (22x11), you would put 2 on the right.
2) Put (a+b) to the left of your result. In (22x11), you would put (2+2) = (4) to the left of 2 to get 42. In (98x11), you
would have 8 at the far right and (9+8) = (17) on the left: 178.
3) If you now have a two-digit number, just stick a on the end and you're set. In (22x11), put 2 to the left of (42) and get 242--the answer!.
But if your number has three digits, be more careful: add a to the hundreds place (the third digit from the left).
For example, in (98x11), your result so far is 178; add 9 to the hundreds digit (or add 900) to get 1078.
100a + 10(a+b) + b
(the numerical equivalent
of the above steps)
(10*1 + 1)(10a + b)=(100a + 10b + 10a + b)
=100a + 10(b+a) + b
=100a + 10(a+b) + b
The number eleven is just really exciting like that.
(see also: How to multiply two digit numbers by 11 in your head)
14 13 11
x17 x19 x19
---- ---- ----
238 247 209
Eleven is a teen; in fact, so is ten, but there is another rule for multiplying by ten. (Add a zero.) This algorithm holds
true while both ab and cd are between 10 and 19:
1) Add the first number (ab) to the ones digit of the second (d). For example, in (13x15), add 13 to 5 to get 18.
2) Multiply by ten (put a zero on the end): in (13x15), (13+5) = (18) becomes 180.
3) Add the product of the ones digits (b x d) to this result. For example, with (13x15), (3x5) = (15) is added to the previous sum (180) to get 195.
10(10a + b + d) + bd
(this form is equivalent
to what we get from the steps above)
Given: a = c = 1
(10a + b)(10c + d)=100ac + 10bc + 10ad + bc
=100a + 10b + 10d + bd
(substituting in 1 for a and c, but not everywhere)
=10(10a + b + d) + bd
Now that's interesting. We can reduce this particular identity even further, to
but this does us little good. In terms of mental calculating ability
, it's easier to perform
10((10a+b) + d) than 100+10b+10d, so the simple three-step solution
above remains the best.
The "Tens, Tens" Pattern
46 55 91
x44 x55 x99
---- ---- ----
2024 3025 9009
This one is my favorite. My calc teacher showed it to me, and the proof is trivial, but it's loads of fun.
If you have two two-digit numbers with the same tens digit and ones digits that add to ten (say, 13 and 17, or 82 and 88),
you can rapidly calculate their product. It's so easy:
1) Multiply the ones digits. Leave this product as it is. For example, in (23x27), you would get 3x7 = 21.
2) If the product of the ones digit is less than 10, add a zero to the left. If it is not, leave it alone. There is only one pair of ones digits with which you need this step: in (11x19), you should get 1x9 = 09.
3) Take the tens digit and multiply it by (itself plus one). In (23x27), you would perform (2 x (2+1)) = (2x3) = 6.
4) Put the result from (3) in front of the result from (2). In (23x27), you would stick 6 in front of 21 to get 621. In (11x19), stick (1 * (1+1)) = 2 in front of 09 to get 209--the answer!
100a(a+1) + bc
(the numerical equivalent of the above steps)
Given: c = 10 - b
ab x ac = (10a + b)(10a + c)
= 100a^2 + 10ab + 10ac + bc
= 100a^2 + 10ab + 10a(10-b) + bc
= 100a^2 + 10ab + 100a - 10ab + bc
= 100a^2 + 100a + bc
(hooray for terms that add to zero)
= 100a(a+1) + bc
And that's that.
64 38 91
x44 x78 x11
---- ---- ----
2816 2964 1001
Suppose we have the same ones digit, but tens digits that add to ten, of the form (ab x cb). Can we use a pattern similar
to the one we used before? Yes!
1) Square the ones digit. Example: (37x77) -> (7x7) = (49).
2) If the square of the ones digit is less than 10 (b < 3), stick a zero in front. Example: in (12x92), (2x2) = 04.
3) Multiply the first tens digit by the second (a x c), and add the ones digit (b). For example, in (12x92), this result
is (1x9) + (2) = 11.
4) Now, put the result from (3) in front of the result from (2). For (12x92), we would get 1104.
100(ac+b) + b^2
(this is again the numerical equivalent of the above procedure)
Given: c = 10-a
(10a + b)(10c + b)=100ac + 10bc + 10ab + b^2
=100ac + 10b(10-a) + 10ab + b^2 (substituting for c)
=100ac + 100b - 10ab + 10ab + b^2
=100ac + 100b + b^2 (more terms that add to zero!)
=100(ac + b) + b^2
From the knowledge of the forward and backward "tens" rules, one can derive any number of exciting shortcuts that spring into existence when the ones or tens places are both 5. Since these are of limited use, they are left as an exercise to the reader.
All in all, these handy simplifications will serve you well in life if you know them. They make for good parlor tricks and, if nothing else,
good, clean math.
Extra-special source material for you overachievers who want a little more:
http://220.127.116.11/search?q=cache:smard.cqu.edu.au/Uploads/June_1998/AlgebraMagic.doc -- "teen" number information
http://library.thinkquest.org/17888/prn_ns.shtml -- backwards "tens, tens"
http://mathforum.org/k12/mathtips/beatcalc.html -- a very impressive array of tips for honing one's innate arithmetic ability. Warning: memorization may inspire criminal insanity.