I always tell my students that I am not (nor are any of their teachers) smarter than they are. I do not believe in a single static intelligence quotient, in any event. We were not born knowing the things that we have learned; rather, somebody showed them to us. Obviously, in higher level proof classes, significant preunderstandings are required. But for mental math, there are so many tricks that we can use. Some here are much better, no doubt, than I am at mental math. However, many of my students, at first, are rather surprised at what one can do in one's head. First, they observe it. Then, I show them the way that I did it. I point out that I simply learned easy tricks (or figured them out) ; it has nothing at all to do with me.

So, let us look at some mental math tricks:

Again, I promise you that if I can do this, so can you. And there are a number of people here who are better at mental math than I am. Please feel free to share your mental math tricks here.

Obviously, I am not going to list every trick that I know. I will just post a few. I hope that this will help you and that you have some tricks to show me.

First, let us multiply by 11 in our head.

17 * 11 = 170 (10 seventeens) + 17 (1 more seventeen) = 187

16 * 11 = 160 (10 sixteens) + 16 (1 more sixteen) = 176

Next, let us multiply by 9 in our head :

17 * 9 = 170 (10 seventeens) - 17 (1 seventeen) = 153

16 * 9 = 160 (10 sixteens) - 16 (1 sixteen) = 144

Now, let us multiply by 25 in our head:

24 * 25 = 2400 (100 twentyfours) / 4 = 600

Let us multiply by 5 in our head :

18 * 5 = 180 (10 eighteens) / 2 = 90

Let us multiply again in our head :

21 * 19 = 20 * 20 - 1 = 400 -1 = 399

why ? (x+1) * (x-1) = x^2 -1

In the above case, x was 20

22 * 18 = 20 * 20 - 4 = 400 - 4 = 396

why ? (x+2) * (x-2) = x^2 -4

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If a number has two prime factors that when multiplied together give the number itself, then the number has no other factors besides those prime factors and 1 and the number itself

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Let x be an element of R+. Suppose that x has no factors smaller than the ceiling of the square root of x (ie rounding the square root up to the next positive whole number). Then, the number is prime. (Otherwise, one would be multiplied two numbers together, both of which are greater than the square root of the number, yielding a number greater than x).

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Divisibility :

2 - is it even ?

3 - is the sum of the digits divisible by 3 ?

4 - are the final two digits (the tens and ones) divisible by 4 ?

5 - does the number end in 5 or 0 ?

6 - is it divisible by both 2 and 3 ?

9 - is the sum of the digits divisible by 9 ?

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