THE BOOK--Playing The Percentages In Baseball

There is the story about the famous mathematician (I think it was Gauss) who was punished for acting up in school when he was quite young (maybe first grade level).  According to the story, he wasn’t allowed to go to recess until he summed all the numbers from 1 to 100.  Gauss instantly realized the total was 5,050, because the numbers to be added consisted of 50 pairs of numbers that each summed to 101.  So he didn’t miss recess.

(At least, I think that’s how the story went.)

I do some of these tricks too.  One of my favorites is figuring out squared numbers.

The difference between n^2 and (n+1)^2 is just 2n+1 which also means the difference grows by 2 each time (12^2 = 144, 13^2 = 169 (a difference of 25 (which is 12+13))

Therefore 14^2 is 27 (13+14) more than 169 which I find to be a much quicker way than trying to do the actual multiplication - especially when you get to a high number.

There is a chapter in Surely You’re Joking, Mr. Feynman where Feynman (one of the best scientists ever, and probably the funniest) talks about beating an abacus salesman who was trying to peddle his wares using little tricks.  One of the stand out parts of an awesome biography.

Craig: I like that little trick.  So, if we know 50-squared (2500), then 49-squared is 99 less than that, or 2401.  Neat.

I hope this isn’t obvious - I like multiplying by eleven.  For 2 digit numbers the product starts with the first digit of whatever you’re multiplying by eleven.  Then, you add the digits together for the middle number, then end with the last digit:

25*11 = 2 2+5(7) 5

or, 275.

whenever 2 digits sum > 9, you have to remember to keep te extra 1:

78*11 = 7 7+8(15) 8

or, 858 - the 1 from the 15 adds to the initial 7. 

works for any size number as long as you can keep track:

12345*11 = 135795

I like to enter the number 5318008 into my calculator and then turn it upside down.

When I was 10 or so, I figured out the trick for adding the numbers from 1 to n, too. I was thinking about that again about a year ago, and I decided I needed to come up with a way to make it work with any starting point (not just 1) and with any interval between numbers.

I don’t remember exactly how, but I came up with this:

x+(x+z)+(x+2z)...+y
=
(y^2-x^2+yz+xz)/2z

So for example:

-1+0.5+2+3.5+5+6.5+8+9.5+11
=
(11^2-(-1)^2+11*1.5+(-1)*1.5)/2*1.5
(121-1+16.5+(-1.5)/3
135/3
45

Maybe not that quick, but useful if you have enough numbers in the sequence.

Squaring multiples of 5:
x5 ^2 = (x*x+1)25
So,
25*25 = 625 (6 = 2*3)
35 * 35 = 1225 (12 = 3*4)
etc…

This pretty much sets the standard: http://www.ted.com/talks/arthur_benjamin_does_mathemagic.html

Fun show, even if he didn’t get 100% of them correct.  If you read the comments, you see he made two mistakes.  Not only him, but the “checkers” confirmed the wrong answers as well, which goes to show how crappy random checkers with calculator in hand are.

Pick a number, reverse the digits, subtract the two, divide by 9 - voila! A palindrome.

a*10^n + b*10^n-1...+y*10^1 + z*10^0
- z*10^n + y*10^n-1...+b*10^1 + a*10^0
= (a-z)*10^n + (b-y)*10^n-1...+(y-b)*10^1 + (z-a)*10^0
= (a-z)*[10^n-10^0] + (b-y)*[10^n-1-10^1] + ...

Each of the terms in square brackets will be divisible by 9, and by grouping terms we’ve gone from n terms to n/2 terms.  Upon dividing, the term in brackets will be of the form have the first i digits equal to one, and the rest equal to zero, where i is its position in the sum.  The summation thus gives a palindrome.

Randall Patrick Munroe’s check to verizon:
http://xkcd.com/verizon/

If you want to add up the first “n” cubes, just do the n(n+1)/2 trick and square it: sum of 1 to 5 is 15, sum of 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1+8+27+64+125 = 225 = 15^2. Weird, but really impressive.

OK, if you want to know the first n squares, it’s n(n+1)(2n+1)/6. A bit harder to remember than the cubes trick.

If you want to apply any of these formulas to a sequence not beginning at 1, just subtract off what you don’t want: say, all the cubes from 6 to 10, figure (10*11/2)^2, and subtract off the ones from 1 to 5, which we know is 225. So, 55^2 - 225 = 2500 + 250 + 250 + 25 - 225 = 2800.

If you want to know the sum of a series like 3+6+9+12+15+...+30, note that it’s just 3 times the sum from 1 to 10, so 10*11/2 = 55, 3*55 = 165.

I used a square root trick in high school. I’d be able to approximate a square root fairly closely to two or three decimal places.

The basic idea was this:
1) Figure out which two perfect square roots the number is between.
2) For that “segment”, the square root function is almost linear, so use that to figure out a starting guess.
3) Adjust the guess up a little bit - the function is always concave down, so the graph always lies a little above the linear approximation. The closer you are to zero, adjust up more. If you are using two non-consecutive perfect roots (for large numbers), adjust up more.

For instance, 134:
1) is between 121 and 144 (11^2 and 12^2).
2) 13 (134-121) divided by 23 (144-121) is a bit below 0.6 - guess around 0.57.
3) My guess is 11.59. (checks Excel) Actual is 11.576.

You can take the difference of squares thing a little bit further. If ever you need to multiply two numbers that are equidistant from an easily squarable number, you simply square that easily squarable number, then subtract the square of the distance that the easily squarable number is away from the other two numbers. Sorry, I know that probably didn’t sound very good but I have no other way of expressing it. Example, 54 and 46 are both 4 away from 50:

54 * 46 = 50^2 - 4^2
50^2 - 4^2 = 2500 - 16
54 * 46 = 2484

Learned that during an SAT prep course way back when and I’ve since wow’ed a few people with that pretty simple trick.

Reading through this thread just reiterates to me that math, outside of baseball stats, really isn’t much fun ... even if you’re good at it.

Now, using numbers to make dirty words on a calculator ... now that’s where it’s at.

CC: I had the exact opposite reaction.  I love this stuff.

If you want to find the remainder of a number divided by 27, you can apply the following methods:

Method 1:  Break the number up into groups of 3 digit numbers, starting from the one’s digit.  For example, 9457399488590197261 would become
9 457 399 488 590 197 261

Add all of these numbers together to get a new number.  In the above example that sum is 9 + 457 + 399 + 488 + 590 + 197 + 261 = 2401.  This new number will have the same remainder when divided by 27 as the original number.

If the original number is negative, then each group of 3-digit numbers should be treated as negatives.  For example, -2849950541 would break into -2 -849 -950 -541
And the sum would be -2432.  So -2849950541 and -2432 will have the same remainder when divided by 27. 

Method 2: 

Partition the number into 3 sets as follows:

Set A includes A1, A4, A7, etc.
Set B includes A2, A5, A8, etc.
Set C includes A3, A6, A9, etc.

Where A1 is the one’s digit, A2 is the ten’s digit, A3 is the hundred’s digit, and so forth.  If the number is negative then each of the digits A1, A2, A3, etc must be treated as negative numbers. 

Separately calculate the sum of the members of each set.
‘Sum A’ will be the sum of the members of Set A
‘Sum B’ will be the sum of the members of Set B
‘Sum C’ will be the sum of the members of Set C

Calculate a value ‘w’ using this equation:

w = x*(Sum A) + y*(Sum B) + z*(Sum C)

where x = 1 or -26
y = 10 or -17
z = 19 or -8

It doesn’t matter what combination of choices you make for x, y, and z from that list.  The number ‘w’ will have the same remainder when divided by 27 as the original number. 

Example:  45031109
Sum A = 9 + 1 + 5 = 15
Sum B = 0 + 3 + 4 = 7
Sum C = 1 + 0 = 1

w = 1*(15) + 10*(7) - 8*(1) = 77

So 45031109 and 77 have the same remainder when divided by 27.

So to find the remainder of any number divided by 27, repeat these two methods in any order and as many times as needed until you get a number between 0 and 26, inclusive. 

Example:  9457399488590197261

Method 1:  9 + 457 + 399 + 488 + 590 + 197 + 261 = 2401

Apply Method 1 to the number ‘2401’: 2 + 401 = 403

Apply Method 2 to the number ‘403’:  w = 1*(3) + 10*(0) -8*(4) = -29

Apply Method 2 to the number ‘-29’:  w = 1*(-9) - 17*(-2) = 25

The remainder is 25.  Furthermore, we know that 9457399488590197261, 2401, 403, -29, and 25 all have the same remainder when divided by 27. 

And Method 2 can get the job done by itself

Example:  9457399488590197261

Method 2:  ‘Sum A’ = 1 + 7 + 0 + 8 + 9 + 7 + 9 = 41
‘Sum B’ = 6 + 9 + 9 + 8 + 9 + 5 = 46
‘Sum C’ = 2 + 1 + 5 + 4 + 3 + 4 = 19

w = 1*(41) + 10*(46) - 8*(19) = 349

Reapply Method 2 to the number ‘349’:  w = -26*(9) + 10*(4) + 19*(3) = -137

Reapply Method 2 to the number ‘-137’: w = 1*(-7) – 17*(-3) + 19*(-1) = 25

The remainder is 25.  Furthermore, we know that 9457399488590197261, 349, -137, and 25 all have the same remainder when divided by 27.

17/
That’s easier than long division?
There should be a limit in the number of steps involved to be able to still call it a “trick”

Yes it is easier than long division for really long numbers, though it doesn’t come with the benefit of a quotient.

Circle/15

What the 7734 is wrong with you?!?!

10/

Are there more restrictions to this one? I tried out working out the proof as described and ran into a problem.

Using the method provided, I wind up with a number of the format

(a-z)*10^(n-1) + ((a+b)-(y+z))*10^(n-2) + ((a+b+c)-(x+y+z))*10^(n-3) + ... + ((a+b+c)-(x+y+z))*10^2 + ((a+b)-(y+z))*10^1 + (a-z)*10^0

While the terms are mirrored as a palindrome should, it brings up a question. What if the coefficient to any digit is greater than 9?
For example, take the number 897344. (8+9+7) - (3+4+4) yields 13. Just going through it as written -

897344 - 443798 = 453546
453546/9 = 50394, Not a palindrome

Similarly, try it with a number like 8911.

8911 - 1198 = 7713
7713/9 = 857

Am I missing something vital?

It’s probably the oldest trick in the book, but I have done it obsessively for years:

Add all digits in a number together. If they tally to a number divisible by three, then the original number is divisible by three. Same story with nine. My OCD will never forgive MLB if it gets rid of he 162-game season; the part of me that wants this habit broken wants a 154-game season again.

Tango’s originally post notes that the sum of odd numbers less than N is N/2 squared. There’s an even easier way (for me, at least) to remember that: the sum of the first N odd numbers is N squared.

I don’t recall learning or reading about this, but I noticed the pattern when I was still in HS:
1 = 1^2
1+3 = 2^2
1+3+5 = 3^3, etc.
I did a quick proof by induction to show it in general.

When applying to colleges I was nominated for a prestigious scholarship at the University of North Carolina which included a separate interview process. One of the questions the committee asked me was if I had ever had an original thought. I said no, saying there had been billions of people with many thoughts each, so the chances that any of my thoughts were truly original seemed exceedingly small. Then, being a smart-alecky 17 year old, I added “besides, if I said I’d ever had an original thought, you’d ask me what it was, and I don’t want to have to answer that question.”

But I then changed definitions a bit - saying that if by “original” they meant uninfluenced by the outside world, then sure, I’d had some, and I gave as an example the observation that the sum of the first N odd natural numbers is N^2, along with my description of the proof by induction.

I didn’t get the scholarship.

Math trick to get someone’s phone number.

1. Grab a calculator
2. Key in the first three digits of your phone number, not the area code
3. Multiply by 80
4. Add 1
5. Multiply by 250
6. Add the last 4 digits of your phone number
7. Add the last 4 digits of your phone number again
8. Subtract 250
9. Divide number by 2

Suppose 1/A = 1/B + 1/C and you know B and C.
This is an annoying problem that comes up a lot; it’s annoying because it is much too easy to make a mistake.

There’s a trick: A = B*C/(B+C).  This means I can solve parallel-resistor problems in my head faster than my students can on their calculators.

25: That’s actually James’ Power-Speed Number.

Fun Fact: James was trying to have three, instead of two, numbers to combine.  Like, I dunno, HR, BB, SB or something.  Well, all we have to do is add 1/BB to the sequence, and we have what we want.

@ Tango

No doubt. You guys are much more “math dudes” than I am. I’m interested in math primarily because it’s included in baseball stats. Without baseball stats, my interest in math goes bye-bye.

I think it’s cool what can be done with math, but not passionate about it. Much respect to math, just not my favorite hobby.

-------------------------------

At 20, I spend too much time thinking about 58008.

Craig Glaser said:

There is a chapter in Surely You’re Joking, Mr. Feynman where Feynman (one of the best scientists ever, and probably the funniest) talks about beating an abacus salesman who was trying to peddle his wares using little tricks.  One of the stand out parts of an awesome biography.

I was most impressed by the chapter entitled “You Just Ask Them?”, but that’s just me.

You sold yourself short on the “divide by 7” trick. The six decimal places just repeat infinitely, so you not only have the answer to six decimal places, you have it exactly (in an infinitely repeating sense).

So 1/7 = 0.142857 142857 142857 ....
and
4/7 = 0.57 142857 142857 142857 ....

Even cooler!