Squares: 1-digit, 2-digit, 3-digit numbers

1^2 = 1

2^2 = 4

3^2 = 9

4^2 = 16

5^2 = 25

6^2 = 36

7^2 = 49

8^2 = 64

9^2 = 81

10^2 = 100 ..... 21^2 = 441
11^2 = 121 ..... 22^2 = 484

12^2 = 144 ..... 23^2 = 529

13^2 = 169 ..... 24^2 = 576

14^2 = 196 ..... 25^2 = 625

15^2 = 225 ..... 26^2 = 676

16^2 = 256 ..... 27^2 = 729

17^2 = 289 ..... 28^2 = 784

18^2 = 324 ..... 29^2 = 841

19^2 = 361 ..... 30^2 = 900

20^2 = 400 ..... 31^2 = 961

n-body problem

http://www.whenwilliusemath.com/didyouknow/newdiscoveries/n-bodyproblem

Looking for postive integers such that... (Reverse)

9 + 9 = 18 and 9 * 9 = 81 ........... 81 is the reverse of 18

3 + 24 = 27 and 3 * 24 = 72 .... 72 is the reverse of 27

with a 3-digit number,

2 + 497 = 499 and 2 * 497 = 994 ... we see 994 is the reverse of 499.

Could you find another 3-digit number whose product is the reverse of the sum?

How about a 4-digit number? a 5-digit number, a 6-digit number, etc.?

Mathematics Genealogy Project

http://genealogy.math.ndsu.nodak.edu/index.php

Maths in the movies

Looking for postive integers such that

The positive integers are the numbers

http://mathworld.wolfram.com/PositiveInteger.html

Here I'm interested to find all positive integers which can be written in the form

(1) (a + b + c)^2 /(abc)
(2) (a + b - c)^2 /(abc)
(3) (a - b + c)^2 /(abc)
(4) (a - b - c)^2 /(abc)

where a, b, c are positive integers

Maths Tricks

FASTEST WAY OF MULTIPLYING 3-DIGIT BY 3-DIGIT NUMBERS


Squaring a number







(1) From a shuffled deck of cards, pick a number between 1 and 10, a count up to find your first chosen card.
(2) The value of your chosen card is your new number. (Face cards are worth 5)
(3) Use your new number to find your next card




Fantastic Math Tricks

http://www.angelfire.com/me/marmalade/mathtips.html