Square remainders
| N | N2 | rem |
N | N2 | rem |
|---|
| 1 | 1 | 1 | 7 | 49 | 1 |
| 2 | 4 | 4 | 8 | 64 | 4 |
| 3 | 9 | 9 | 9 | 81 | 9 |
| 4 | 16 | 4 | 10 | 100 | 4 |
| 5 | 25 | 1 | 11 | 121 | 1 |
| 6 | 36 | 0 | 12 | 144 | 0 |
"All squares, on division by twelve, leave a remainder which is also a
square."
rem = remainder mod twelve
The statement is also true, by the way, if we replace the word
"twelve" by "five", "sixteen", or "twenty-four".
Can you find other solutions?
Quick Squares
(NB "10" means any number base)
A handy rule for squaring numbers
(mental, or pencil & paper)
- Find the difference, "d" between the number, "n", and the next multiple of "10" (above or below n)
- If you needed to add the difference, d, to get the multiple, also form the
number n-d;
if you needed to subtract the difference, also form the number n+d.
- Multiply these numbers (n+d) and (n-d) and then add the square of
the difference, d2, to this answer.
Which says that: n2 = (n-d)(n+d) + d2
Here are some examples in numbers:
- (base ten)
n = "38"; next multiple of 10 is 38+2, i.e. 40; this is n+d, so d = 2
also form n-d: 38 - 2 = 36.
Square of 38 is 36 x 40 + 22 = 1440+4 = 1444
- (base seven)
n = "36" ; next multiple is 40, or 36 + 1
also form 36 - 1 = 35.
Square of 36 is 35 x 40 + 12 = 2060 + 1 = 2061.
- (any base greater than 4)
n = "12" ; nearest multiple is 10, or 12 - 2;
also form 12 + 2 = 14.
Square of 12 is 14 x 10 + 22 = 140 + 4 = 144
And the proof? [ (n-d)(n+d) ] +d2 =[ n2 -nd +nd + d2 ] + d2
= [n2 - d2] + d2 = n2.
