**Number Curiosities**

We ended last month’s Curiosities by getting numbers into shape:

A number that is the sum of a set {1, 2, 3, …, n} is called triangular. We can illustrate the reason for this name by diagramming 1 + 2 + 3 + 4.

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So ten is a triangular number.

The triangular set begins {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, and so on}. Knowing these totals is a big help with puzzles like Sudoku and Ken-Ken.

**Square Numbers and Arrays.** To calculate a second power, we multiply a number by itself. The second power of 5 is 5x5, and 25 belongs to the set of square numbers. To Indicate a second power I’ll use a caret

5 ^ 2 = 5 times 5 = 25.

Squares happen to be the sum of two consecutive triangular numbers.

1 + 3 = 4; 3 + 6 = 9; 6 + 10 = 16, and so on.

What’s going on? We’ll take that array of 10 seen at the top and add the next triangular number to it – upside down.

## + |
## = |

10 + 15 = 25, the square of 5.

We can also find the squares by adding up consecutive odd numbers.

odds: 1 3 5 7 9...

squares: 4 9 16 (and 9 more is 25)...

We look at one more picture to see why this works.

We get six sixes from five fives by adding two 5s and a one, i.e. the odd number 11. In general algebraically it’s

n^2 + 2n + 1 = (n + 1)^2

**Final Digits.** The square of any number must end in one of the digits 0, 1, 4, 5, 6, and 9. The square of 48,693 must end with a 9 (3x3).

The square of any number, ending in 2 must end in a 4. Algebraically its

(10n + 2) times (10n + 2) equals (10n times 10n) + (20 times n) + (20 times n) + (2 times 2),

the four partial products. The first three products have no impact on the final digit.

In symbols: (10n + 2)(10n + 2) = 100n^2 + 40n + 4. Example: 23^2 = 400 + 120 + 4.

**Tasks**

1. Take the eight consecutive squares of 11 through 18, {121, 144, 169, etc.} and partition them into two groups of four that have the same total.

2. Use the partial product algebra above to show that any eight consecutive squares can be so partitioned.

**Head Scratchers**

3. {10, 11, 12} is a consecutive triplet the squares of which total the same as the squares of {13, 14}. Verify this fact. The next similar occurrence will be a sum of four consecutive squares being equal to the sum of the very next three squares. Can you find them? Incredibly, there is no end to these ever-growing strings of square numbers.

For your convenience we link to a table of squares.

4. Martin Gardner once wrote that any squares that end in a double digit must end in -00 or -44. Can you find a reason for this? Can you find some -44 squares? The algebra shown above may be a help.

Next month:** Third Powers – The Cubes**

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