# Mathemagics

## Doomsday ruleโ

The Doomsday rule is an algorithm devised by John Conway to quickly calculate the weekday of any date in history. The algorithm works on the following principle:

There is a set of dates (called doomsdays), that for any year, all fall on the same day (called the anchor day of the year).

Using the anchor, we can find out the weekday for any date using simple arithmetic.

The doomsdays we need to remember in order to use the algorithm are:

MM/DDFull dateMnemonic
1/3 (1/4 for leap years)Jan 3 / Jan 4-
2/28 (2/29 for leap years)Feb 28 / Feb 29-
3/14Mar 14Pi day
4/4, 6/6, 8/8, 10/10, 12/12April 4, Jun 6, Aug 8, Oct 10, Dec 12Even months except Feb
5/9, 9/5, 7/11, 11/7May 9, Sept 5, Jul 11, Nov 79-to-5 at 7-11

Other memorable doomsdays are Jul 4 (Independence day), Oct 31 (Halloween) and Dec 26 (Boxing day).

In total, there are 52 doomsdays in a year. In leap years, the doomsdays in the month of January and February are shifted by one day. However, the total remains the same.

The algorithm involves the following steps.

1. Find the anchor day for the century (required for step 2).
2. Find the anchor day for the year. All doomsdays fall on this anchor day.
3. Count forward/backward from the nearest doomsday to the specified date.

For calculations, we assign an index for each weekday. Starting from Monday ($1$) to Sunday ($7$), the index equals $rank\mod 7$.

DayIndexMnemonic
Monday1One-day
Tuesday2Twos-day
Wednesday3Threes-day
Thursday4Fours-day
Friday5Five-day
Saturday6Sixtur-day
Sunday0None-day
##### Example

Jul 27, 1987

1. Anchor day for the century (1900s) = 3 (Wednesday)

2. Anchor day for the year (1987) = 6 (Saturday)

3. Nearest doomsday to Jul 27 is Jul 11 = Saturday (from step 2).

July 25 = July 11+7+7 = Saturday

July 27 = Monday.

### Finding the anchor day for the centuryโ

Given a year $y$,

Let $c$ = the first 2 digits of the year (or $\left\lfloor\frac{y}{100}\right\rfloor$)

#### Method 1โ

Anchor = $5\times(c\mod 4)\mod 7$ + Tuesday

#### Method 2โ

Value of $c\mod 4$Anchor
0Tuesday
1Sunday
2Friday
3Wednesday
##### Example

1700s

$17\mod 4 = 1 \implies$ Sunday

Or

$5\times(17\mod 4)\mod 7$ + Tuesday = $5$ + Tuesday = Sunday

The following table lists the anchor days for 1500s to 2600s:

CenturyAnchor day (Index)
1500s, 1900s, 2300sWednesday (3)
1600s, 2000s, 2400sTuesday (2)
1700s, 2100s, 2500sSunday (0)
1800s, 2200s, 2600sFriday (5)

Mnemonics for recent centuries:

• 1900s: We-in-dis-day (most living people were born in that century)
• 2000s: Twos-day or Y-Tue-K
• 2100s: Twenty-one-day is Sunday (2100 is the start of the next century)

### Finding the anchor day for the yearโ

Given a year $XXYY$,

#### Method 1โ

Let $a$ = Anchor day of the century

$b = \left\lfloor\frac{YY}{12}\right\rfloor$

$c = YY\mod 12$

$d = \left\lfloor\frac{c}{4}\right\rfloor$

Anchor day of the year = $(a+b+c+d)\mod 7$

#### Method 2โ

Anchor = $\left(YY+\left\lfloor\frac{YY}{4}\right\rfloor\right)\mod 7$

#### Method 3: The "odd+11" methodโ

1. Let $t = YY$
2. If $t$ is odd, add $11$.
3. $t = \frac{t}{2}$
4. If $t$ is odd, add $11$.
5. $t = 7-(t\mod 7)$
6. Count forward $t$ days from the century's anchor day to get the year's anchor day.
##### note

Each common year advances the anchor day by one day. Each leap year advances it by two days.

##### Example

1987

Let $a = 3$

$b = \left\lfloor\frac{87}{12}\right\rfloor = 7$

$c = 87\mod 12 = 3$

$d = \left\lfloor\frac{3}{4}\right\rfloor = 0$

Anchor = $(3+7+3+0)\mod 7 = 13\mod 7 = 6$ = Saturday

It can be useful to use your index, middle, and ring fingers and the pinkie to store the values for $a$, $b$, $c$, and $d$ respectively during mental calculations.

Or using "odd+11" method:

$87$ -> $98$ -> $49$ -> $60$ -> $7-(60\mod 7)$ = $7-4$ = $3$ -> Wednesday + 3 = Saturday

## Magic squaresโ

To create a 4x4 magic square that adds upto $p$, let $c = p-33$ and then replace the value of $c$ in the following square:

$14$$c$$12$$7$
$11$$8$$13$$c+1$
$5$$10$$c+2$$16$
$c+3$$15$$6$$9$

## Rapid cube rootโ

This trick relies on the fact that the cubes of single digit whole numbers end in unique digits.

$0^3 = 0$
$1^3 = 1$
$2^3 = 8$
$3^3 = 27$
$4^3 = 64$
$5^3 = 125$
$6^3 = 216$
$7^3 = 343$
$8^3 = 512$
$9^3 = 729$

Example: Cube root of $389017$ ($AAABBC$) $= 73$

Steps:

1. Cube just lower than $AAA\ (389) = 343 = 7^{3}$
2. Cube ending in $C\ (7) = 27 = 3^{3}$

## Divisibility rulesโ

NumberTestExamples
$2$The last digit is even$0, 2, 4$
$3$The sum of digits is divisible by 3$357: 3+5+7=15/3=5$
$4$The last 2 digits form number that is divisible by 4$732: 32/4=8$
$5$Ends in 0 or 5$7330, 85$
$6$Is divisible by 2 and 3$72$
$7$The alternating sum of blocks of three from right to left gives a multiple of 7$1,369,851: 851โ369+1=483=7\times69$
$8$The last three digits form a number that is divisible by 8$28,152: 152=8\times19$
$9$The sum of the digits form a number that is divisible by 9$2880: 2+8+8+0=18=2\times9$
$10$The ones digit is 0$270, 50$
$11$The alternating sum of the digits is divisible by 11$918,082: 9โ1+8โ0+8โ2=22=2\times11$
$12$Is divisible by 3 and 4$336$
$13$The alternating sum of blocks of three from right to left gives a multiple of 13$2,911,272: 272-911+2=-637=13\times-49$
$14$Is divisible by 2 and 7$238$
$15$Is divisible by 3 and 5$415$

## Miscellaneousโ

• Sum of 10 consecutive fibonacci numbers is always equal to the 7th term in the series times 11.

$15+20+35+55+90+145+\underline\textbf{235}+380+615+995$

$=235\times11=2585$