## Zeller’s Rule: Day on any date in the calendar

Zeller’s Rule : With this technique named after its founder Zeller, you can solve any ‘Dates and Calendars’ problems.

Zellers rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.

Zeller’s Rule Formula:

F = K + [(13xM – 1)/5] + D + [D/4] + [C/4] – 2C

K = Date => for 25/3/2009, we take 25
In Zeller’s rule months start from march.
M = Month no. => Starts from March.
March = 1, April = 2, May = 3
Nov. = 9, Dec = 10, Jan = 11
Feb. = 12
D = Last two digits of the year, using previous year for January and February. Thus for 2009 = 09
C = The first two digits of century => for 2009 = 20

Example: 25/03/2009

F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 – (2 x 20)
= 25 + 12/5 + 09 + 09/4 + 20/4  – 2×20
=25+2+09+2+5-40 [ We will just consider the integral value and ignore the value after decimal]
= 43 – 40 =

Replace the number with the day using the information given below.
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
7 or 0 = Sunday

So it’s Wednesday on 25th march, 2009.

If the number is more than 7, divide the no. by 7. The remainder will give you the day.

I shall be very grateful if anyone of you can provide me the java script for this formula, so that I can post it here for everyone’s convenience.

Thanks in advance. I hope you will find the above method very useful.

you may find another post named, cyclicity very interesting.

Vineet Patawari – PGDM, ACA, B.COM (H)

## Speed Multiplication by 111 : Vedic Maths

Multiplication of a number consisting of only ones with another number becomes very easy using Vedic Maths techniques. You must see the earlier post on shortcut for multiplying a number by 11

MULTIPLYING A NUMBER BY 111
To multiply a two-digit number by 111, add the two digits and if the sum is a single digit, write this digit TWO TIMES in between the original digits of the number. Some examples:

36×111= 3996
54×111= 5994

The same idea works if the sum of the two digits is not a single digit, but you should write down the last digit of the sum twice, but remember to carry if needed. So

57×111= 6327
because 5+7=12, but then you have to carry the one twice.

For 3 digit numbers
Carry if any of these sums is more than one digit.
Thus 123×111 = 1 | 3 (=1+2) | 6 (=1+2+3) | 5 (=2+3) | 3

Similarly,
241×111 = 26751

For an example where carrying is needed

Say, 352×111=3 | 8 (=3+5) | 10 (=3+5+2)| 7 (=5+2)| 2
= 3 | 8 | 10 | 7 | 2 = 3 | 9 | 0 | 7 | 2
= 39072

(Because of the carries, it may be easier to do the sums and write the answer down from right to left.)

Let me know, if you liked this Vedic Maths trick.

## Vedic Multiplication by 11

Speed Vedic Multiplication Trick

Vedic Multiplication by 11

Step 1.

Assume that there are two invisible 0 (zeroes), one in front and one behind the number to be multiplied with 11

say if the number is 234, assume it to be  0 2 3 4 0

Step 2.

Start from the right, add the two adjacent digits and keep on moving left

02340

Add the last zero to the digit in the ones column (4), and write the answer below the ones column. Then add 4 with digit on the left i.e. 3. Next add 3 with 2. Next 2 with 0.

0+4 = 4

4+3 = 7

3+2 = 5

2+0 = 2

Similarly,

36 x 11 = 0+3   |   3+6   | 6+0  = 396

74 x 11 =0+ 7 |  7+4 |  4+0 =  7  | 11 |  4 = 814   (1 of 11 is carried over and added to next digit, so 7+1 = 8 )
6349 x 11 = (0+6)  |  (6+3)   |   (3+4)   |   (4+9)  |   9+0 =  69839

This method works for all the number, no matter how long or short, times 11. Just try it yourself and get amazed at the simplicity of the concept.

In the next post will learn Vedic Multilplication by 111, 1111, 11111, and so on.

## Vedic Maths Subtraction

#### Learn Amazingly Fast Vedic Mathematics Subtraction

Very often we have to deduct a number from numbers like 1000, 10000, 100000 and so on.

This Vedic Maths Subtraction method found as sutra in ancient vedas, is given below is very useful for such subtractions.

Memory Trick: ALL FROM 9 AND THE LAST FROM 10

Use the formula all from 9 and the last from 10, to perform instant subtractions.

For example 1000 – 357 = ?      (subtraction from 1000)

We simply take each figure in 357 from 9 and the last figure from 10.
Step 1. 9-3 = 6
Step 2. 9-5 = 4
Step 3. 10-7 = 3

So the answer is 1000 – 357 = 643
And that’s all there is to it!

This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.
Similarly 10,000 – 1049 = 8951      (subtraction from 10000)

9-1 = 8
9-0 = 9
9-4 = 5
10-9 = 1

For 1000 – 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083.
So 1000 – 83 becomes 1000 – 083 = 917

Corollary: If last term is 0, keep that last term as 0 and subtract the last non Zero term from 10 .

Illustration: 10000 – 920 = 10000 – 0920 = (9-0) (9-9) (10-2) 0 =9080

Illustration: 100000 – 78010 = (9-7) (9 – 8 ) (9- 0) (10 – 1) 0 = 21990

## Multiply 2 numbers, sum of whose unit places is 10

Vedic Multiplication: This method of multiplication which is from Vedic Maths will make it very easy to multiply two numbers when sum of the last digits is 10 and previous parts are the same

You will get the answer in two parts.

First part, to get left hand side of the answer: multiply the left most digit(s) by its successor

Second part, to get right hand side of the answer: multiply the right most digits of both the numbers.

Example

First part: 4 x (4+1)

Second part: (4 x 6)

Combined effect:  (4 x 5)  | (4 x 6) = 2024

*| is just a separator. Left hand side denotes tens place, right hand side denotes units place

More Examples

37 x 33 = (3 x (3+1)) |  (7 x 3) = (3 x 4) | (7 x 3) = 1221

11 x 19 = (1 x (1+1)) |  (1 x 9) = (1 x 2)  | (1 x 9) = 209

As you can see this method is corollary of  “Squaring number ending in 5”

It can also be extended to three digit numbers like :

E.g. 1: 292 x 208.

Here 92 + 08 = 100, L.H.S portion is same i.e. 2

292 x 208 = (2 x 3) x 10 | 92 x 8  (Note: if 3 digit numbers are multiplied, L.H.S has to be multiplied by 10)

60 | 736 (for 100 raise the L.H.S. product by 0) = 60736.

E.g. 2: 848 X 852

Here 48 + 52 = 100,

L.H.S portion is 8 and its next number is 9.

848 x 852 = 8 x 9 x 10 | 48 x 52 (Note: For 48 x 52, use methods shown above)

720 | 2496

= 722496.

[L.H.S product is to be multiplied by 10 and 2 to be carried over because the base is 100].

Eg. 3: 693 x 607

693 x 607 = 6 x 7 x 10 | 93 x 7 = 420 / 651 = 420651.

Note: This Vedic Maths method can also be used to multiply any two different numbers, but it requires several more steps and is sometimes no faster than any other method. Thus try to use it where it is most effective

## How to Quickly Find Square of Any Number Ending in 5

Finding square of any number with unit’s digit being 5 is the most common, yet very interesting trick of Vedic Maths.  Using this technique you can find the square of any number ending in 5 very easily.  Also explore a quick method of squaring numbers ending in 9. Given below is the step by step explanation of this Vedic Maths Method.

Let us take a 2 digit number in generic form, say the number is a5 (=10a+5), where a is the digit in ten’s place

Square of a5= a x (a+1) | 25

That means a is multiplied by the next higher number, i.e. (a+1). Now let’s take example of a real number ending in 5, say 45.

452 = Left hand side of the answer will be 4 multiplied by its successor i.e. 5 and the right hand side part will always be 25 for squares of numbers of which the unit’s digit is 5.

Giving the answer a x (a+1) | 25 ( |     stands for concatenation}

i.e. 4  x  (4+1) | 25 = 4 x 5 | 25 = 2025

Similarly we can proceed for 3 digit numbers ending in 5

Few more examples:

952=9 x 10 | 25 =9025

1252 = 12 x 13 | 25 = 15625

5052 = 50 x 51 | 25 = 255025

Test yourself

Find out the square of 85, 245, 145, 35, 15, and 95?

Answer: 7225, 60025, 21025, 1225, 225, 9025

Please let us know if you like this Vedic Maths trick