Welcome to Universe of Maths!

What are large numbers after crores called?

noreply@blogger.com (Shruthi) — Thu, 19 Nov 2009 19:46:00 +0000

Recently my friend had asked me about how the large numbers are called in India? I was stupefied after mentioning Lakhs and Crores. Thanks to google, I could find names of the large numbers after crores. Below is the table of how large numbers are called in the Indian Numbering System.

Number	Figure	Indian Name	American Name
10⁵	1,00,000	Lakh	One hundred thousand
10⁷	1,00,00,000	Crore	Ten million
10⁹	1,00,00,00,000	Arawb	One billion
10¹¹	1,00,00,00,00,000	Kharawb	One hundred billion
10¹³	1,00,00,00,00,00,000	Neel	Ten trillion
10¹⁵	1,00,00,00,00,00,00,000	Padma	One quadrillion
10¹⁷	1,00,00,00,00,00,00,00,000	Shankh	One hundred quadrillion
10¹⁹	1,00,00,00,00,00,00,00,00,000	Mahashankh	Ten quintillion

At present, the Indian numbering system is used in India, Pakistan, Bangladesh, Nepal and Myanmar (Burma) and unlike American and European Numbering system, which is based on grouping by three decimal places, Indian Numerical system is based on grouping by two decimal places.

You can know more on the history of large numbers and sanskrit names at http://en.wikipedia.org/wiki/History_of_large_numbers

Number 9 table and patterns

noreply@blogger.com (Shruthi) — Wed, 28 Oct 2009 02:12:00 +0000

Is memorizing 9 table tough for you? If yes, then after reading this article you would be able to easily write the table.

Just have a look at the 9 table below
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10= 90

Now, carefully look at the product/results. Did you notice anything? Firstly,the Tens place of the result is increasing order of 0 to 8 and the Units place of the result is nothing but decreasing order of 9 to 0. Secondly, the products are digital reverse of each other. Look at the table below.

9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54 (reverse of 45)
9 x 7 = 63 (reverse of 36)
9 x 8 = 72 (reverse of 27)
9 x 9 = 81 (reverse of 18)
9 x 10= 90 (reverse of 09)

Last but not the least, sum the digits of the result is always 9. Look at the below table for reference.
9 x 1 = 9 | 0 + 9 = 9
9 x 2 = 18 | 1 + 8 = 9
9 x 3 = 27 | 2 + 7 = 9
9 x 4 = 36 | 3 + 6 = 9
9 x 5 = 45 | 4 + 5 = 9
9 x 6 = 54 | 5 + 4 = 9
9 x 7 = 63 | 6 + 3 = 9
9 x 8 = 72 | 7 + 2 = 9
9 x 9 = 81 | 8 + 1 = 9

Now that you know how to write the 9 table, check out some cool patterns that can be generated with number 9.

9 x 0 + 1 = 1
9 x 1 + 2 = 11
9 x 12 + 3 = 111
9 x 123 + 4 = 1111
9 x 1234 + 5 = 11111
9 x 12345 + 6 = 111111
9 x 123456 + 7 = 1111111
9 x 1234567 + 8 = 11111111

9 x 9 + 7 = 88
9 x 98 + 6 = 888
9 x 987 + 5 = 8888
9 x 9876 + 4 = 88888
9 x 98765 + 3 = 888888
9 x 987654 + 2 = 8888888
9 x 9876543 + 1 = 88888888

1/9 = 0.111111111111
1/99 = 0.010101010101
1/999 = 0.001001001001
1/9999 = 0.000100010001
1/99999 = 0.000010000100001
so on.

Multiply the numbers close to the power of 10s in just 5 seconds! - Part 2 and Part 3

noreply@blogger.com (Shruthi) — Fri, 05 Jun 2009 00:25:00 +0000

In my last article "Multiply the numbers close to the power of 10s in just 5 seconds! - Part 1", I have mentioned a simple Vedic mathematics method of how to multiply numbers that are below to the power of 10. In this article, I would mention about - Part 2: Multiplying numbers greater than the power of 10s. and Part 3: Multiplying numbers of which one is greater than power of 10 and the other is less than power of 10.

As mentioned in previous article, in Vedic mathematics this method is known as "Nikhilam Navatashcaramam Dashatah - All from 9 and the last from 10"

Part 2: Multiplying numbers greater than power of 10s

Example 1: 12 X 15

Step 1: Identify the base (power of 10) to which the number is close. Here, 12 and 15 are close to 10. Therefore, base is 10

Step 2: Identify by how much both the numbers are greater than the base. Here, 12 is more by 2 and 15 is more by 5.

Step 3: Write the original numbers along with the number by which it is greater than 10 as below.

12 + 2

15 + 5

Step 4: Now, Cross add the original number and the surplus number. That is add 12 to 5 or add 15 to 2. So, we get

12 + 2

15 + 5

-----------

Step 5: Now, vertically multiply the two surplus numbers. Here, the surplus numbers are 2 and 5. Therefore, we have

12 + 2

15 + 5

-----------

17 / 0

-----------

Step 6: Now, since the base is 10, the right hand most digit is units and can have only one digit. Therefore, keep the 0 of the 10 on the right hand side and carry the 1 over the left and change the 17 into 18.

12 + 2

15 + 5

-----------

17 / 0 = 180

-----------

Note: If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to left hand side of the answer.

Therefore we have 12 X 15 = 180

Example 2: 101 X 109

Step 1: Identify the base (power of 10) to which the number is close. Here, 101 and 109 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are greater than the base. Here, 101 is more by 1 and 109 is more by 9.

Step 3: Write the original numbers along with the number by which it is greater than 100 as below.

101 + 1

109 + 9

Step 4: Now, Cross add the original number and the surplus number. That is add 101 to 9 or add 109 to 1. So, we get

101 + 1

109 + 9

-----------

110

-----------

Step 5: Now, vertically multiply the two surplus numbers. Here, the surplus numbers are 9 and 1. Therefore, we have

101 + 1

X |

109 + 9

-----------

110 / 9

-----------

Step 6: Now, since the base is 100, the right hand most digit is tens and units and should have two digit. Therefore, fill the vacancies by adding zero. thus

101 + 1

X |

109 + 9

-----------

110 / 09 = 11009

-----------

Therefore we have 101 X 109 = 11009

Note: If the Right hand side contains less number of digits than the number of zeros in the base, the remaining digits are filled up by giving zero or zeroes on the left side of the right hand side digit.

Example 3: 125 X 101

Step 1: Identify the base (power of 10) to which the number is close. Here, 125 and 107 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are greater than the base. Here, 125 is more by 25 and 101 is more by 1.

Step 3: Write the original numbers along with the number by which it is greater than 100 as below.

125 + 25

101 + 1

Step 4: Now, Cross add the original number and the surplus number. That is add 125 to 1 or add 101 to 25. So, we get

125 + 25

101 + 1

-----------

126

-----------

Step 5: Now, vertically multiply the two surplus numbers. Here, the surplus numbers are 25 and 1. Therefore, we have

125 + 25

101 + 1

-----------

126 / 25 = 12625

-----------

Therefore we have 125 X 101 = 12625

Part 3: Multiplying a number which is greater than power of 10, with a number which is less than power of 10

Example 1: 9 X 15

Step 1: Identify the base (power of 10) to which the number is close. Here, 9 and 15 are close to 10. Therefore, base is 10

Step 2: Identify by how much the numbers are greater than or less than the base. Here, 9 is less by 1 and 15 is more by 5.

Step 3: Write the original numbers along with the number by which it is greater / less than 10 as below.

15 + 5

9 - 1

Step 4: Now, based on the signs, cross add and cross subtract the numbers accordingly. That is add 9 to 5 or subtract 15 by 1. So, we get

15 + 5

9 - 1

-----------

Step 5: Now, vertically multiply the surplus and deficit numbers along with the signs. Here, the numbers are 1 and 5. Therefore, we have

15 + 5

9 - 1

-----------

14 / -5

-----------

Step 6: Now, to get rid of the minus sign, subtract the right hand side by the base and less the left hand side by 1. Here, we have base as 10 and right hand side number as 5. therefore we have 10 - 5 =5. Hence,

15 + 5

9 - 1

-----------

14 / -5 = 135

-----------

Therefore we have 9 X 15 = 135

Example 2: 101 X 97

Step 1: Identify the base (power of 10) to which the number is close. Here, 101 and 97 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are greater than and less than the base. Here, 101 is more by 1 and 97 is less by 3.

Step 3: Write the original numbers along with the number by which it is greater than and less than 100 as below.

101 + 1

97 - 3

Step 4: Now, based on the signs, cross add and cross subtract the numbers accordingly. That is subtract 101 by 3 or add 97 to 1. So, we get

101 + 1

97 - 3

-----------

Step 5: Now, vertically multiply the surplus and deficit numbers along with the signs. Here, the numbers are 3 and 1. Therefore, we have

101 + 1

X |

97 - 3

-----------

98 / -3

-----------

Step 6: Now, to get rid of the minus sign, subtract the right hand side by the base and less the left hand side by 1. Here, we have base as 100 and right hand side number as 3. therefore we have 100 - 3 =97. Hence,

101 + 1

X |

97 - 3

-----------

98 / -3 = 9797

-----------

Therefore we have 101 X 97 = 9797

Example 3: 10006 X 9999

10006 + 6

9999 - 1

-----------

10005 / -6 = (10005 - 1) / (10000 - 6) = 100049994

-----------

Therefore we have 10006 X 9999 = 100049994

I am sure with little bit of practice, you can successfully perform multiplications very fast when compared to the traditional method. Also, please note that these methods should be applied accordingly for which it is designed. This method may not give an effective result, if you are trying to multiply 67 and 72. Vedic mathematics defines many methods by which calculations can be made simple, but you need to use them accordingly.

Hope this method was useful for you. If you have any other methods do share and let me know if you have any questions regarding these methods.

1089 pattern and facts.

noreply@blogger.com (Shruthi) — Thu, 04 Jun 2009 18:02:00 +0000

* Multiplying the number 1089 by the integers from 1 to 9 produces a pattern: multipliers adding up to 10 give products that are the digit reversals of each other.

1089 x 1 = 1089

1089 x 2 = 2178

1089 x 3 = 3267

1089 x 4 = 4356

1089 x 5 = 5445 (Palindrome)

1089 x 6 = 6534 (reverse of 4356)

1089 x 7 = 7623 (reverse of 3267)

1089 x 8 = 8712 (reverse of 2178)

1089 x 9 = 9801 (reverse of 1089)

Also note the patterns within each column. First column is increasing order of 1 to 9, second column is increasing order of 0 to 8, third column is reverse of second column and fourth column is reverse of first column.

* Another cool aspects are

- 1/1089 = 0.00 09 18 27 36 45 54 63 72 81 ..... remember the 9 table

- 1/9801 = 0.00 01 02 03 04 05 06 07 08 09 10 11 12 13 14...

Multiply the numbers close to the power of 10s in just 5 seconds! - Part 1

noreply@blogger.com (Shruthi) — Wed, 03 Jun 2009 22:31:00 +0000

Multiplication of numbers that are close to the power of 10s (i.e 99999 x 99997 ; 99982 99992) can be easily done in just 5 seconds. Tough to believe it right, but it is possible and all that you need to know is basic subtraction, addition and multiplication tables till 5. Just, follow the below examples with their explanation and by the end of the article you would be all set.

Part 1: Multiplying numbers below to the power of 10s

Example 1: 9 X 6

Step 1: Identify the base (power of 10) to which the number is close. Here, 9 and 6 are close to 10. Therefore, base is 10

Step 2: Identify by how much both the numbers are less than the base. Here, 9 is less by 1 and 6 is less by 4.

Step 3: Write the original numbers along with the number by which it is less than 10 as below.

9 - 1

6 - 4

Step 4: Now, Cross subtract the original number and the deficit number. That is subtract 9 by 4 or subtract 6 by 1. So, we get

9 - 1

6 - 4

-----------

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficit numbers are 1 and 4. Therefore, we have

9 - 1

X |

6 - 4

-----------

5 / 4 = 54 (since base is 10 we can have 1 digit on the right hand side which is 4)

-----------

Therefore we have 9 X 6 = 54

Example 2: 7 X 6

Step 1: Identify the base (power of 10) to which the number is close. Here, 7 and 6 are close to 10. Therefore, base is 10

Step 2: Identify by how much both the numbers are less than the base. Here, 7 is less by 3 and 6 is less by 4.

Step 3: Write the original numbers along with the number by which it is less than 10 as below.

7 - 3

6 - 4

Step 4: Now, Cross subtract the original number and the deficit number. That is subtract 7 by 4 or subtract 6 by 3. So, we get

7 - 3

6 - 4

-----------

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 3 and 4. Therefore, we have

7 - 3

X |

6 - 4

-----------

3 / 2

-----------

Step 6: Now, since the base is 10, the right hand most digit is units and can have only one digit. Therefore, keep the 2 of the 12 on the right hand side and carry the 1 over the left and change the 3 into 4.

7 - 3

X |

6 - 4

-----------

3 / 2 = 42

-----------

Therefore we have 7 X 6 = 42

Note: If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to left hand side of the answer.

Example 3: 91 X 99

Step 1: Identify the base (power of 10) to which the number is close. Here, 91 and 99 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are less than the base. Here, 91 is less by 9 and 99 is less by 1.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

91 - 9

99 - 1

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 99 by 9 or subtract 91 by 1. So, we get

91 - 9

99 - 1

-----------

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 9 and 1. Therefore, we have

91 - 9

X |

99 - 1

-----------

90 / 9

-----------

Step 6: Now, since the base is 100, the right hand most digit is tens and units and should have two digit. Therefore, fill the vacancie by adding zero. thus

91 - 9

X |

99 - 1

-----------

90 / 09 = 9009

-----------

Therefore we have 91 X 99 = 9009

Example 4: 93 X 97

Step 1: Identify the base (power of 10) to which the number is close. Here, 93 and 97 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are less than the base. Here, 93 is less by 7 and 97 is less by 3.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

93 - 7

97 - 3

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 93 by 3 or subtract 97 by 7. So, we get

93 - 7

97 - 3

-----------

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 7 and 3. Therefore, we have

93 - 7

X |

97 - 3

-----------

90 / 21 = 9021

-----------

Therefore we have 93 X 97 = 9021

Example 5: 883 X 997

Step 1: Identify the base (power of 10) to which the number is close. Here, 883 and 997 are close to 1000. Therefore, base is 1000.

Step 2: Identify by how much both the numbers are less than the base. Here, 883 is less by 117 and 997 is less by 3.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

883 - 117

997 - 003

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 883 by 3 or subtract 117 by 7. So, we get

883 - 117

997 - 003

---------------

880

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 117 and 3. Therefore, we have

883 - 117

X |

997 - 003

---------------

880 / 351 = 880351 (since base is 1000 we can have 3 digits on the right hand side)

-----------

Therefore we have 883 X 997 = 880351

This method is known as "Nikhilam Navatashcaramam Dashatah - All from 9 and the last from 10" in Vedic Mathematics.

Note:

1. This method can be easily applied to the numbers that are near to the base 10s i.e., 10,100, 1000....

2. If one number is near base of 10s and the other is not, then also it can easily be applied - Ex 88 X 99 or 25 X 98 and above Example 5.

3. The deficit numbers can easily written by following the rule that all the digits (of the original number) are to be subtracted from 9 but the last (right hand most one) one should be subtracted from 10. Ex deficit of 63 is 37 (6-9 =3 / 10-3=7).

With little bit of practice by following the above method, I am sure you can do multiplication of the numbers near to power of 10s in just 5 seconds. Wondering that above method is good for the numbers that are below base 10s but what about numbers which are above it? Don't worry, Vedic mathematics has a solution for it too, I shall discuss the same in my next article, till then practice this one and let me know if you have any questions.

Maths Trick - Guess the number

noreply@blogger.com (Shruthi) — Wed, 03 Jun 2009 16:29:00 +0000

1. Ask your friend to think of any number between 1 to 10.

Lets say, s/he had thought of 3.

2. Ask him/her to double the number.

3. Ask him/her to add 10 to the result (step 2).

6 + 10 = 16

4. Ask him/her to divide the new number with 2.

16 / 2 = 8

5. Ask him/her to tell the result and tell them you would tell the number s/he has thought.

Thinking how you would tell him what number s/he has thought. It's very simple, just subtract 5 from the number he has told and you would get the original number he has thought.

Here it is, 8 - 5 = 3

Simple right! Now want to know how it works, just see the below explanation

In the entire process, you are indirectly adding the original number by 5 and hence by knowing the end result you can guess the original number by just subtracting it by 5

Step 1 - Say the original number what your friend has thought is a.

Step 2 - Double the number - 2a

Step 3 - Add 10 to the new number - 2a + 10

Step 4 - Divide the new number by 2 (2a + 10)/2 = (2a/2) + (10/2) = a + 5

Step 5 - New number - a + 5

Therefore, Original number = New Number - 5 = a + 5 - 5 = a

Note: You can also use 2 and 3 digit numbers to make it harder!

Have fun playing this trick on your friends.

Maths Trick - Guess the result - 1089.

noreply@blogger.com (Shruthi) — Wed, 03 Jun 2009 05:45:00 +0000

1. Ask your friend to think of or write down any 3 digit number such that the digits are in ascending order (highest to lowest) and that they are not repetitive.

Lets say, s/he has written 765. While your friend is writing down the number, you write down 1089 and don't show it to your friend.

2. Ask him/her to reverse the number and subtract it from the original number (step1).

765 - 567 = 198

3. Ask him/her how many digits the result has. If it has 3 digits, ask him to reverse the result and if it has only 2 digits, then ask him to add a zero in front of it and then reverse the digit. Ex Number - 099 reverse number - 990

Here it is 3 digits, and hence reverse number is 891

4. Ask him/her to add this number (from step 3) to its previous number (from step 2 - obtained by doing subtraction) and ask him/her not to tell you the result.

891 + 198 = 1089

5. Now you can reveal the number (written by you in step 1) which is an exact match of their result (step 4)

Curious to know how it works? It is simple, just read the below explanation.

By asking your friend to subtract the original number from the reversed number, you are making the number indirectly multiple of 99.

Say the Original number is xyz = 100x + 10y + z

Therefore, Reversed number is zyx = 100z + 10y +x

Subtracting these both, xyz - zyx = (100x + 10y + z) - (100z + 10y +x) = 99x -99y = 99 (x-y)

Now, if you see the below 99 table you can see that (A.) Products are digit reversals of each other, that is 99 x 6 = 594 which is reverse of 495 = 99 x5. Such is the case with others. Therefore (B.) the result of adding the number with its reversed number is always 1089.

99 x 2 = 198 + 891 = 1089

99 x 3 = 297 + 792 = 1089

99 x 4 = 396 + 693 = 1089

99 x 5 = 495 + 594 = 1089

------------------------------------------

99 x 6 = 594 + 495 = 1089

99 x 7 = 693 + 396 = 1089

99 x 8 = 792 + 297 = 1089

99 x 9 = 891 + 198 = 1089

99 x 10 = 990 + 099 = 1089

Have fun playing the trick on your friends.

Maths Trick - Guess the missing number

noreply@blogger.com (Shruthi) — Wed, 20 May 2009 23:16:00 +0000

1. Ask your friend to think of or write down any number that contains more than 3 digits.

Lets say, s/he has written 18274.

2. Ask him/her to add all the digits.

1 + 8 + 2 + 7 + 4 = 22

3. Ask him/her to subtract this number (from step 2) from the original number (from step1).

18274 - 22 = 18252

4. Ask him/her to select any digit from this new number (step 3) and strike it out.

Lets say, s/he selected digit 5. So, resulted number is 1822

5. Ask him/her to add the digits of the resulted new number (that is resulted by striking out).

1 + 8 + 2 + 2 = 13.

6. Ask him/her to tell you the sum and that you would tell the strike out / selected digit.

With the help of this number, you can guess the digit your friend has strike out / selected. Just subtract the number from the next multiple of 9. Here, next multiple of 9 is 18 and hence the digit that your friend had strike out is 5 (=18 - 13).

Wondering how it works? The trick is very simple and it uses the property of 9 - a number whose digits sums up to 9, is multiple of 9.In the above trick,

Step 1 - 3: By subtracting the number with sum of its digits, we are making the number a whole multiple of 9. Here, 9 x 2028 = 18252.
Step 4 - 5: We have two cases, resultant number obtained by striking out being multiple of 9 or not multiple of 9 as in the example here.

Non Multiple of 9: By striking out a digit (5) from the number (18252, we are making the resultant number (1822) to be not divisible 9 and difference of, sum of the digits, of the new number and the old number will be exactly less by the striked out digit. i.e., (1+8+2+2) - (1+8+2+5+2) = 5.
Multiple of 9: This is possible only when the number striked out is 0 or 9. So, in such cases you can simply tell your friend that s/he has striked out either 9 or 0.

Have fun playing the trick on your friends.

Maths Calendar Trick - Guess the dates!

noreply@blogger.com (Shruthi) — Sun, 19 Apr 2009 10:50:00 +0000

You must have read in my earlier post about the calendar trick - to calculate sum of 9 numbers in 1 sec. Here is one more, you can predict the dates that your friend has selected based on their sum.

1. Ask your friend to select 4 days that form a square as shown in the adjacent figure. Let's say he has selected the 4 days as shown in the figure 5, 6, 12 and 13.

2. Ask him/her to sum up all the 4 days and tell you only the sum. Here it is 5 + 6 + 12 + 13 = 36.

Now, To guess the dates, all you need to do is to divide the sum by 4 and then subtract the result by 4. This will give the the First Date. (36/4) - 4 = 5, the reset can be calculated as below
Second Date = First Date + 1 = 5 + 1 = 6
Third Date = First Date + 7 = 5 + 7 = 12
Fourth Date = Third Date + 1 = 12 + 1 = 13

Did you understand why did it work? It is simple, say the first date is n, therefore the second is n+1, third is n+7 and fourth is n+8.
Now, by adding all the dates we have, n + (n +1) + (n + 7) + (n + 8) = 4n + 16 = 4(n + 4)
In the above example, the sum of the dates was 36. Therefore, we have
=> 4 (n + 4) = 36
=> n + 4 = 36 / 4 = 9
=> n = 9 - 4 = 5, which is nothing but the First Date.

So, now did you understand how the trick works? Let me summarize for you -
Your friend adds any 4 dates that form a square on the calendar and tells you the sum. You divide the sum by four and then subtract the result by 4 to get the first date. You add 1, 7, and 8 to get the other dates.

Have fun by trying the trick on your friends!

Maths Calendar Trick - Sum of 9 numbers

noreply@blogger.com (Shruthi) — Thu, 16 Apr 2009 15:54:00 +0000

The Calendar is something that we often look at almost everyday, for either planning our exams / meetings or holidays. But did you ever notice that you can play many cool tricks with it. One such trick is to calculate the sum of all 9 numbers selected on a calendar, in just 1 second. Let's see how it goes -

1. Draw a 3x3 box around any 9 numbers in the calendar. Let's say you had selected the 9 numbers as shown in the adjacent picture.

2. Now, multiply the middle number with 9. Here, it is 15 x 9 = 135

This result is nothing but sum of all the 9 numbers in the 3 x 3 box. That is 7 + 8 + 9 + 14 + 15 + 16 + 21 + 22 + 23 = 135

This trick works with any 9 numbers grouped in 3x3 box of the calendar. So, next time, ask you friend to select any 3x3 box in the calendar and amaze him by telling the sum of numbers while he is still calculating either manually or by calculator.

12345679 pattern

noreply@blogger.com (Shruthi) — Thu, 16 Apr 2009 14:49:00 +0000

The number 12345679 (did you notice it has all numbers from 1 to 9 except 8) can be used to generate a cool pattern. Want to know how? Just multiply 12345679 with your favorite digit, say 5 and then again multiply the result (12345679 x 5 = 61728395) with 9. The end result will contain only your favorite digits, here it is 555555555.

12345679 x 1 x 9 = 111111111
12345679 x 2 x 9 = 222222222
12345679 x 3 x 9 = 333333333
12345679 x 4 x 9 = 444444444
12345679 x 5 x 9 = 555555555
12345679 x 6 x 9 = 666666666
12345679 x 7 x 9 = 777777777
12345679 x 8 x 9 = 888888888
12345679 x 9 x 9 = 999999999

Did you get the logic behind the pattern? It's simple, you are indirectly multiplying your favorite digit with 111111111 - . product of 12345679 and 9.

Addition of Odd Consecutive Numbers

noreply@blogger.com (Shruthi) — Thu, 16 Apr 2009 11:14:00 +0000

Did you ever notice that by playing with numbers we can generated some cool patterns. One such pattern is generation of a square number by adding any number of consecutive odd integers, starting with one. Below is the pattern..

1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 + 7 + 9 = 25 = 5²
1 + 3 + 5 + 7 + 9 + 11 = 36 = 6²
1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 7²
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 8²
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 = 9²
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100 = 10²
------
In general 1 + 3 + 5 + ........ + (2n-1)= n², where n is a positive integer.

Is it not cool! if you have any such patterns that you can generate by playing with numbers, then do share.

Perfect Numbers

noreply@blogger.com (Shruthi) — Wed, 15 Apr 2009 20:59:00 +0000

We all know that the word 'Perfect' means 'being complete of its kind and without defect'. You must have come across some people who are so perfect but did you know that there are numbers too which are complete that is perfect. Yes, long back, Pythagorus has defined a number to be Perfect if it equals to the sum of all of its divisors (factors), excluding the number itself . For example,
6 - divisors are 1, 2, 3; 1+2+3 =6 Perfect Number!
28 - divisors are - 1, 2, 4, 7, 14 ; 1+2+4+7+14 = 28 Perfect Number!

The major study on Perfect Numbers was done by Euclid , Nichomachus, Ibn al-Haytham, and Euler and some of the important conclusions drawn are
1. As per Euclid, perfect numbers can be generated by the formula 2ⁿ⁻¹(2ⁿ − 1)
2. As per Nichomachus, all Perfect Numbers are even. (a theory that is yet to be disproved).
2. As per Euler, all even Perfect numbers are of the form 2ⁿ⁻¹(2ⁿ − 1), where 2ⁿ − 1 is a Mersenne prime.

3. Every even perfect number ends with 6 or 8.

Now, that you know about Perfect Numbers, you must be curious to know some amazing facts associated with these numbers -
1. There is a belief of religious significance being associated with the Perfect Numbers 6 and 28. That is - 6 is the number of days taken by God to create the world, and it was believed that the number was chosen by him because it was perfect. Again God chose the next perfect number 28 for the number of days it takes the Moon to travel round the Earth. (http://www.gap-system.org/~history/HistTopics/Perfect_numbers.html).
2. There are only 4 (6, 28, 486, 8128) perfect numbers below 10,000.
3. As of September 2008, only 46 Mersenne primes are known, which means there are 46 perfect numbers known. Be first to find 47th.
4. Till date, people have not yet defined the benifit of Perfect Numbers in real time and also they could not find an Odd Perfect Number. Mathematicians are still working on these and if interested you can join them.

You can find more on Perfect Numbers at - http://en.wikipedia.org/wiki/Perfect_number and http://www-users.cs.york.ac.uk/~susan/cyc/p/perfect.htm

All about Palindrome Numbers

noreply@blogger.com (Shruthi) — Sun, 12 Apr 2009 19:35:00 +0000

I am quite sure, that many of us know about 'palindrome' words – LEVEL, RADAR, MADAM, that can be read the same way in either direction. The same exists with numbers, just have a look at numbers: 878, 6886, 97679, 123321, 4345434
The above numbers remain same when their digits are reversed, such numbers are known as Palindromic Numbers or Number Palindromes and Prime Numbers that are palindrome are called as Palindromic Prime or Palprime. First few palprimes are - 2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311….

Apart from the digit pattern in the palindrome numbers, there are some other amazing facts of Palindrome Numbers as follows
1. All palprimes have odd number of digits except the number 11. ( from above list you can see that there are no 2 or 4 digit numbers)
2. Palindromes that have consecutive digits and even-highest middle digit are always divisible by 11 and the resultant quotient is also a palindrome. Ex.121 is divisible by 11 and quotient is 11 which is a palindrome. It has Even Highest Middile digit as 2.

121-(2)-11

1234321-(4)-112211

12345654321-(6)-1122332211

1223456787654321-(8)-11223344332211

3. Palindromes can be formed from a number that is not a palindrome. By adding the original non-palindrome number with the number formed by reversing the digits, we get a number that is palindrome either in one step or subsequent sets. This process is called as Lychrel Process. Example
One Step:
521 - Non Palindrome number
+ 125 - Reverse of Non Palindrome number
-------
646 - Palindrome !
-------
More than One Step:
529 - Non Palindrome number
+ 925 - Reverse of Non Palindrome number
---------
1454 - Non Palindrome
+ 4541 - Reverse of Non Palindrome number
---------
5995 - Palindrome !
---------
4. If a number cannot form a palindrome through the iterative process of repeatedly reversing the digits and adding the resulting numbers as mentioned above (step 3), then such a number is called as Lychrel Number. Till date no Lychrel numbers are known, though many numbers are suspected Lychrels, the smallest being 196. You can find more on Lychrel Numbers at http://en.wikipedia.org/wiki/Lychrel_number.

If you have any other cool facts about palindromes, do share.

What are large numbers after billion called?

noreply@blogger.com (Shruthi) — Thu, 09 Apr 2009 05:07:00 +0000

Every one of us know that 10⁶ is called as million, 10⁹ as billion and also 10¹² as trillion but beyond it, I guess very few people are aware. Here goes how the large numbers are called as in the American System and in European System.

Number	American Name	European Name
10⁹	Billion	Milliard
10¹²	Trillion	Billion
10¹⁵	Quadrillion	Billiard
10¹⁸	Quintillion	Trillion
10²¹	Sextillion	Trilliard
10²⁴	Septillion	Quadrillion
10²⁷	Octillion	Quadrilliard
10³⁰	Nonillion	Quintillion
10³³	Decillion	Quintilliard
10³⁶	Undecillion	Sextillion
10³⁹	Duodecillion	Sextilliard
10⁴²	Tredecillion	Septillion
10⁴⁵	Quattuordecillion	Septilliard
10⁴⁸	Quindecillion	Octillion
10⁵¹	Sexdecillion	Octilliard
10⁵⁴	Septendecillion	Nonillion
10⁵⁷	Octodecillion	Nonilliard
10⁶⁰	Novemdecillion	Decillion
10⁶³	Vigintillion	Decilliard

Want to know more on larger numbers? Just refer the sites - http://en.wikipedia.org/wiki/Names_of_large_numbers, http://www.unc.edu/~rowlett/units/large.html and http://www.uni-bonn.de/~manfear/numbers_names.php . Before just ending this article I would like to introduce you to the terms Googol and Googoplex.

Googol is a number one followed by 100 zeroes i.e, 10¹⁰⁰ and Googolplex is a number one with a googol of zeroes i.e, 10^googol 10^10¹⁰⁰.

That's all about the large numbers...hope this article was informative...do leave your comments / questions if any.

Multiply any number with 11 or 12 or 19 in just 5 seconds!

noreply@blogger.com (Shruthi) — Wed, 01 Apr 2009 23:25:00 +0000

You need not scratch your head, if you don't know 11 table. You can do this multiplication in 5 seconds, if you know simple additions. Surprised! yes, we would get the product of the numbers using addition.

Note that we would add the digits present in the number from right to left and also, imagine a 0 before and after the number i.e., if the number is 123 then consider it as 01230.

Let us calculate 789 x 11
Step 1: Consider a 0 before and after the number. ie, 07890
Step 2: Now for this number, add last two digits from right, 9 + 0 = 9. Therefore 789 x 11 = ___9
Step 3: Add the next two digits, here 8+9 =17. Write the last digit - 7 of the sum and take the carry - 1 to next step. Therefore 789 x 11 = __79
Step 4: Add the next two digits and the carry in step2, here (7 + 8 = 15) + 1 = 16. Write the last digit - 6 of the sum and take the carry - 1 to next step. Therefore 789 x 11 = _679
Step 5: Add the next digit and the carry in step 3, here 0 + 7 + 1 = 8. Write the sum against the result.
Therefore 789 x 11 = 8679

Now let us try calculating 12345 x 11
Step 1 : 0123450
Step 2 : 12345 x 11 = _____5 (5 + 0 = 5)
Step 3 : 12345 x 11 = ____95 (4 + 5 = 9)
Step 4 : 12345 x 11 = ___795 (3 + 4 = 7)
Step 5 : 12345 x 11 = __5795 (2 + 3 = 5)
Step 6 : 12345 x 11 = _35795 (1 + 2 = 3)
Step 7 : 12345 x 11 = 135795 (0 + 1 + no carry = 1)
Therefore 12345 x 11 = 135795

Let us try 4567 x 11
Step 1: 045670
Step 2: 4567 x 11 = ____7 (7 + 0 = 7)
Step 3: 4567 x 11 = ___37 (6 + 7 = 13, 3 with carry 1)
Step 4: 4567 x 11 = __237 (5 + 6 = 11 + 1 = 12, 2 with carry 1)
Step 5: 4567 x 11 = _0237 (4 + 5 = 9 + 1 = 10, 0 with carry 1)
Step 6: 4567 x 11 = 50237 (0 + 4 + 1 = 5)
Therefore 4567 x 11 = 50237

Simple right. In Vedic Mathematics the above method is called as Adding to the Neighbor.

Easy right...and if I am not wrong you must be thinking it would be so nice if so some method is present for multiplying numbers with 12, 13, 14....19. So that we need not by heart the tables.

We can extend the above method with slight modification. All you need to do is that when multiplying the number with 12, double the digit before adding to the neighbor. For 13, just triple the digit before adding to the neighbor. For 14 its 4 times and so on you can do for all the numbers till 19. Lets try few examples .

Calculate 7654 x 12
Step 0: 076540
Step 1: 7654 x 12 = ____8 (4 x 2 = 8 + 0 = 8)
Step 2: 7654 x 12 = ___48 (5 x 2 = 10 + 4 = 14, 4 with carry 1)
Step 3: 7654 x 12 = __848 (6 x 2 = 12 + 5 = 17 + 1 = 18, 8 with carry 1)
Step 4: 7654 x 12 = _1848 (7 x 2 = 14 + 6 = 20 + 1 = 21, 1 with carry 2)
Step 5: 7654 x 12 = 91848 (0 x 2 = 0 + 7 + 2 = 9)
Therefore 7654 x 12 = 91848

Calculate 12345 x 13
Step 0: 0123450
Step 1: 12345 x 13 = _____5 (5 x 3 = 15 + 0 = 15, 5 with carry 1)
Step 2: 12345 x 13 = ____85 (4 x 3 = 12 + 5 = 17 + 1 = 18, 8 with carry 1)
Step 3: 12345 x 13 = ___485 (3 x 3 = 9 + 4 = 13 + 1 = 14, 4 with carry 1)
Step 4: 12345 x 13 = __0485 (2 x 3 = 6 + 3 = 9 + 1 = 10, 0 with carry 1)
Step 5: 12345 x 13 = _60485 (1 x 3 = 3 + 2 = 5 + 1 = 6, no carry)
Step 6: 12345 x 13 = 160485 (0 x 3 = 0 + 1 = 1)
Therefore 12345 x 13 = 160485

You can try multiplying the numbers with 14, 15, 16, 17, 18 and 19 and check the answers with either standard method or calculators.
Hope this simple method of multiplying has helped you in speeding up your calculations. If you have any queries or if you have any other simpler method to solve, then do let us know.

Calculate square of 33333 in just 5 seconds!

noreply@blogger.com (Shruthi) — Sun, 22 Jun 2008 16:51:00 +0000

Wondering, how can this be done…..you would do it all by your self after reading this article.

The trick is simple and can be applied for calculating square of number containing all threes.

Just observe the pattern in the result obtained below
33 x 33 = 1089
333 x 333 = 110889
3333 x 3333 = 11108889
33333 x 33333 = 1111088889

The result

Always contains digits – 1, 0, 8 and 9 in the order specified.
Always starts with 1 and ends with 9.
With only variation in repetition of the digits 1 and 8. The repetition value is nothing but one less then the number of digits of the squared number.
The digit 0 is always between the digits 1 and 8

Now lets see square of 33333

Step 1 – Number of digits for which the square has to be calculated is 5. Therefore, 1 and 8 would repeat for 4 times (5-1).
Step 2 – Start by writing 1 four times- 1111
Step 3 – Append 0 – 11110
Step 4 – Append 8 four times - 111108888
Step 5 – End by appending 9 to the number - 1111088889

Therefore, 33333 x 33333 = 1111088889

Note: Total number of the digits of the result will always be double of the digits of the squared number.(Ex. Squared Num = 33333, # of digits in the squared num = 5, # of digits in the result = 10 (5+5), Result =1111088889).

I think now you can easily calculate square of 333333333 also in just 5 seconds!

Do share with others and if you have any other method to solve in simpler way….share it with us…