We Love Maths

145 - A number with rare property

2012-01-20T19:16:00.002+05:30

A factorion is a natural number that equals the sum of the factorials of its decimal digits. For example, 145 is a factorion because

1! + 4! + 5!
= 1 + 24 + 120
= 145.

There are just four factorions (in base 10) and they are 1, 2, 145 and 40585 (rare property)

1 = 1!

2 = 2!

145 = 1! + 4! + 5!

40585 = 4! + 0! + 5! + 8! + 5!

Links : Wikipedia, Wolfram MathWorld

Math Trick - Base Multiplication using Vedic Maths

2011-12-31T19:59:00.001+05:30

Wonder Of Maths

2011-12-30T21:40:00.001+05:30

Proof:

Hexadecimal Math Puzzle

2011-12-30T18:49:00.000+05:30

If only DEAD people understand hexadecimal, how many people understand hexadecimal?

What is sum of all digits from 1 to 100?

2011-12-29T18:41:00.000+05:30

like (1)+(2)+(3)+(4)+......+ (9+7) + (9+8) + (9+9) + (1+0+0)

Easy Puzzle

2011-12-28T18:25:00.000+05:30

What is the remainder when the product (1402)x(1776)x(1912)x(99) is divided by 5?

Google's Love for Mathematicians

2011-12-27T21:30:00.003+05:30

I posted this thing on Facebook page on Dec 6, but missed to post it here.

When you search on Google with " (sqrt(cos(x))*cos(200x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01, sqrt(9-x^2), -sqrt(9-x^2) from -4.5 to 4.5 ", you will get such Graph

Also type "(sqrt(cos(x))*cos(200x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01, sqrt(9-x^2), -sqrt(9-x^2)" to another cool graph

Math Puzzle

2011-12-27T18:19:00.000+05:30

Use the digits 1, 2, 3, 4, 5 and 6 once only, in this multiplication sum to make it correct.

?? x ? = ???

Source

Indian PM declares 2012 'National Mathematical year' as a tribute to Ramanujan

2011-12-26T18:14:00.000+05:30

Declaring 2012 as the 'National Mathematical year' as a tribute to maths wizard Srinivasa Ramanujan, Prime Minister Manmohan Singh today voiced concern over the ''badly inadequate'' number of competent mathematicians in the country.
He also said that the perception that pursuit of mathematics does not lead to attractive career possibilities "must change."

"It is a matter of concern that for a country of our size, the number of competent mathematicians that we have is badly inadequate", he said at a function to here mark the 125th birth anniversary of Ramanujan.

Singh also declared December 22, the birthday of Ramanujan, as 'National Mathematics Day.'

Students have not pursued mathematics at advanced levels over more than three decades, which has resulted in a decline in quality of mathematics teachers at schools and colleges, Singh who is on a two-day visit to the state, told a galaxy of academics at Madras University.

"There is a general perception in our society that the pursuit of mathematics does not lead to attractive career possibilities. This perception must change. This perception may have been valid some years ago, but today there are many new career opportunities available to mathematics and the teaching perception itself has become much more attractive in recent years", Singh said.

The Prime Minister said the mathematical community has a duty to find out "ways and means" to address the shortage of top quality mathematicians and reach out to the public, especially in the modern context, where mathematics has tremendous influence on every kind of human endeavour.
Noting that the Central government has pursued a policy of encouraging scientific activities of diverse kinds, the Prime Minister said, "Given our traditions, we naturally attach special importance to mathematics...in many ways, mathematics can be regarded as the mother science".

He said Ramanujan overcame formidable difficulties to reach the pinnacle of greatness, illustrating the inadequacy of University evaluation system in the early decades of the last century, while at the same time showing the system displayed enough flexibility to take care of mavericks like him.

"There have been many reforms since those days but there would still be talent which would elude proper evaluation. Our institutions of higher learning must be sensitive to this problem."

"A genius like Ramanujan would shine bright even in the most adverse of circumstances, but we should be geared to encourage and nurture good talent which may not be of the same calibre as that of Ramanujan", Singh said.

Honouring Professor Robert Kanigel, who has written a biography of Ramanujan, Singh said this book has made Ramanujan well known to the public at large all over the world.
He said the country was proud of Ramanujan and Tamil Nadu has a special claim on him for he was a Tamilian.

"Along with C V Raman and Subramanyam Chandrashekhar (both Nobel laureates), he is among the three great men of science and mathematics that Tamil Nadu and India have given to the world of modern times", he said.

Source

Determine the largest prime divisor of 87! + 88!

2011-12-18T21:08:00.001+05:30

87! + 88!
= 87! (1 + 88) [ n! = n x (n-1)! ]
= 87! x 89

Here, Highest divisor is 89 and also it is Prime

So, the largest prime divisor of 87! + 88! is 89

[Proof] if n is odd, then n^2 - 1 is divisible by 8

2011-12-06T21:10:00.002+05:30

Method 1

Here, n is odd. so we can take n = 2k +1 where k=0,1,2,3........so on

n² - 1
= (2k+1)² - 1
= 4k² + 4k + 1 -1
=4 ( k² + k)

Here, k is any number (may be odd or even)

If k is even, n² - 1 = 4(4j² + 2j) = 8(2j² + j)

If k is odd, n² - 1 = 4((2j + 1)² + (2j + 1)) = 4(4j² + 4j + 1 + 2j + 1) = 4(4j² + 6j + 2) = 8(2j² + 3j + 1)

So, for all values of k, n² - 1 =4 ( k² + k) is divisible by 8

Method 2

(it was posted on our Facebook page by Charbel Sakr)

if n is odd it is congruent to either 1 3 5 or 7 mod 8

in each case n² is congruent to 1 mod 8 ( 1 9 25 and 49)

so that n² -1 congruent to 0 mod 8 in other words n² - 1 divisible by 8

How can you make 101 with {a,a,b,b} using only {×,÷,-,+} , where a & b are whole single-digit numbers?

2011-11-30T21:27:00.000+05:30

((9+2)*9 ) +2 = 101

if you have another answer, you can post here in comment

Math Puzzle : Number Series

2011-11-27T20:07:00.001+05:30

‎1, 2, 4, 7, 28, 33, __, __

Math Puzzle

2011-11-26T20:00:00.001+05:30

Using the 4 twice, and using no other numbers, can you create an equation that equals 12?

How can you make 124 from {a,a,b,b} using only {×,÷,-,+} , where a & b are single digit numbers.

2011-11-25T21:23:00.006+05:30

If you have another answer, you can post it in comment

Happy Number

2011-11-24T21:15:00.001+05:30

Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1.

For Example , 19 is happy number, as the associated sequence is:

1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

Read More >>>

List of Happy Numbers :

1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100, 103, 109, 129, 130, 133, 139, 167, 176, 188, 190, 192, 193, 203, 208, 219, 226, 230, 236, 239, 262, 263, 280, 291, 293, 301, 302, 310, 313, 319, 320, 326, 329, 331, 338, 356, 362, 365, 367, 368, 376, 379, 383, 386, 391, 392, 397, 404, 409, 440, 446, 464, 469, 478, 487, 490, 496

[Proof] d(e^x)/dx = e^x

2011-11-24T16:18:00.002+05:30

Method 1

Let y=e^x

We can write, ln(y)=x from above

By differentiating both sides we get:

^y'/_y=1

So y'= y

Therefore (e^x)' = e^x

Method 2

With the Maclaurin formula we can write

e^x = 1 + x/1! + x²/2! + ... + xⁿ/n! + ...upto infinity

Now differentiating this with respect to x,

d(e^x)/dx = d(1)/dx + d(x)/dx + d(x²/2!)/dx + d(x³/3!)/dx....... upto infinity

d(e^x)/dx = 0 + 1 + 2x/2! + 3x²/3! + .....

d(e^x)/dx = 1 + x + x²/2! + .....

here, R.H.S. is equal to e^x according to Maclaurin formula

Thus, d(e^x)/dx = e^x

Determine the largest prime divisor of 51! + 52!

2011-11-23T21:02:00.005+05:30

51! + 52!
= 51! (1 + 52) [ n! = n x (n-1)! ]
= 51! x 53

Here, Highest divisor is 53 and also it is Prime

So, the largest prime divisor of 51! + 52! is 53

Prime Number

2011-11-22T14:03:00.000+05:30

1709,
175709,
17575709,
1757575709,
175757575709,
17575757575709,
1757575757575709 and
175757575757575709 are all prime,

but 17575757575757575709 = 232433 × 75616446785773

isn't it amazing........

Something Unique about Year 2011

2011-11-21T20:48:00.000+05:30

Please click on Facebook like button or Google +1 button here.

Google Doodle on the 410th birthday of Pierre de Fermat

2011-11-21T13:32:00.000+05:30

Math Puzzle

2011-11-21T01:25:00.002+05:30

Use each of the digits 1 through 7 once to write a four- digit number plus a three-digit number that gives the greatest possible sum?

How many 0's are there in between 1 to 1000?

2011-11-20T13:24:00.000+05:30

192 zeros

10,20,.....,990 = 99 zeros

100,200,300,......,900 = 9 zeros (extra 9 zeros are already counted in 10,20,30,....,990)

1000 = 3 zeros

101,102,...,109 and 201,202,..,209 upto 901,902,....,909 = 81 zeros

99+9+3+81 = 192

6174 - Kaprekar's constant

2011-11-20T01:25:00.000+05:30

‎6174 is known as Kaprekar's constant after the Indian mathematician D. R. Kaprekar.
This number is notable for the following property:

Take any four-digit number, using at least two different digits. (Leading zeros are allowed.)
Arrange the digits in ascending and then in descending order to get two four-digit numbers, adding leading zeros if necessary.
Subtract the smaller number from the bigger number.
Go back to step 2.

The above process, known as Kaprekar's routine, will always reach its fixed point, 6174, in at most 7 iterations. Once 6174 is reached, the process will continue yielding 7641 – 1467 = 6174.

For example, choose 3524:

5432 – 2345 = 3087
8730 – 0378 = 8352
8532 – 2358 = 6174

The only four-digit numbers for which Kaprekar's routine does not reach 6174 are repdigits such as 1111, which give the result 0 after a single iteration. All other four-digit numbers eventually reach 6174 if leading zeros are used to keep the number of digits at 4:

2111 – 1112 = 0999
9990 – 0999 = 8991 (rather than 999 – 999 = 0)
9981 – 1899 = 8082
8820 – 0288 = 8532
8532 – 2358 = 6174

Taken from Wikipedia

Math Puzzle

2011-11-19T13:19:00.001+05:30

Find three consecutive odd integers such that the sum of the first , two times the second, and three times the third is 82.

Number Sequence

2011-11-19T01:22:00.001+05:30

‎18, 46, 94, 63, 52, __

Six weeks = 10! seconds

2011-11-18T13:17:00.000+05:30

I saw this relationship on twitter by @standupmaths

Six weeks = 10! seconds

Because,

6 weeks × 7 days × (8×3) hours × (4×√9×5) mins × (2×√9×10) secs

= 1×2×3×4×5×6×7×8×9×10

=10! seconds

Love letter of Mathematician

2011-11-18T01:18:00.000+05:30

Darling Simplify this equation:

9x-7i > 3(3x-7u)

Solution:
9x-7i > 9x-21u
-7i > -21u
7i < 21u

i <3 u
i ♥ u

Math Joke

2011-11-17T01:05:00.000+05:30

How 2 address ur Mom's Elder n Younger sisters Mathematically...?
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.
.
.
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Maxi-Mum and Mini- Mum

Relationship between Squares and Odd numbers

2011-11-16T12:48:00.000+05:30

1² = 1 = 1

2² = 4 = 1+3

3² = 9 = 1+3+5

4² = 16 = 1+3+5+7

5² = 25 = 1+3+5+7+9

6² = 36 = 1+3+5+7+9+11 and so on

3w1 series

2011-11-16T00:53:00.000+05:30

First seven numbers of 3w1 series are prime. The numbers 31, 331, 3331, 33331, 333331, 3333331, and 33333331 are all prime. It was thought that every number of the form 3w1 would be prime. But next numbers in this sequence are composite

for example, 333333331 = 17 * 19607843

Funny Maths

2011-11-15T12:42:00.000+05:30

Math Puzzle : What are last 3 digit of 21^15 ?

2011-11-15T00:45:00.000+05:30

Answer : 301

Method : use binomial theorem (20+1)^15

now take first three term of expansion --> 15C0 + 15C1*20 + 15C2 * (20)^2 = 301

all other terms will have more than 3 zeros at end,, so it will not affect last 3 digit

Circulation of Number

2011-11-14T00:39:00.000+05:30

Only Number with such property

142857*1=142857

142857*2=285714

142857*3=428571

142857*4=571428

142857*5=714285

142857*6=857142

But , 142857*7=999999

9376 : Automorphic number

2011-11-13T22:07:00.001+05:30

9376 is only four digit number such that last four digits of its square is same as that of number.

9376 x 9376 = 8790 9376

Such numbers are called Automorphic number.

"The number n is called an Automorphic number if (the decimal expansion of) n² ends with n. These numbers are also called curious."

Other such numbers are 1, 5, 6, 25, 76, 376, 625, 9376, 90625, 109376, 890625,....

Beauty of Mathematics

2011-11-13T13:01:00.003+05:30

1 x 8 + 1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

123456 x 8 + 6 = 987654

1234567 x 8 + 7 = 9876543

12345678 x 8 + 8 = 98765432

123456789 x 8 + 9 = 987654321