Tuesday, November 30, 2010

Finding prime numbers such that....

2, 3, 5 and 7 are prime numbers.
5 - 2 = 3, 5 - 3 = 2, 7 - 2 = 5 and 7 - 5 = 2 produce prime numbers.
Now, I'm interested to find all pairs of primes (x, y) for which
(5^x - 2^x)(5^y - 2^y) /(xy) is an integer
(5^x - 3^x)(5^y - 3^y) / (xy) is an integer
(7^x - 2^x)(7^y - 2^y) /(xy) is an integer
(7^x - 5^x)(7^y - 5^y) / (xy) is an integer

Then, after that extend the research to other prime numbers

Monday, November 29, 2010

Integer as sum of cubes

Could you find an integer X that can be expressed in two different ways as the sum of three perfect cubes, that is to say

X = a^3 + b^3 + c^3 = d^3 + e^3 + f^3


And, could you find an integer Y that can be expressed in three different ways as the sum of two perfect cubes, that is to say,

Y = a^3 + b^3 = c^3 + d^3 = e^3 + f^3


Note that 1729 is the smallest number that can be expressed as the sum of two cube numbers in two different ways

1729 = 1^3 + 12^3 = 9^3 + 10^3

Arithmetic Progression with squares

An arithmetic progression (AP) is



1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81

10^2 = 100 ..... 21^2 = 441
11^2 = 121 ..... 22^2 = 484
12^2 = 144 ..... 23^2 = 529
13^2 = 169 ..... 24^2 = 576
14^2 = 196 ..... 25^2 = 625
15^2 = 225 ..... 26^2 = 676
16^2 = 256 ..... 27^2 = 729
17^2 = 289 ..... 28^2 = 784
18^2 = 324 ..... 29^2 = 841
19^2 = 361 ..... 30^2 = 900
20^2 = 400 ..... 31^2 = 961

32^2 = 1024 ..... 49^2 = 2401 ..... 66^2 = 4356 ..... 83^2 = 6889
33^2 = 1089 ..... 50^2 = 2500 ..... 67^2 = 4489 ..... 84^2 = 7056
34^2 = 1156 ..... 51^2 = 2601 ..... 68^2 = 4624 ..... 85^2 = 7225
35^2 = 1225 ..... 52^2 = 2704 ..... 69^2 = 4761 ..... 86^2 = 7396
36^2 = 1296 ..... 53^2 = 2809 ..... 70^2 = 4900 ..... 87^2 = 7569
37^2 = 1369 ..... 54^2 = 2916 ..... 71^2 = 5041 ..... 88^2 = 7744
38^2 = 1444 ..... 55^2 = 3025 ..... 72^2 = 5184 ..... 89^2 = 7921
39^2 = 1521 ..... 56^2 = 3136 ..... 73^2 = 5329 ..... 90^2 = 8100
40^2 = 1600 ..... 57^2 = 3249 ..... 74^2 = 5476 ..... 91^2 = 8281
41^2 = 1681 ..... 58^2 = 3364 ..... 75^2 = 5625 ..... 92^2 = 8464
42^2 = 1764 ..... 59^2 = 3481 ..... 76^2 = 5776 ..... 93^2 = 8649
43^2 = 1849 ..... 60^2 = 3600 ..... 77^2 = 5929 ..... 94^2 = 8836
44^2 = 1936 ..... 61^2 = 3721 ..... 78^2 = 6084 ..... 95^2 = 9025
45^2 = 2025 ..... 62^2 = 3844 ..... 79^2 = 6241 ..... 96^2 = 9216
46^2 = 2116 ..... 63^2 = 3969 ..... 80^2 = 6400 ..... 97^2 = 9409
47^2 = 2209 ..... 64^2 = 4096 ..... 81^2 = 6561 ..... 98^2 = 9604
48^2 = 2304 ..... 65^2 = 4225 ..... 82^2 = 6724 ..... 99^2 = 9801



The squares 1, 25 and 49 are in arithmetic progression, with a common difference of 24.

Find others.

Note: we can have only three squares in arithmetic progression, that is to say, it is not possible to have 4 squares in arithmetic progression. A proof will be needed.

Digits

Here I'll be looking at each square number and I'll add this number to its digits in order to find a square number.

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16, ..... 16 + 1 + 6 = 23
5^2 = 25, ..... 25 + 2 + 5 = 32
6^2 = 36, ..... 36 + 3 + 6 = 45
7^2 = 49, ..... 49 + 4 + 9 = 62
8^2 = 64, ..... 64 + 6 + 4 = 74
9^2 = 81, ..... 81 + 8 + 1 = 90

So far, the results did not give me any square number.
So, I'll continue with the next square numbers.

10^2 = 100 ..... 21^2 = 441
11^2 = 121 ..... 22^2 = 484
12^2 = 144 ..... 23^2 = 529
13^2 = 169 ..... 24^2 = 576
14^2 = 196 ..... 25^2 = 625
15^2 = 225 ..... 26^2 = 676
16^2 = 256 ..... 27^2 = 729
17^2 = 289 ..... 28^2 = 784
18^2 = 324 ..... 29^2 = 841
19^2 = 361 ..... 30^2 = 900
20^2 = 400 ..... 31^2 = 961

100 + 1 + 0 + 0 = 101
121 + 1 + 2 + 1 = 125
144 + 1 + 4 + 4 = 153
169 + 1 + 6 + 9 = 185
196 + 1 + 9 + 6 = 212
225 + 2 + 2 + 5 = 234
256 + 2 + 5 + 6 = 269
289 + 2 + 8 + 9 = 308
324 + 3 + 2 + 4 = 333
361 + 3 + 6 + 1 = 371
400 + 4 + 0 + 0 = 404
441 + 4 + 4 + 1 = 450
484 + 4 + 8 + 4 = 500
529 + 5 + 2 + 9 = 545
576 + 5 + 7 + 6 = 594
625 + 6 + 2 + 5 = 638
676 + 6 + 7 + 6 = 695
729 + 7 + 2 + 9 = 747
784 + 7 + 8 + 4 = 803
841 + 8 + 4 + 1 = 854
900 + 9 + 0 + 0 = 909
961 + 9 + 6 + 1 = 977

No squares found.

32^2 = 1024 ..... 49^2 = 2401 ..... 66^2 = 4356 ..... 83^2 = 6889
33^2 = 1089 ..... 50^2 = 2500 ..... 67^2 = 4489 ..... 84^2 = 7056
34^2 = 1156 ..... 51^2 = 2601 ..... 68^2 = 4624 ..... 85^2 = 7225
35^2 = 1225 ..... 52^2 = 2704 ..... 69^2 = 4761 ..... 86^2 = 7396
36^2 = 1296 ..... 53^2 = 2809 ..... 70^2 = 4900 ..... 87^2 = 7569
37^2 = 1369 ..... 54^2 = 2916 ..... 71^2 = 5041 ..... 88^2 = 7744
38^2 = 1444 ..... 55^2 = 3025 ..... 72^2 = 5184 ..... 89^2 = 7921
39^2 = 1521 ..... 56^2 = 3136 ..... 73^2 = 5329 ..... 90^2 = 8100
40^2 = 1600 ..... 57^2 = 3249 ..... 74^2 = 5476 ..... 91^2 = 8281
41^2 = 1681 ..... 58^2 = 3364 ..... 75^2 = 5625 ..... 92^2 = 8464
42^2 = 1764 ..... 59^2 = 3481 ..... 76^2 = 5776 ..... 93^2 = 8649
43^2 = 1849 ..... 60^2 = 3600 ..... 77^2 = 5929 ..... 94^2 = 8836
44^2 = 1936 ..... 61^2 = 3721 ..... 78^2 = 6084 ..... 95^2 = 9025
45^2 = 2025 ..... 62^2 = 3844 ..... 79^2 = 6241 ..... 96^2 = 9216
46^2 = 2116 ..... 63^2 = 3969 ..... 80^2 = 6400 ..... 97^2 = 9409
47^2 = 2209 ..... 64^2 = 4096 ..... 81^2 = 6561 ..... 98^2 = 9604
48^2 = 2304 ..... 65^2 = 4225 ..... 82^2 = 6724 ..... 99^2 = 9801

1024 + 1 + 0 + 2 + 4 = 1031 .......... 4356 + 4 + 3 + 5 + 6 = 4374
1089 + 1 + 0 + 8 + 9 = 1107 .......... 4489 + 4 + 4 + 8 + 9 = 4514
1156 + 1 + 1 + 5 + 6 = 1169 .... ..... 4624 + 4 + 6 + 2 + 4 = 4640
1225 + 1 + 2 + 2 + 5 = 1235 .......... 4761 + 4 + 7 + 6 + 1 = 4779
1296 + 1 + 2 + 9 + 6 = 1314 .......... 4900 + 4 + 9 + 0 + 0 = 4913
1369 + 1 + 3 + 6 + 9 = 1388 .......... 5041 + 5 + 0 + 4 + 1 = 5051
1444 + 1 + 4 + 4 + 4 = 1457 .......... 5184 + 5 + 1 + 8 + 4 = 5202
1521 + 1 + 5 + 2 + 1 = 1530 .......... 5329 + 5 + 3 + 2 + 9 = 5348
1600 + 1 + 6 + 0 + 0 = 1607 .......... 5476 + 5 + 4 + 7 + 6 = 5498
1681 + 1 + 6 + 8 + 1 = 1697 .......... 5625 + 5 + 6 + 2 + 5 = 5643
1764 + 1 + 7 + 6 + 4 = 1782 .......... 5776 + 5 + 7 + 7 + 6 = 5801
1849 + 1 + 8 + 4 + 9 = 1871 .......... 5929 + 5 + 9 + 2 + 9 = 5954
1936 + 1 + 9 + 3 + 6 = 1955 .......... 6084 + 6 + 0 + 8 + 4 = 6102
2025 + 2 + 0 + 2 + 5 = 2034 .......... 6241 + 6 + 2 + 4 + 1 = 6254
2116 + 2 + 1 + 1 + 6 = 2126 .......... 6400 + 6 + 4 + 0 + 0 = 6410
2209 + 2 + 2 + 0 + 9 = 2222 .......... 6561 + 6 + 5 + 6 + 1 = 6579
2304 + 2 + 3 + 0 + 4 = 2313 .......... 6724 + 6 + 7 + 2 + 4 = 6743
2401 + 2 + 4 + 0 + 1 = 2408 .......... 6889 + 6 + 8 + 8 + 9 = 6920
2500 + 2 + 5 + 0 + 0 = 2507 .......... 7056 + 7 + 0 + 5 + 6 = 7074
2601 + 2 + 6 + 0 + 1 = 2610 .......... 7225 + 7 + 2 + 2 + 5 = 7241
2704 + 2 + 7 + 0 + 4 = 2717 .......... 7396 + 7 + 3 + 9 + 6 = 7421
2809 + 2 + 8 + 0 + 9 = 2828 .......... 7569 + 7 + 5 + 6 + 9 = 7596
2916 + 2 + 9 + 1 + 6 = 2934 .......... 7744 + 7 + 7 + 4 + 4 = 7766
3025 + 3 + 0 + 2 + 5 = 3035 .......... 7921 + 7 + 9 + 2 + 1 = 7940
3136 + 3 + 1 + 3 + 6 = 3149 .......... 8100 + 8 + 1 + 0 + 0 = 8109
3249 + 3 + 2 + 4 + 9 = 3267 .......... 8281 + 8 + 2 + 8 + 1 = 8300
3364 + 3 + 3 + 6 + 4 = 3380 .......... 8464 + 8 + 4 + 6 + 4 = 8486
3481 + 3 + 4 + 8 + 1 = 3497 .......... 8649 + 8 + 6 + 4 + 9 = 8676
3600 + 3 + 6 + 0 + 0 = 3609 .......... 8836 + 8 + 8 + 3 + 6 = 8861
3721 + 3 + 7 + 2 + 1 = 3734 .......... 9025 + 9 + 0 + 2 + 5 = 9041
3844 + 3 + 8 + 4 + 4 = 3863 .......... 9216 + 9 + 2 + 1 + 6 = 9234
3969 + 3 + 9 + 6 + 9 = 3996 .......... 9409 + 9 + 4 + 0 + 9 = 9431
4096 + 4 + 0 + 9 + 6 = 4115 .......... 9604 + 9 + 6 + 0 + 4 = 9623
4225 + 4 + 2 + 2 + 5 = 4238 .......... 9801 + 9 + 8 + 0 + 1 = 9819

No squares found here.

In fact, we cannot find. A formal proof will be required.



Sum of powers of digits

There are many numbers that can be expressed as the sum of powers of their digits.

2^3 + 4^2 = 24,

4^2 + 3^3 = 43,

6^2 + 3^3 = 63,

8^1 + 9^2 = 89,

1^1 + 3^2 + 5^3 = 135,

1^3 + 5^3 + 3^3 = 153,

etc.

Find others.

Sum of divisors

A divisor of an integer n, also called a factor of n, is an integer which divides n without leaving a remainder

Here, I take the sum of the divisors of integers, then I'll be searching for squares and cubes such that the sum of its divisors is a square and a cube respectively.

Check my old post for 3-digit, 4-digit, 5-digit and 6-digit square numbers:

The first square numbers are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, ...

Here is a list of perfect cube numbers: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, ...

Now, I write their divisors and find their sums.

The factors of 4 (=2^2) are: 1 2 4
1 + 2 + 4 = 7
The factors of 8 (=2^3) are: 1 2 4 8
1 + 2 + 4 + 8 = 15
The factors of 9 (=3^2) are: 1 3 9
1 + 3 + 9 = 13
The factors of 16 (=4^2) are: 1 2 4 8 16
1 + 2 + 4 + 8 + 16 = 31
The factors of 25 (=5^2) are: 1 5 25
1 + 5 + 25 = 31
The factors of 27 (3^3) are: 1 3 9 27
1 + 3 + 9 + 27 = 40
The factors of 36 (=6^2) are: 1 2 3 4 6 9 12 18 36
1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91
The factors of 49 (=7^2) are: 1 7 49
1 + 7 + 49 = 57

64 is an interesting number because it is both a square (8^2) and a cube (4^3)
The factors of 64 are: 1 2 4 8 16 32 64
1 + 2 + 4 + 8 + 16 + 32 + 64 = 127

The factors of 81 (=9^2) are: 1 3 9 27 81
1 + 3 + 9 + 27 + 81 = 121 = 11^2

So, 81 is the first square whose sum of its divisors is a square.
I need to find a cube.
The search continues ...

Sunday, November 28, 2010

Divisibility problem

x and y are integers, such that

x > 1, y > 1


For all n > 0, (x^n - 1) is divisible by (y^n - 1)

Show that x = y^k, with k being a positive integer.

Saturday, November 27, 2010

Magic of numbers

Could you explain the following results:


7 * 7 = 49
67 * 67 = 4489
667 * 667 = 444889
6667 * 6667 = 44448889
66667 * 66667 = 4444488889
666667 * 666667 = 444444888889
6666667 * 6666667 = 44444448888889



4 * 4 = 16
34 * 34 = 1156
334 * 334 = 111556
3334 * 3334 = 11115556
33334 * 33334 = 1111155556
333334 * 333334 = 111111555556


9 * 9 = 81
99 * 99 = 9801
999 * 999 = 998001
9999 * 9999 = 99980001
99999 * 99999 = 9999800001
999999 * 999999 = 999998000001
9999999 * 9999999 = 99999980000001


7 * 9 = 63
77 * 99 = 7623
777 * 999 = 776223
7777 * 9999 = 77762223
77777 * 99999 = 7777622223
777777 * 999999 = 777776222223


1 * 7 + 3 = 10
14 * 7 + 2 = 100
142 * 7 + 6 = 1000
1428 * 7 + 4 = 10000
14285 * 7 + 5 = 100000
142857 * 7 + 1 = 1000000
1428571 * 7 + 3 = 10000000
14285714 * 7 + 2 = 100000000
142857142 * 7 + 6 = 1000000000
1428571428 * 7 + 4 = 10000000000
14285714285 * 7 + 5 = 100000000000
142857142857 * 7 + 1 = 1000000000000


7 * 15873 = 111111
14 * 15873 = 222222
21 * 15873 = 333333
28 * 15873 = 444444
35 * 15873 = 555555
42 * 15873 = 666666
49 * 15873 = 777777
56 * 15873 = 888888
63 * 15873 = 999999



91 * 1221 = 111111

900991 * 123321 = 111111111111

Rational number and geometric progression

To show that a real number q is rational if and only if there are three distinct integers, a, b, c, such that q + a, q + b, q + c forms a geometric progression

A geometric progression, also known as a geometric sequence, is a...


Rational Right Triangle

Let A > 0 be a rational number.

A is the area of a right triangle with rational sides if and only if there exist three rational numbers u, v and w, so that

u^2 - v^2 = v^2 - w^2 = A

Arithmetic, Population, Energy

Dr. Albert A. Bartlett
Professor Emeritus
Department of Physics
University of Colorado at Boulder
http://en.wikipedia.org/wiki/Albert_Bartlett


The greatest shortcoming of the human race is our inability to understand the Exponential Function.

The Exponential function is used to describe the size of anything that growing steadily,
for example, 5% per year.

The exponential function

We are talking about a situation where the time that is required for the growing quantity
to increase by a fixed fraction is constant.

If it takes a fixed length of time to grow 5%, then it follows that it takes a longer fixed length of time to grow by 100%.

This longer time is called the doubling time.

Then, we need to calculate the doubling time. Watch the video.

You take 70 divided by the % growth per unit time
So, if your growth rate is 5% per year, then the doubling time is

70/5 = 14 years

the growing size will double every 14 years.

Doubling Time Formula



How do you get 70?

In the video, the professor says, "it's approximately the natural logarithm of 2"

ln(2) = 0.6931471805599
70 = 100 * ln (2)

If tripled instead of double, you use the natural logarithm of 3
getting,
ln(3) = 1.0986122886681

The natural logarithm

Read more: the rule of 72, the rule of 70 and the rule of 69

Related topic: Solving for the period needed to double money

The video and those links do not show clearly how this is done.

In the case of doubling the amount of money, we use simple compound interests.

Since the initial principal is to be doubled we have the equation (formula for compound interests):

2 = (1 + r/100)^n

Taking the log of both sides:

log 2 = n * log(1 + r/100)

or

n = log 2 / [log(100 + r) - 2]

Then we look for a number, x, such that when divided by r, the result is approximately n.
Hence,
x / r = log 2 / [log(100 + r) - 2]

x = 0.301 * r / [log(100 + r) - 2]

Suppose r = 5%

0.301 * 5 = 1.50500
log(100 + r) = log(105)
log(105) - 2 = 0.0211892991

So, x = 1.50500 / 0.0211892991 = 71.0264173 (approximately 70)

Friday, November 26, 2010

Euler's Conjecture

It is closely related to Fermat's Last theorem.

Fermat's Last Theorem

Euler, in 1769 proposed that there are no sets of numbers such that

a^4 + b^4 + c^4 = d^4

or

a^5 + b^5 + c^5 + d^5 = e^5


The conjecture was disproved in 1966 by Lander and Parkin who found counterexample
for n=5:

27^5 + 84^5 + 110^5 + 133^5 = 144^5

Another counterexample was found by Noam Elkies in 1988:

2682440^4 + 15365639^4 + 18796760^4 = 20615673^4

Roger Frye who found the smallest possible n=4 solution

95800^4 + 217519^4 + 414560^4 = 422481^4

Try combinations for n > 5. No known solutions are found so far!

Composite numbers

http://mathworld.wolfram.com/CompositeNumber.html

A composite (positive integer) is a product a*b with a and b not necessarily distinct integers.

We can express every composite as

xy + xz + yz + 1,

with x, y, z positive integers

Thursday, November 25, 2010

Two consecutive whole cubes

That is, (a + 1)^3 and a^3
Let's consider the difference between the two
(a + 1)^3 - a^3
we get
a^3 + 3a^2 + 3a + 1 - a^3
= 3a^2 + 3a + 1

I'm interested in finding the set of numbers of this difference being a square, that is
(a + 1)^3 - a^3 = n^2
3a^2 + 3a + 1 = n^2

For example, (8^3) - (7^3) = 169 = 13^2

Then, we can show that

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

for some integer k.

We need to determine "a" such that 3a^2 + 3a + 1 = n^2

This is given in the OEIS

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, ...

The recurrence for this sequence is a(n) = 14a(n-1) - a(n-2) + 6.

And,
(104+1)^3 - (104^3) = 32761 = 181^2
(1456^3) - (1455^3) = 6355441 = 2521^2

Giving us the sequence A001570

Numbers n such that n^2 is simultaneously square and centered hexagonal.

A centered hexagonal number, or hex number, is

Since n^2 = 3a^2 + 3a + 1
then 4n^2 = 12a^2 + 12a + 4
And 4n^2 - 1 = 12a^2 + 12a + 3
4n^2 - 1 = 3(2a + 1)^2

If p is prime and p|(2a + 1)^2,
then p|2n + 1 or p|2n − 1,
but p cannot divide both (2n + 1) and (2n - 1)
Otherwise, p|(2n+1)−(2n−1) = 2, which cannot be since (2a+1)^2 is odd.

We now have two possible cases (u and v are integers):
2n + 1 = u^2 and 2n - 1 = 3v^2,
or 2n + 1 = 3u^3 and 2n - 1 = v^2

The former case implies
2n = u^2 - 1 = 3v^2 + 1 or u^2 = 3v^2 + 2 = 2 (mod 3)
which is impossible since for any integer u, u^2 = 0 or 1 (mod 3)
Therefore,
2n - 1 = v^2 for some odd positive integer v.

Let k = (v - 1)/2.
Then
k^2 + (k + 1)^2 = 2k^2 + 2k + 1 = {(2k + 1)^2 + 1}/2 = (v^2 + 1)/2 = n.

Palindromic num3ers: Searching for 3-digit numbers such as

Here we search for 3-digit numbers that are palindromic in other bases, base 2, 3, 4, 5, 6, 7, 8 and 9.

The tool I'm using is :

For example,

Base 10 .................. Palindromic in base(s)
100 .......................... 1 0201 (base 3), 202 (base 7), 121 (base 9)
101 .......................... NO
102 .......................... NO
103 .......................... NO
104 .......................... 404 (base 5), 252 (base 6)
105 .......................... 1221 (base 4), 151 (base 8)
106 .......................... NO
107 .......................... 110 1011 (base 2), 212 (base 7)
108 .......................... NO
109 .......................... 414 (base 5), 131 (base 9)
110 .......................... NO
111 ........................... 303 (base 6)
112 ........................... 11011 (base 3)
113 ........................... NO
114 ........................... 424 (base 5), 222 (base 7)
115 ........................... NO
116 ........................... NO
117 ........................... 313 (base 6)
118 ........................... 141 (base 9)
119 ........................... 1110111 (base 2), 434 (base 5)
120 ........................... NO
121 ............................ 11111 (base 3), 232 (base 7), 171 (base 8)
122
123
124


Could you show how to find such numbers without going thru all 3-digit numbers?

Interesting factorization

How would you factorize

a^3 (b - c) + b^3 (c - a) + c^3 (a - b)

Wednesday, November 24, 2010

Rational Number



A rational number is any number that can be expressed as the quotient a/b of two integers, with the denominator b not equal to zero. Since b may be equal to 1, every integer is a rational number.

OR

can be defined by (a,b) ~ (c,d) if, and only if, a*d - b*c = 0
("~" is the equivalence relation)

that is to say, a/b = c/d <=> a*d - b*c = 0


Can you show that every positive rational number can be written as a quotient of products of factorials of (not necessarily distinct) prime numbers?

For example, 10/9 = (2!*5!) / (3!*3!*3!)

How would you write 5/7 ?

Word Arithmetic

(1) Now + Here = Nowhere

(2) S + laughter = slaughter

(3) It is curious that the word 'Lie" is the exact middle of the word "Believe"

Tuesday, November 23, 2010

Finding Prime Numbers in Binary such as

How many primes among the positive integers, written as usual in base 10, are alternating 1’s and 0’s, beginning and ending with 1?

Tools:
Base 10 .................................... Base 2

2 ................................................. 10
3 ................................................. 11
5 ................................................. 101 (Yes)
7 ................................................. 111
11 ................................................ 1011
13 ................................................ 1101
17 ................................................ 10001 (the 0's and 1's do not alternate)
19 ................................................ 10011
23 ................................................ 10111
29 ................................................ 11101
31 ................................................ 11111
37 ................................................ 100101 (not quite)
41 ................................................ 101001 (not quite)
43 ................................................ 101011
47 ................................................ 101111
53 ................................................ 110101
59 ................................................ 111011
61 ................................................ 111101
67 ................................................ 1000011
71 ................................................ 1000111
73
79
83
89
97
101
103
107
109
113
127
131
137
139
149
151
157
163
167
173
179
181
191
193
197
199
211
223
227
229
233
239
241
251
257
263
269
271
277
281

Math Quotes

Religious differences breed wars, mathematical differences breed new ideas.

Maths in the news

Doing the Math on a Groupon Deal



'Dancing with the Stars' math made easy



Climate Change Math in Treaties Flawed by Suspect Calculations



A culture of Mathematics






Club gives Waterloo West students a hands-on approach to math



Klein Bottle Opener


Maths formula proves giraffes can swim


Mathematics has proven that giraffes can swim - even though they wouldn't be very good at it and nobody has ever seen them do it.


Child geniuses: What happens when they grow up?



Why thinking about nothing is energy-

draining


Mathematicians studying the curious phenomenon of people trying to think about nothing have made an interesting discovery about how our brains work. It turns out that stopping a thought burns energy in the same way that thinking does.

Why Thinking of Nothing Can Be So Tiring



Math, sex and tattoos: Cal prof stirs it up with short film


Secret of Big Caves Revealed by Math



The Increasing Importance of Math for Journalists :

Post stories that just don't add up



Books: 'Old Dogs, New Math' an aid for parents when it doesn't add up



Virginia high school uses poker to teach math

Poker at Falls Church high school stirs debate


Conrad Wolfram: Teaching kids real math with computers

"We have a real problem with math education right now," says technologist Conrad Wolfram in his TEDTalk.




'Eureka machine' can discover laws of nature

The machine formulates laws by observing the world and detecting patterns in the vast quantities of data it has collected

http://www.guardian.co.uk/science/video/2009/apr/02/eureka-machine-artificial-intelligence

Math in the News Archive


http://www.ams.org/news/math-in-the-media/math-in-the-media


Maths in the news

Doing the Math on a Groupon Deal


It all adds up for number nuts drawing on comic side of maths




Arithmetic Games

(1) Show that every positive integer is a sum of one or more numbers of the form 2^r * 3^s, where r and s are nonnegative integers and no summand divides another

For example,

1 = 2^0 * 3^0, r = s = 0
2 = 2^1 * 3^0, r = 1, s = 0
3 = 2^0 * 3^1, r = 0, s = 1
4 = 2^2 * 3^0, r = 2, s = 0

How would you write 5?

23 = 9 + 8 + 6

Actually, we can write, 5 = 2 + 3 = 1 + 4

So, 5 = 2 + 3 = (2^1 * 3^0) + (2^0 * 3^1)
and, 5 = 1 + 4 = (2^0 * 3^0) + (2^2 * 3^0)

Naturally, a formal proof will be needed.

The formal proof: we proceed by induction.
That is to say, we suppose all integers less than n-1 can be represented.
Then we face two cases:
(1) If n is even, then, ...
(2) If n is odd, then, ...

P.S. This problem is originally due to Paul Erdos. Note that the representations need not be unique: for instance,

11 = 2 + 9 = 3 + 8


(2) Find the smallest prime p such that the digits of p (in base 10) add up to a prime number greater than 10.


Monday, November 22, 2010

My Favorite Unsolved problem: (7) Factorial problem

Factorial


0! is a special case that is explicitly defined to be 1

1! = 1
2! = 2 * 1 = 2
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24
5! = 5 * 4 * 3 * 2 * 1 = 120
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800
etc.

Notice that
4! + 1 = 24 + 1 = 25 = 5^2
5! + 1 = 120 + 1 = 121 = 11^2
But
6! + 1 = 720 + 1 = 721 is not a square number

7! + 1 = 5040 + 1 = 5041 = 71^2
8! + 1 = 40320 + 1 = 40321 is not a square number
9! + 1 = 362880 + 1 = 362881 is not a square number
10! + 1 = 3628800 + 1 = 3628801 is not a square number

So the unsolved problem is: can we find an integer n larger than 7, such as n! + 1 = x^2 ?

My Favorite Unsolved problem: (6)

It is known that the number 3 can be expressed in two ways

1^3 + 1^3 + 1^3 = 1 + 1 + 1 = 3

and also

4^3 + 4^3 + (-5)^3 = 64 + 64 - 125 = 128 - 125 = 3


Can we find other ways to express 3 as the sum of three positive or negative cubes?

My Favorite Unsolved problem: (5)

Does there exist any number that can be expressed as a sum of 2 positive 5th powers in at least 2 different ways, that is to say, can we find a positive integer such,

a^5 + b^5 = c^5 + d^5

Math in the movies

The professor and his beloved Equation (2005)





Saturday, November 20, 2010

Numbers that are equal to the sum of cubes of its third parts



153 = 1^3 + 5^3 + 3^3

370 = 3^3 + 7^3 + 0^3

371 = 3^3 + 7^3 + 1^3

407 = 4^3 + 0^3 + 7^3

165033 = 16^3 + 50^3 + 33^3

221859 = 22^3 + 18^3 + 59^3

336700 = 33^3 + 67^3 + 00^3

336701 = 33^3 + 67^3 + 1^3

340067 = 34^3 + + 00^3 + 67^3

341067 = 34^3 + 10^3 + 67^3

407000 = 40^3 + 70^3 + 00^3

407001 = 40^3 + 70^3 + 1^3

444664 = 44^3 + 46^3 + 64^3

487215 = 48^3 + 72^3 + 15^3

982827 = 98^3 + 28^3 + 27^3

983221 = 98^3 + 32^3 + 21^3

166500333 = 166^3 + 500^3 + 333^3

296584415 = 296^3 + 584^3 + 415^3

333667000 = 333^3 + 667^3 +000^3

333667001 = 333^3 + 667^3 +001^3

334000667 = 334^3 + 000^3 + 667^3

710656413 = 710^3 + 656^3 + 413^3

828538472 = 828^3 + 538^3 + 472^3

Numbers that are equal to the sum of squares

1233 = 12^2 + 33^2

990100 = 990^2 + 100^2

94122353 = 9412^2 + 2353^2

Math Education / Games

Singapore Math Explained To The Community


Nice Game: Jumping Frogs


Fido Puzzle: Pick a 3-digit or a 4-digit number

Could you figure out how it's done?

Friday, November 19, 2010

Math sites & tools

(1) Interactive Mathematics

Interesting Property of the cubes



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
10^3 = 1000
11^3 = 1331
12^3 = 1728
.......
.......

Now,
1^3 + 2^3 = 9 = 3^2
1^3 + 2^3 + 3^3 = 36 = 6^2
1^3 + 2^3 + 3^3 + 4^3 = 10^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 = 15^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 = 441 = 21^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 = 784 = 28^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 = 1296 = 36^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 = 2025 = 45^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 + 10^3 = 3025 = 55^2
1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 + 10^3 + 11^3 = 4356 = 66^2
etc.

Could you prove that the sum of consecutive perfect cubes is always a square number?

Letter sequences

From the alphabet

A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Take a sequence of seven consecutive letters, and change one letter in that sequence to a vowel, that is,

A, E, I, O, U (and sometimes Y)

and rearrange the result to name something familiar, something you might find around the house.

Which sequence of letters and which vowel will be the optimum?




Tuesday, November 16, 2010

Odd numbers

                                               1
                                          3       5
                                     7       9      11
                                13     15     17     19
                             21   23     25     27     29
                        31     33     35     37     39     41
                    43     45     47     49     51     53     55
                57    59     61     63     65     67     69     71
           73     75     77     79     81     83     85     87     89
       91     93     95     97     99    101   103   105   107   109
  111   113   115   117   119   121    123    125   127   129   131

Notice the following:

Row #1 : 1 = 1^3
Row #2 : 3 + 5 = 8 = 2^3
Row #3 : 7 + 9 + 11 = 27 = 3^3
Row #4 : 13 + 15 + 17 + 19 = 64 = 4^3
Row #5 : 21 + 23 + 25 + 27 + 29 = 125 = 5^3
Row #6 : 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3
Row #7 : 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 = 7^3
Row #8 : 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512 = 8^3
Row #9 : 73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89 = 729 = 9^3
Row #10 : 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 = 1000 = 10^3
Row #11 : 111 + 113 + 115 + 117 + 119 + 121 + 123 + 125 + 127 + 129 + 131 = 1331 = 11^3

So, in row #n, we would have the sum equal n^3.

Of course, we would need to establish a proof. Would anyone like to prove it?