Posted in Astronomy | Tagged 2009, Astronomy, international, international year of astronomy, IYA, unesco, universe | 2 Comments »
5 040 – Greek philosopher Plato gave this number as the population of an ideal city.
This number has several factors, which implies an efficient division of population for the sake of land distribution, taxes, war etc.
In fact,
5, 040 = 7! = (1)(2)(3)(4)(5)(6)(7)
5, 040 = (24)(32)(51)(71)
Thus, 5 040 has (4+1)(2+1)(1+1)(1+1) = 60 factors
Posted in Mathematics, Philosophy | Tagged 5040, city, fact, factor, greek, idea, ideal city, numbers, philosopher, Plato, population | 2 Comments »
Compute in (less than) 30 seconds without using calculator.
My technique:
Since 2007 is the arithmetic mean of the numbers 2004, 2006, 2008 and 2010, then: Let x = 2007
So, x-3 = 2004, x-1 = 2006, x+1 = 2008, x+3 = 2010
Hence, the expression is equal to
Simplifying,
= x2 – 5 = (2007)2 – 5 = 4, 028, 044
Posted in Mathematics | Tagged algebra, calculator, compute, Math, Mathematics, numbers, technique, without calculator | Leave a Comment »
When the polynomial x2007 + 1 is divided by x2 – 3x +2, the remainder can be expressed in the form ax+b. Find the value of a.
My Solution:
Let k be the quotient, then:
x2007 + 1 = k(x2 – 3x +2) + (ax+b)
Since x2 – 3x +2 = (x-2)(x-1),
x2007 + 1 = k(x-2)(x-1) + (ax+b)
To omit k and to solve for the value of a easily, set x=2 and x=1:
For x=2
(2)2007 + 1= k(2-2)(2-1) + (2a+b)
22007 + 1= 2a+b (*)
For x=1
(1)2007 + 1 = k(1-2)(1-1) + (a+b)
2 = a+b (**)
To solve for the value of a, perform elimination using (*) and (**).
..Which in turns give: a=22007-1
Posted in Mathematics | Tagged algebra, division, elimination, factor, factor theorem, Math, Mathematics, polynomial, remainder, remainder theorem | 2 Comments »
Compute the number of squares between 49 and 94
My short solution: The technique is to express these two numbers into perfect squares so that it’s easy to compute the number of squares between these two numbers.
49 = (22)9 = (29)2 = 5122 ;
94 = (32)4 = (34)2 = 812
Therefore, the number of squares between 49 and 94 is (512-81)-1 = 430
Posted in Mathematics | Tagged combinatorics, exponent, law of exponent, Math, Mathematics, numbers, numerical analysis, perfect square, square | Leave a Comment »
Five digit numbers containing all digits from 1-5 are arranged from highest to lowest. What is the 25th number?
(Examples of these numbers are 54 321; 23 451; 41 532)
My Solution:
First, it’s impractical to list these (5)(4)(3)(2)(1)=120 numbers from highest to lowest and then determine the 25th number in the list. So, we can instead do some technique and investigate the distribution of these numbers.
Since it’s from highest to lowest, the first step is to determine the number of these five digit numbers which have a ten thousands digit of 5.
For thousands digit: 4 numbers to choose.
For hundreds digit: 3 numbers to choose.
For tens digit: 2 numbers to choose.
For unit digit: 1 number to choose.
Thus, the number of these 5 digit numbers which have a ten thousands digit of 5 is (4)(3)(2)(1)=24. Then these 24 numbers are the first 24 numbers in the list. Hence, the 25th number is the highest number which has a ten thousands digit of 4. And this 25th number is 45 321.
Posted in Mathematics | Tagged arrangement, combinatorics, digits, five digit number, Math, Mathematics, numbers, numerical analysis | 1 Comment »
Prove that: , for any
Solution: Since , then in order to produce a short
proof, the expression should be express in the form
Doing so,
Hence,

Posted in Mathematics | Tagged calculus, differential calculus, e, limit, Math, Mathematics, number e, proof, proving | Leave a Comment »
At what angle of projection is the range maximum? Prove it mathematically.
To solve for the required angle of projection ( ), we need first to find a general expression for the range.
To find range, set y = 0 first and use:
,
Solving for the general expression for range,
Regardless of the value of , the maximum value of range would be attain if
is maximum. Since 1 is the maximum value of sine function, then,
should be equal to 1. Since sin 90º = 1, therefore
(angle of projection) should be equal to 45º
Posted in Physics | Tagged 45 degrees, angle, classical mechanics, kinematics, maximum, newtonian mechanics, Physics, projectile, proof, proving, range, trajectory, trigonometry | Leave a Comment »

“A physical theory must possess mathematical beauty.”
In the most basic explanation, mathematical beauty is related to simplicity and surprise.
- Paul Dirac, quantum physicist



Posted in Mathematics, Physics | Tagged beauty, Dirac, Math, mathematical beauty, Mathematics, Paul Dirac, physical theory, simplicity, surprise, theory | 5 Comments »
To celebrate the beauty of our planet on Earth Day, NASA released its 10 favorite photos taken by astronauts on the International Space Station.
Click here: NASA’s top 10 views of Earth
Posted in Astronomy | Tagged Astronomy, beauty, Earth, Earth day, international space station, NASA, photos, planet | Leave a Comment »
Video by pauleycamerieexodus
Posted in Physics | Tagged CNN, general theory of relativity, Physics, ronald mallett, Theoretical Physics, time, time machine, time travel, video | 2 Comments »
Researchers Prove Existence of New Basic Element for Electronic Circuits — ‘Memristor’ from PhysOrg.com
HP today announced that researchers from HP Labs have proven the existence of what had previously been only theorized as the fourth fundamental circuit element in electrical engineering.
This scientific advancement could make it possible to develop computer systems that have memories that do not forget, do not need to be booted up, consume far less power and associate information in a manner similar to that of the human brain.
In a paper published in today’s edition of Nature, four researchers at HP Labs’ Information and Quantum Systems Lab, led by R. Stanley Williams, presented the mathematical model and a physical example of a “memristor” – a blend of “memory resistor” – which has the unique property of retaining a history of the information it has acquired.
Leon Chua, a distinguished faculty member in the Electrical Engineering and Computer Sciences Department of the University of California at Berkeley, initially theorized about and named the element in an academic paper published 37 years ago. Chua argued that the memristor was the fourth fundamental circuit element, along with the resistor, capacitor and inductor, and that it had properties that could not be duplicated by any combination of the other three elements…..
Read More: [...]
Posted in Physics | Tagged circuit, computer, computer science, electrical engineering, electronic circuit, electronics, engineering, HP labs, memory resistor, memristor, nanoelectronics, Physics | 2 Comments »
(AIME 1994) Find the positive integer n for which
[log2 1] + [log2 2] + [log2 3] +…+ [log2 n] = 1994
where [x] denotes the greatest integer less than or equal to x (for example [Π] = 3)
Solution:
Observe that:
[log2 1] = 0(20)
[log2 2] + [log2 3] = 1(21)
[log2 4] + [log2 5] + [log2 6] + [log2 7] = 2(22)
[log2 8] + [log2 9] +…+ [log2 15] = 3(23)
[log2 16] + [log2 17] +…+ [log2 31] = 4(24) ……
** [log2 c] + [log2 (c+1)] +…+ [log2 (2c-1)] = b(2b); where 2b = c, for positive integers b and c.
So,
0(20) + 1(21) + 2(22) + 3(23) + 4(24) + 5(25) + 6(26) + 7(27) + 456 = 1994
**7(27) is the sum of [log2 128] + [log2 129] + [log2 130] +…+ [log2 255]. Hence, 456 = [log2 256] + [log2 257] + [log2 258] +…+ [log2 n].
Since [log2 256]=[log2 257]=[log2 258]=…=[log2 n] = 8
So, 456/8 = 57; Then n = 256 + (57-1).
Therefore, n = 312
Posted in Mathematics | Tagged AIME, algebra, American Invitational Mathematics Examination, logarithm, Math, Mathematics | 2 Comments »
There was a young fellow from Trinity (Cambridge University?)…..
…..Who took √∞ ….. But the number of digits….. Gave him the fidgets…..
He dropped Math and took up Divinity…
Hehehehe … ∞
This is the phrase at the bottom of title page of the book One, Two, Three…Infinity by George Gamow.
I enjoyed reading it, since it is about facts and speculations of Science!
Posted in Mathematics | Tagged book, divinity, George Gamow, infinity, Math, math student, Mathematics, number of digits, square root | 2 Comments »
Grigori Yakovlevich Perelman

sometimes known as Grisha Perelman, is a Russian mathematician who has made landmark contributions to Riemannian geometry and geometric topology. In particular, he has proved Thurston’s geometrization conjecture. This solves in the affirmative the famous Poincaré conjecture, posed in 1904 and regarded as one of the most important and difficult open problems in mathematics until it was solved.
In August 2005, Perelman was awarded the Fields Medal, for “his contributions to geometry and his revolutionary insights into the analytical and geometric structure of the Ricci flow“. The Fields Medal is widely considered to be the top honor a mathematician can receive. However, he declined to accept the award or appear at the congress.
On December 22, 2006, the journal Science recognized Perelman’s proof of the Poincaré Conjecture as the scientific “Breakthrough of the Year,” the first such recognition in the area of mathematics
In 1982, as a member of the USSR team competing in the International Mathematical Olympiad, an international competition for high school students, he won a gold medal, achieving a perfect score.
Perelman was best known for his work in comparison theorems in Riemannian geometry. Among his notable achievements was the proof of the soul conjecture.
More about Perelman:
Perelman on Wikipedia
Perelman’s Homepage
On New York Times
The Poincaré Conjecture–Proved
Eccentric Genius
Fields Medal
Quote:
“My Mathematics has nothing to do with awards or public recognition, the prize is the joy of solving a difficult problem” – Grigori Yakovlevich Perelman
Posted in Mathematics, Physics | Tagged Fields Medal, Geometric Topology, Grigori Perelman, Math, Mathematician, Mathematics, Poincare Conjecture, Riemannian Geometry | 2 Comments »
Let P and P’ denote points inside rectangles ABCD and A’B'C’D', respectively. If PA = a+b, PB=a+c, PC = c+d, PD =b+d, P’A=ab, P’B'=ac, P’C'=cd, prove that P’D'=bd.
(Pi Mu Epsilon Journal – 4(Spring 1967)258, proposed by Stanley Rabinowitz)
————————————————————–
*Let sides AB and CD are the horizontal sides, and sides BC and AD are the vertical sides.
*Draw 2 perpendicular lines (vertical and horizontal) inside rectangle ABCD such that their intersection is point P. It follows that the the vertical line segment is perpendicular to sides AB and CD, and the horizontal line segment is perpendicular to sides BC and AD.
*Let:
point a1 – intersection of the v.l.segment and side AB
point a2 - intersection of the h.l.segment and side BC
point a3 – intersection of the v.l.segment and side CD
point a4 – intersection of the h.l.segment and side AD
*Let: u=distance bet. A and a1= distance bet. D and a3
w=distance bet. B and a1= distance bet. C and a3
x=distance bet. A and a4= distance bet. B and a2
y=distance bet. D and a4= distance bet. C and a2
*Then, by Pythagorean theorem
(PA)2 = x2 + u2, (PB)2 = w2 + x2 , (PC)2 = w2 + y2, (PD)2 = y2 + u2
Since (y2 + u2) + (w2 + x2) = (x2 + u2) + (w2 +y2).
Therefore, (PD)2 + (PB)2 = (PA)2 + (PC)2,
(b+d)2 + (a+c)2 = (a+b)2 + (c+d)2,
(ac)2 + (bd)2 = (ab)2 + (cd)2 – - – (1)
*Same procedure to A’B'C’D',
(P’D')2 = (P’A')2 + (P’C')2 – (P’B')2,
(P’D')2 = (ab)2 + (cd)2 – (ac)2
*Substitute (1):
(P’D')2 = [(ac)2 + (bd)2] – (ac)2
Therefore, P’D’ = bd
Quote: “As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection” - Joseph-Louis Lagrange
Posted in Mathematics | Tagged Euclidean Geometry, Geometry, Math, Math proof, Mathematics, Pi Mu Epsilon Journal, points, proving, Pythagorean Theorem, rectangle, Stanley Rabinowitz | 2 Comments »
(AHSME 1996) When n standard 6 sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same probability of obtaining a sum of S. What is the smallest possible value of S?
————————————————————–
Observe that for any natural number a:
For 2a dice (even):
prob. of obtaining a sum of 12a (maximum sum) = prob. of obtaining a sum of 2a (minimum sum)
prob. of obtaining a sum of 12a -1 = prob. of obtaining a sum of 2a+1
prob. of obtaining a sum of 12a -2 = prob. of obtaining a sum of 2a+2
.
.
prob. of obtaining a sum of 12a-(5a-1) = prob. of obtaining a sum of 2a+(5a-1)
prob. of obtaining a sum of 12a-(5a) = prob. of obtaining a sum of 2a+(5a)
For 2a+1 dice (odd):
prob. of obtaining a sum of 12a+6 (maximum sum) = prob. of obtaining a sum of 2a+1 (minimum sum)
prob. of obtaining a sum of (12a+6)-1 = prob. of obtaining a sum of (2a+1)+1
prob. of obtaining a sum of (12a+6)-2 = prob. of obtaining a sum of (2a+1)+2
.
.
prob. of obtaining a sum of (12a+6)-(5a+1) = prob. of obtaining a sum of (2a+1)+(5a+1)
prob. of obtaining a sum of (12a+6)-(5a+2) = prob. of obtaining a sum of (2a+1)+(5a+2)
Based on these observations,
*In order to minimize S, 1994 should be one of the largest (or maximum) sum/s
*Note that the maximum sums, 12a and 12a+6, are both divisible by 6 for any natural number a. Since 1994 is not divisible by 6, it is not the maximum sum
*The nearest natural number which is divisible by 6 and greater than 1994 is 1998. Hence 1998 is the maximum sum. So 12a = 1998 (if even) or 12a+6 =1998 (if odd).
12a = 1998…a =166.5 (not a natural number), 12a+6 = 1998… a = 166 (natural number)
*So there are odd numbers of dice. Since 1998-1994 = 4, then 1994 should be (12a+6)-4. And since prob. of obtaining a sum of (12a+6)-4 = prob. of obtaining a sum of (2a+1) +4, then prob. of obtaining a sum of 1994 is the same prob. of obtaining a sum of [2(166) +1] +4 = 337
Quote:
“There is no more difficult art to acquire than the art of observation, and for some men it is quite as difficult to record an observation in brief and plain language.” – William Osler
Posted in Mathematics | Tagged AHSME, American High School Mathematics Examination, dice, Divisibility, Math, Mathematics, Probability | Leave a Comment »










