Education

Web Courses

We develop college and school curriculum courses - complete with materials, references, problems, tests, certificates and credits. We are in the process of getting them approved by number of universities, schools, Institutes and colleges in many countries. We wish university professors, educationists, and scientists to be involved in this global project. We shall customize courses as per schools curricula and standard. Our charges shall be reasonable, only to keep this electronic remote learning program well funded. But we are committed to keep our charges minimal for students in poor countries and worldwide. We also would like to get our courses approved by recognized bodies in different countries. We believe that knowledge is for universal consumption and not limited to priviledged few. We would like to offer the courses free of charge to all via public libraries. We would like to fund educationists to continue doing research and creating up to date quality education or research materials. We recognize web is the way to reach all learners across the world without any boundary. We would like to offer the courses free to those who deserve anywhere in the world.

Given below is a Link to explore:

We shall be interested in hearing from you if customized courses are to be developed by us.

We shall be very interested to hear from you if you are an educationist and you wish to invest your intellectual assets with us for marketing worldwide. We shall be paying you up to 12% of the total revenue earned from the course. We are committed to publish all education material.

Either way, we shall be interested in signing agreement with you.

We edit the courses for web publishing; we quality check for correctness; we market them globally; and we protect them from being stolen or forgotten.

Contact: info@techvisionsystems.com

Problems

We love solving problems - theoretical or practical. Let us know if you have any problem.

Here is a set of problems for 2006 Ross Summer Course at Ohio State University:

http://education.techvisionsystems.com/problems06.pdf

Here are their resolutions:

[Please try them before looking up the solution]

Ans to 1.

Out of 7 product components:  (a – b)

There can at most be 6 product components (a-b) that are odd numbers in the product; [that is (odd-even) or (even-odd)] and this happens when there are 3 (or 4) odd numbers and rest (4 or 3) are even numbers. In no case, all 7 product components can be odd. The seventh component has to be even number [(odd-odd) or (even-even)]. As a result the entire product is going to be even.

Ans to 2.

(a)    32 is not a nice number

(b)   2p for all p positive is not a nice number.

If N is a Nice number; there is for some m>n, N = Sum( 1 to m) – Sum(1 to n);

That is, N = (m2+m-n2-n)/2 =(m-n)(m+n+1)/2.

Suppose 2p  is a nice number, 2p=(m-n)(m+n+1)/2. Both (m-n) and (m+n+1) are even. Therefore (m+n) is odd. That is either m or n is odd. Then m-n is odd. But we know m-n is even. This is a contradiction.

Therefore 2p is not a nice number.

(c)    1000 is a very nice number. 198+199+200+201+202 (average 200, 5 consecutive number). 28+29+30+31+…39+40+41+….+68 (average 40, 25 consecutive number).

The number is very nice number if it is possible to factorize the number in two or more different manner with at least one factor being odd number [e.g. 200x5=1000 or 40x25=1000]

Ans to 3.

Suppose A B C D E F G H I J K L are the names of 12 students sharing 3 tables each with 4 places.

 X A B C D E F G H I J K L A X 1 1 2 1 1 1 2 1 1 2 2 B X X 1 1 2 2 2 1 1 2 1 1 C X X X 1 1 1 1 1 5 1 1 1 D X X X X 1 2 1 2 1 1 1 2 E X X X X X 2 2 1 1 1 1 2 F X X X X X X 1 1 1 2 1 1 G X X X X X X X 1 1 2 2 1 H X X X X X X X X 1 2 2 1 I X X X X X X X X X 1 1 1 J X X X X X X X X X X 1 1 K X X X X X X X X X X X 2 L X X X X X X X X X X X X Table1 Table2 Table3 Day 1 ABCI DHKL EFGJ Day 2 ADEL BGHJ CFIK Day 3 AGKL BDFJ CEHI Day 4 ADFH BEGK CIJL Day 5 AHJK BEFL CDGI

Given above is the sitting arrangement for 5 days where all students dine with every other student in the same table at least once.

Ans 4.

(a) Proof: Suppose the statement is false.

if A ={1,9     }

B = {3, 11    }

Case 1. Suppose 2 is in A, 1+1+2=4,2+2+1=5,2+2+2=6 are in B, then 12 is in A. but 2+1+9=12. A is TSP. Contradiction.

Case 2. Suppose 2 is in B. B={2,3,11  }, B is not TSP, A is then = {1,9,6,7  } but 1+1+7=9 is in A. A is TSP. Contradiction.

We have contradictions in either case.

Therefore Statement is true.

(b) A={1,9,2}, B={3,4,5,6,7,8,9,10} None of them are TSP.

Ans  5.

(a) We divide the 10x10 array into four 5x5 arrays. One of them must have at least 13 red (or blue). Here are 3 cases for the 5x5 array which has 13 red dots.

Case 1. One of the row (or column) is having 5 reds. Then there is at least one other row having at least 2 reds. These 2 red points and corresponding 2 column points in the row having 5 red points form a monochrome rectangle.

Case 2. One row (or column) has 4 red points. Then there is one other row having 3 red points. These two rows must 2 columns having reds in corresponding rows (since there are only 5 points in a row. Thus a monochrome red rectangle exists.

Case 3. None of the above cases.  Then there are at least 3 rows, A B C, having 3 dots each. Take any 2 rows (A,B) and suppose there is only one column having intersecting column with red points. Consider now the 3rd row (C) with 3 red points. It shall have at least two columns with red dots with corresponding column positions having red dots with at least one of the A or B rows. Hence there is at least one red monochrome rectangle.

(b) Yes. See the above proof for 5x5 arrays.

Ans to 6.

There is balanced 3-wheel.  1, 6, 3, 2, 5, 4 (, 1)

Line segments are: 1 – 2, 6 – 5, 3 – 4

1+6 = 2 + 5; 6+3 = 4+5; 2+3 = 4+1

There is no balanced 4 wheel.

The wheel need be balanced on the diameter running through the point having load 1. The load diametrically opposite to 1, got to be an odd number say 2X+1=y. So that each half should weigh even = 18 (36/2).  Consider the axis (1,y).  y is odd.

Case 1: Both remaining odd numbers belong to one side (say A). This shall have one even number. This would result in 3 even numbers in the opposite side (say B). Sum of one odd and one even in A, is sum of two even numbers in side B. It is impossible.

Case2: one odd number in each side of the axis (1,y). y is 3/5/7. y can’t be 3 or 7. Since 18-(3+1)/2 or 18-(7+1)/2 are both even and can’t sum of two even + one odd number. So y is 5. Neighbors of 1 can’t be 2 or 4, since opposite 1 is 5. Hence neighbors 5 are 2 and 4. Neighbors of 1 are 8 and 6. Opposite 8 is 4. Opposite of 6 is 2. 7 can’t be neighbor of 8. 7 is neighbor of 6. 6+7=2+3 (impossible).

Therefore there are no balanced 4 wheels.

Balanced wheel does not exist for any n that is even.

Balanced wheel is balanced around every axis. There shall be n-1 loads on either side of any axis; n being even, n-1 is odd. Either side must weigh n(2n+1). Average load of each axis is 2n+1 (odd). In order to be balanced all axis either have both even, or both odd loads and no odd-even loads. Balanced wheel got to be symmetric of placement of odd axes and even axis in each diameter. Therefore, each odd and even axis got to be interleaved. This would create only odd sums. We also know each sum has an equal sum pair on the diagonally opposite side. If the starting sum is 2n+1, opposite is also 2n+1. This requires 4 points to plot. the adjacent sum need be 2n+3 and the opposite being same, they too should have two adjacent sums of 2n-1. It takes 4 more points to plot. If n is a multiple of 4, the next points to plot that yield 2n+5 and 2n-3. With any starting number (say 1, and corresponding 2n, and say n+1 and n at the opposite), we shall be creating divergent sequence of points 2n+2,3,2n,1,2n-2, …, n-2,n+1,n,n+3, n+2, … .However we can’t have any number exceeding 2n. Hence it is impossible to draw a balanced wheel for any n even.

Ans to 7.

(a)    45 degree. All the corner triangles shall be isosceles triangles for the Square.  The path of Robbie will form a rectangle.

(b)   No. It shall be always making same initial angle after 4 turns.

(c)    It shall be converging to one of the diagonals.

Ans to 8.

(a) For any N = 2m = 2(m+1) – 2m. N is an alternating sum.

Any number N can be written with binary digits (0 and 1). That is N = 2m1 + 2m2+ …   + 2mp, where m1 > m2 > … > mp.  Each of these terms can be written as alternating sum. Hence N can be written as alternating sum of some increasing sequence of distinct powers of 2.

(b) Suppose p is alternating-sum of some increasing sequence of distinct powers of 2 and it is not a power of 2. It can be written in two different ways.  Take the last binary digit which is not zero (say, at 2m position). Replace it with 2(m+1) – 2m.  This is another expression of alternate sum of p.

Ans to 9.

(a)       È  = 360 (180/5)

(b)       È   = 600, however 0th step shall be 3rd step.  0, 1, 3, 5 all need be co-linear. Point 5 and 1 shall be equidistant from 0 and in the opposite side. Triangle 3 4 0 and triangle 2 3 0 are same. That is È = 900 unless 3 is coincident with 0. If È = 900, it is not possible to go to R from L or vice versa, starting from 0.

(c)       È   = (360/7)0