## Category Archives: IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha)

### Some light moments with an IITJEE foundation math student

I have a bright kid working towards his IITJEE Foundation math. Some days back I had suggested to him to solve some hard word problems based on simultaneous equations from a very old classic text. He started on his own, almost with some guidance from me. Until he attacked very well — those “age” kind of problems. Father’s age vs. son’s age, etc. But, in this case, he suddenly sprang to his feet: He almost yelled, “Sir! Please check my solution! I am getting the answer as husband’s age is 48 and wife’s age is 23!” Actually, I too was a bit shocked; but, I checked his calculations in detail; they were mathematically correct. It suddenly flashed in my head:

“Do you know Vedant? Mathematics is not human! It has no emotions, no feelings, whatsoever! I told him the following quote of Bertrand Russell: ” Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty, cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.””

And, of course, equations//numbers don’t lie! 🙂 🙂 🙂

Nalin Pithwa.

### Petronet Kashmir Super 30: Nine crack IITJEE 2017 exams

(This news is a bit delayed here on this blog, nevertheless, inspirational.)

Reference: DNA, Mumbai print edition, June 15 2017:

Srinagar: Petronet LNG Ltd. (PLL), which started its CSR voyage from western India and traversing South and East, have now reached Northern state of Jammu and Kashmir.

Considering the lack of facilities/faculty in J and K to impart coaching for Engineering Entrance Examination to facilitate admissions to IITs/NITs/other institutions of repute, Petronet LNG Ltd. sponsored 40 underprivileged students in 2016-17 to fulfill their dream of higher engineering education (35 boys and 5 girls) under Petronet Kashmir Super 30 programme in association with Indian Army, CSRL. This 11 months’ residential programme was attended by 40 students (kargil 7, Pulwama 5, Bandipura 6, Baramulla 4, Anantnag 7, Ganderbal 4, Kulgam 1, Tangmarg 2, Ramban 1, Shopian 1, Sopora 1, Nagrota 1).

Despite hardships in Kashmir and educational institutions being closed last year, the project continued with its mission. The declaration of IITJEE results on April 27 2017 saw selection of 28 students (including 2 girls) who were coached under Petronet Kashmir Super 30 cracking the IITJEE mains exams. Six other students will join Regional Engineering Colleges.

Shri Dharmendra Pradhan, Hon’ble Minister of State (I/C) Petroleum and Natural Gas and Mr. Chowdhary Zulfkar Ali ji, Hon’ble Minister for Dept. of Food, Civil Supplies and Consumer Affairs and Information Dept., J and K, interacted with students of Petronet Kashmir Super 30 at Delhi and felicitated them for their hard work and achievement.

Shri Pradhan expressed happiness that students from underprivileged backgrounds from places like Kupwara, Pulwama, Anantnag, Shopian etc. have shown their determination and got selected in prestigious institutes. He also announced on behalf of Petronet LNG Ltd. that 100 students will be sponsored from Kashmir for engineering entrance next year.

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🙂 🙂 🙂 Congratulations to Petronet (Kashmir) LNG Team including their faculties and the students from Nalin Pithwa !

### Arithmetic Puzzle

Following is a very common arithmetic puzzle that you may have encountered as a child:

Express any whole number $latex n&bg=ffffff$ using the number 2 precisely four times and using only well-known mathematical symbols.

This puzzle has been discussed on pp. 172 of Graham Farmelo’s “The Strangest Man“, and how Paul Dirac solved it by using his knowledge of “well-known mathematical symbols”:

$latex displaystyle{n = -log_{2}left(log_{2}left(2^{2^{-n}}right)right) = -log_{2}left(log_{2}left(underbrace{sqrt{sqrt{ldotssqrt{2}}}}_text{n times}right)right)}&bg=ffffff$

This is an example of thinking out of the box, enabling you to write any number using only three/four 2s. Though, using a transcendental function to solve an elementary problem may appear like an overkill.  But, building upon such ideas we can try to tackle the general problem, like the “four fours puzzle“.

This post on Puzzling.SE describes usage of following formula consisting of  trigonometric operation $latex cos(arctan(x)) = frac{1}{sqrt{1+x^2}}&bg=ffffff$ and $latex tan(arcsin(x))=frac{x}{sqrt{1-x^2}}&bg=ffffff$ to obtain the square…

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### Somewhere — find out where !!

You have been left in the middle of an island on which there are two villages. In village A, all of the residents, no matter where they are, always tell the truth. In village B, all of the residents, no matter where they are, always tell lies. After walking a few miles from the middle of the island, you find yourself in a village square where there is a resident of one of the villages sitting on some stone steps. What one question would you ask the resident so that you would know which of  the two villages you were in?

PS: Of course, there is no GPS device with you! Or if you have, there is no connectivity!

Have fun!

Nalin Pithwa.

### Matchstick Problem

Determination is like a match stick !!! 🙂

Translation: How do we move only 1 matchstick to make the equation valid?

(“Not Equals” sign $latex neq$ is not allowed)

This is a really tricky question… Hint: Need to think in Chinese, this is a Chinese joke 😛

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## Motivational Books for Students (Educational)

The Motivational Books recommended in the above post seems to be very popular with readers on this site. Many readers (presumably parents) have bought the books! Buy a Christmas present for yourself this Xmas. 🙂

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### Non-linear equations for IITJEE Mathematics training

(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

Example 1.

Solve the following system of equations:

$x+y+z=13$ $\ldots \ldots \ldots$ Equation I

$x^{2}+y^{2}+z^{2}=65$ $\ldots \ldots \ldots$ Equation II

$xy=10$ $\ldots \ldots\ldots$ Equation III

Solution I:

From II and III, we get $(x+y)^{2} + z^{2}=85$

Plug in $u=x+y$; then this equation becomes $u^{2}+z^{2}=85$.

Also, from I, we get $u+z=13$

hence, we obtain $u=7 \hspace{0.1in}or \hspace{0.1in}u=6$, $z=6 \hspace{0.1in} or \hspace{0.1in}7$.Thus,we have

$x+y=7$, $xy=10$; and $x+y=6$, $xy=10$. Hence, the solutions are

$x=5, \hspace{0.1in} or \hspace{0.1in} 2$; $y=2, \hspace{0.1in} or \hspace{0.1in} 5$; $z=6$;

or,

$x=3 \pm \sqrt{-1}$; $y=3 \mp \sqrt{-1}$; $z=7$.

Example 2:

Solve the following system of equations:

$(x+y)(x+z)=30$ Equation I

$(y+z)(y+x)=15$ Equation II

$(z+x)(z+y)=18$ Equation III

Solution: 2:

Write u, v, w for $y+z$, $z+x$, $x+y$ respectively; thus

$vw=30$, $uw=15$, $vu=18$…..call this equations A

Multiplying these equations together, we have

$u^{2}v^{2}w^{2}=30 \times 15 \times 18 = 8100$. Hence, $uvw=\pm 90$.

Combining this result with each of the equations in A, we have

$u=3, v=6. w=5$; or, $u=-3, v=-6, w=-5$; therefore, we get two sets of linear equations, as follows:

$y+z=3$

$z+x=6$

$x+y=5$

for which the solution set is: $x=4, y=1, z=2$; and the other set is:

$y+z=-3$

$z+x=-6$

$x+y=-5$

for which the solution set if $x=-4, y=-1, z=-2$.

Example 3:

Solve the following system of equations:

$y^{2}+yz+z^{2}=49$…call this equation (1)

$z^{2}+xz+x^{2}=19$…call this equation (2)

$x^{2}+xy+y^{2}=39$…call this equation (3)

Subtracting (2) from (1), we get the following:

$y^{2}-x^{2}+z(y-x)=30$, that is, $(y-x)(x+y+z)=30$…call this equation (4).

Similarly, from (1) and (3), we get the following:

$(z-x)(x+y+z)=10$…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

$\frac{y-x}{z-x}=3$, hence, $y=3x-2z$

Substituting in Equation (3), we obtain $x^{2}-3xz+3z^{2}=13$.

From (2), we get the following: $x^{2}+xz+z^{2}=19$

Solving these homogeneous equations (hint: put $z=mx$, where m is parameter to be found), we get the following:

$x=\pm 2$, $z=\pm 3$; and therefore, $y=\pm 5$

or, $x=\pm \frac{11}{\sqrt{7}}$; $z=\pm \frac{1}{\sqrt{7}}$; and therefore, $y=\mp \frac{19}{\sqrt{7}}$.

Example 4:

Solve the following system of equations:

$x^{2}-yz=a^{2}$

$y^{2}-zx=b^{2}$

$z^{2}-xy=c^{2}$

Solution 4:

Multiply the equations by y, z, and x respectively, and add; then,

$c^{2}x+a^{2}y+b^{2}z=0$…call this equation I

Multiply the equations by z, x and y respectively; and add; then

$b^{2}x+c^{2}y+a^{2}z=0$…call this equation II

From (I) and (II), by cross multiplication,

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{x}{c^{4}-a^{2}b^{2}}=k$, suppose.

Substitute in any one of the given equations; then,

$k^{2}(a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2})=1$; hence, we get the following solution:

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{z}{c^{4}-a^{2}b^{2}}=\pm \frac{1}{\sqrt{a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}}}$.

More stuff in the pipeline,

Nalin Pithwa