Pre RMO Practice Sheet 2

Question 1:

A man walks a certain distance and rides back in 3.75 hours, he could ride both ways in 2.5 hours. How many hours would it take him to walk both ways?

Let the man walk distance x km in time t hours. His walking speed is $\frac{x}{t}$ kmph.

Let him ride distance x km in time T hours. His riding speed is $\frac{x}{T}$ kmph.

First journey:

$\frac{x}{t} + \frac{x}{T} = 3.75$

Second journey:

$\frac{2x}{T} = 2.5$

Hence, $\frac{x}{T} = 1.25$

Using above in first equation: $\frac{x}{t} + 1.25 = 3.75$

Hence, $\frac{x}{t} = 2.50$ = his speed of walking. Hence, it would take him 5 hours to walk both ways.

Question 2:

Positive integers a and b are such that $a+b=\frac{a}{b} + \frac{b}{a}$. What is the value of $a^{2}+b^{2}$?

Given that a and b are positive integers.

Given also $\frac{a}{b} + \frac{b}{a} = a+b$.

Hence, $a^{2}(b-1)+b^{2}(a-1)=0$

As a and b are both positive integers, so $a-1$ and $b-1$ both are non-negative.

So, both the terms are non-negative and hence, sum is zero if both are zero or $a=1$ and $b=1$.

Hence, $a^{2}+b^{2}=2$

Question 3:

The equations $x^{2}-4x+k=0$ and $x^{2}+kx-4=0$ where k is a real number, have exactly one common root. What is the value of k?

$x^{2}-4x+k=0$…equation I

$x^{2}+kx-4=0$…equation II

Let $k \in \Re$ and let $\alpha$ is a common root.

Hence, $\alpha$ satisfies both the equations. So, by plugging in the value of $\alpha$ we get the following:

$\alpha^{2} -4\alpha + k =0$ and $x^{2}+kx-4=0$ and so using these two equations, we get the following:

$(k+4)(1-\alpha)=0$.

Case 1: $\alpha \neq 1$ then $k=-4$. But hold on, we havent’t checked thoroughly if this is the real answer. We got to check now if both equations with these values of alpha and k have only one common root.

Equation I now goes as : $x^{2}-4x-4=0$ so this equation has irrational roots. On further examination, we see that if $\alpha=1$, then k can be any value. So, what are the conditions on k? We get that from equation I: plug in the value of alpha:

$1-4+k=0$ so $k-3=0$ and $k=3$.

So, we now recheck if both equations have only one common root when alpha is 1 and k is 3:

$x^{2}-4x+3=0$ so the roots of first equation are 3 and 1.

$x^{2}+3x-4=0$ so the roots of second equation are -4 and 1.

Clearly so $k=3$ is the final answer. ðŸ™‚

Cheers,

Nalin Pithwa

Pre RMO practice sheet

Question 1:

What is the smallest positive integer k such that $k(3^{3}+4^{3}+5^{3})=a^{n}$ for some positive integers a and n with $n>1$?

Solution 1:

We have $k \times 216= a^{n}$ so that $k \times 6^{3} = a^{n}$ giving k=1.

Question 2:

Let $S_{n} = \sum_{k=0}^{n}\frac{1}{\sqrt{k+1}+\sqrt{k}}$.

What is the value of $\sum_{n=1}^{99} \frac{1}{S_{n}+S_{n-1}}$

Given

$S_{n} = \sum_{k=0}^{n} \frac{1}{\sqrt{k+1}+\sqrt{k}} \times \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}}$

$S_{n} = \sum_{k=0}^{n} (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}$

Similarly,

$S_{n-1} = \sum_{k=0}^{n-1}(\sqrt{k+1}-\sqrt{k}) = \sqrt{n}$

Now, $\sum_{n=1}^{99} \frac{1}{S_{n}+S_{n-1}} = \sum_{n=1}^{99}\frac{1}{\sqrt{n+1}-\sqrt{n}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$

which in turn is equal to $\sum_{n=1}^{99} (\sqrt{n+1}-\sqrt{n}) = (\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+ (\sqrt{4}-\sqrt{3}+ \ldots + (\sqrt{100}-\sqrt{99})) = \sqrt{100} -\sqrt{1}=10-1=9$

Question 3: Homework:

It is given that the equation $x^{2}+ax+20=0$ has integer roots. What is the sum of all possible values of a?

Cheers,

Nalin Pithwa

A series question : pre RMO, RMO, IITJEE

Question 1:

Let $x_{1}, x_{2}, \ldots, x_{2014}$ be real numbers different from 1, such that

$x_{1}+x_{2}+\ldots+x_{2014}=1$ and

$\frac{x_{1}}{1-x_{1}} + \frac{x_{2}}{1-x_{2}} + \ldots + \frac{x_{2014}}{1-x_{2014}} = 1$ also.

Then, what is the value of

$\frac{x_{1}^{2}}{1-x_{1}} + \frac{x_{2}^{2}}{1-x_{2}} + \frac{x_{3}^{2}}{1-x_{3}} + \ldots + \frac{x_{2014}^{2}}{1-x_{2014}}$ ?

Solution 1:

Note that

$\sum \frac{x_{i}^{2}}{1-x_{i}} = \sum \frac{x_{i} + (x_{i} - x_{i}^{2})}{1-x_{i}} = \sum (\frac{x_{i}}{1-x_{i}} - x_{i}) = \sum \frac{x_{i}}{1-x_{i}} -\sum x_{i} = 1 - 1 = 0$ which is required answer.

Note that the maximum index 2014 plays no significant role here.

Question 2:

Let f be a one-to-one function from the set of natural numbers to itself such that

$f(mn) = f(m)f(n)$ for all natural numbers m and n.

What is the least possible value of $f(999)$ ?

From elementary number theory, we know that given f is a multiplicative function and hence, the required function is such that if p and q are prime, then

$f(pq)=f(p)f(q)$

That is we need to decompose 999 into its unique prime factorization.

So, we have $999 = 3 \times 333 = 3^{2} \times 111 = 3^{3} \times 97$ where both 3 and 97 are prime.

We have $f(999) = f(3)^{3} \times f(97)$ and we want this to be least positive integer. Clearly, then f(3) cannot be greater than 97. Also, moreover, we need both f(3) and f(97) to be as least natural number as possible. So, $f(3)=2$ and $f(97)=3$ so that required answer is 24.

Question 3:

HW :

What is the number of ordered pairs (A,B) where A and B are subsets of $\{ 1,2,3,4,5\}$ such that neither $A \subset B$ nor $B \subset A$ ?

Cheers,

Nalin Pithwa

Basic algebra facts of HCF and LCM

1. If x and y are two natural numbers, HCF(x,y) x LCM (x,y) = xy
2. If a, b, c are three natural numbers, then the product abc=HCF(a,b,c) x LCM(ab,bc,ca) and also abc=HCF(ab, bc,ca) x LCM(a,b,c).

Symmetric difference is associative

We want to show that : $A \triangle (B \triangle C) = (A \triangle B) \triangle C$

Part I: TPT: $LHS \subset RHS$

Proof of Part I: let $x \in LHS$

Then, $x \in A \triangle (B \triangle C)$. By definition of symmetric difference,

$x \in \{ A - (B \triangle C)\} \bigcup \{(B \triangle C) -A \}$

Hence, $x \in A$, $x \notin (B \triangle C)$, OR $x \in B \triangle C$, $x \in A$

That is, $x \notin A \bigcap (B \triangle C)$.

Hence, $x \notin A$, and $x \notin B \triangle C$

Hence, $x \notin A$ and $x \in B \bigcap C$.

Hence, $x \in (B \bigcap C)-A$.

Hence, $x \in B$, $x \in C$, but $x \notin A$.

Therefore, $x \in B$, but $x \notin A \bigcap C$. —- Call this I.

Consider $y \in (A \triangle B) \triangle C$.

Therefore, $y \notin (A \triangle B) \bigcap C$.

Therefore, $y \notin C$, and $y \notin A \triangle B$

Therefore, $y \notin C$, and $y \in A \bigcap B$.

Therefore, $y \in (A \bigcap B)-C$.

Hence, $y \in A$, $y \in B$, but $y \notin C$.

That is, $y \in B$, $y \notin A \bigcap C$. Call this II.

From I and II, $LHS \subset RHS$.

Part II: TPT: $RHS \subset LHS$.

Quite simply, reversing the above steps we prove part II.

QED.

Cheers,

Nalin Pithwa

Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in $m+n$ ways. More generally, if $E_{i}$ ($i=1,2,\ldots,k$) are k events such that no two of them can occur at the same time, and if $E_{i}$ can occur in $n_{i}$ ways, then one of the k events can occur in $n_{1}+n_{2}+\ldots+n_{k}$ ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, \$if $E_{1}$ can occur in $n_{1}$, $E_{2}$ can occur in $n_{2}$ ways (no matter how $E_{1}$ occurs), $E_{3}$ can occur in $n_{3}$ ways (no matter how $E_{1}$ and $E_{2}$ occur), $\ldots$, $E_{k}$ can occur in $n_{k}$ ways (no matter how the previous $k-1$ events occur), then the k events can occur simultaneously in $n_{1}n_{2}n_{3}\ldots n_{k}$ ways.

)3. Definitions and some basic relations:

Suppose X is a collection of nÂ distinctÂ objects and r is a nonnegative integer less than or equal to n. An r-permutationÂ of X is a selection of r out of the n objects but the selections areÂ ordered.Â

An n-permutation of XÂ is called a simply aÂ permutation of X.

The number of r-permutations of a collection of n distinct objects is denoted by $P(n,r)$; this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of $(n-1)$ objects can be chosen to occupy the second position in $(n-1)$ ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in $n(n-1)$ ways,….and all r positions can be filled in

$P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}!$ ways.

In particular, $P(n,n) = n!$

Note: AnÂ unordered selectionÂ of r out of the n elements of X is called an r-combination of X.Â In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by $n \choose r$ (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary $(n-r)$-subset, whence the important relation ${n \choose r}$ = $n \choose {n-r}$.

To evaluate $n \choose r$, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule:

$P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)$

The number of terms on the right is the number of r-subsets of X. That is, $n \choose r$. Thus, $P(n,r)=P(r,r) \times {n \choose r}$=$r! \times {n \choose r}$.

The following is our summary:

1. $P(n,r) = \frac{n!}{(n-r)!}$
2. ${n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}$=$n \choose {n-r}$

4. The Pigeonhole Principle: Basic Version:

If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters).

5.Â The number of ways in which $m+n$ things can be divided into two groups containing m and n equal things respectively is given by : $\frac{(m+n)!}{m!n!}$

Note: If $m=n$, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is $\frac{(2m)!}{2!m!m!}$

6.Â The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: $\frac{(m+n+p)!}{m!n!p!}$

Note: If we put $m=n=p$, we obtain $\frac{(3m)!}{m!m!m !}$ but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in $\frac{(3m)!}{m!m!m!3!}$ ways.

7.Â The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: $\frac{n!}{p!q!r!}$

8.Â The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: $n^{r}$. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-))

9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : $2^{n}-1$

10.Â The total number of ways in which it is possible to make a selection by taking some or all out of $p+q+r+\ldots$ things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : $(p+1)(q+1)(r+1)\ldots-1$.

Regards,

Nalin Pithwa.

III. Tutorial problems. Symmetric and alternating functions. RMO and IITJEE math

1. Simplify: $(b^{-1}+c^{1})(b+c-a)+(c^{-1}+a^{-1})(c+a-b)+(a^{-1}+b^{-1})(a+b=c)$
2. Simplify: $\frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$
3. Simplify: $\frac{b^{2}+c^{2}-a^{2}}{(a-b)(a-c)} + \frac{c^{2}+a^{2}-b^{2}}{(b-c)(b-a)} + \frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}$
4. Simplify: $\frac{b-c}{1+bc} + \frac{c-a}{1+ca} + \frac{a-b}{1+ab}$
5. Simplify: $\frac{a(b-c)}{1+bc} + \frac{b(c-a)}{1+ca} + \frac{c(a-b)}{1+ab}$
6. Factorize: $(b-c)^{2}(b+c-2a)+(c-a)^{2}(c+a-2b)+(a-b)^{2}(a+b-2c)$. Put $b-c=x, c-a=y, a-b=z$and $b+c-2a=y-z$
7. Factorize: $8(a+b+c)^{2}-(b+c)^{2}-(c+a)^{2}-(a+b)^{2}$. Put $b+c=x, c+a=y, a+b=z$.
8. Factorize: $(a+b+c)^{2}-(b+c-a)^{2}-(c+a-b)^{2}+(a+b-c)^{2}$
9. Factorize: $(1-a^{2})(1-b^{2})(1-c^{2})+(a-bc)(b-ac)(c-ab)$
10. Express the following substitutions as the product of transpositions: (i) $\left(\begin{array}{cccccc}123456\\654321\end{array}\right)$ (ii) $\left(\begin{array}{cccccc}123456\\246135\end{array}\right)$ (iii) $\left(\begin{array}{cccccc}123456\\641235\end{array}\right)$

Regards,

Nalin Pithwa.

II. tutorial problems. Symmetric and alternating functions. RMO, IITJEE math

Reference: Higher Algebra by Bernard and Child.Â

Exercises: (based on the earlier blogged chapter from the above reference):

Prove the identities from problem 1 to 5 given below where $\Sigma{\alpha}$, $\Sigma{\alpha\beta}$ etc. denote symmetric functions of $\alpha$, $\beta$, $\gamma$, $\delta$. Also verify by putting $\alpha=\beta=\gamma=\delta=1$:

1 $(\alpha+\beta+\gamma+\delta)(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}) = \Sigma{\alpha^{2}}+\Sigma{\alpha^{2}\beta}$

2. $(\alpha+\beta+\gamma+\delta)(\beta\gamma\delta+\gamma\delta\alpha+\delta\alpha\beta+\alpha\beta\gamma) = \Sigma{\alpha^{2}\beta\gamma}+4\alpha\beta\gamma\delta$

3. $(\beta\gamma\delta+\gamma\delta\alpha+\delta\alpha\beta+\alpha\beta\gamma)^{2}=\Sigma{\alpha^{2}}{\beta^{2}}{\gamma^{2}}+2\Sigma{\alpha\beta\gamma\delta}\Sigma{\alpha\beta}$

4.$(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)^{2}=\Sigma{\alpha^{2}\beta^{2}}+2\Sigma{\alpha^{2}\beta\gamma}+6\alpha\beta\gamma\delta$

5. $\Sigma{\alpha\beta}.\Sigma{\alpha\beta\gamma}=\Sigma{\alpha^{2}\beta^{2}\gamma}+3\alpha\beta\gamma\delta.\Sigma{\alpha}$

Regards,

Nalin Pithwa.

Tutorial problems. I. Symmetric and Alternating functions. RMO/IITJEE Math

Exercises:

1. Show that $(bc-ad)(ca-bd)(ab-cd)$ is symmetric with respect to a, b, c, d.
2. Show that the following expressions are cyclic with respect to a, b, c, d, taken in this order: $(a-b+c-d)^{2}$ and $(a-b)(c-d)+(b-c)(d-a)$
3. Expand the expression using $\Sigma$ notation: $(y+z-2x)(z+x-2y)(x+y-2z)$
4. Expand the expression using $\Sigma$ notation: $(x+y+z)^{2}+(y+z-x)^{2}+(z+x-y)^{2}+(x+y-z)^{2}$
5. Prove that $(\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+\alpha^{2}\beta^{2})(\alpha+\beta+\gamma)= \Sigma\alpha^{2}\beta^{2}+\alpha\beta\gamma\Sigma\alpha\beta$
6. Prove that $(\alpha-\beta)(\alpha-\gamma)+(\beta-\gamma)(\beta-\alpha)+(\gamma-\alpha)(\gamma-\beta)=\Sigma{\alpha^{2}}-\Sigma{\alpha}{\beta}$
7. Prove that $(\beta-\gamma)(\beta+\gamma-\alpha)+(\gamma-\alpha)(\gamma+\alpha-\beta)+(\alpha-\beta)(\alpha+\beta-\gamma)=0$
8. Prove that : $\alpha(\beta-\gamma)^{2}+\beta(\gamma-\alpha)^{2}+\gamma(\alpha-\beta)^{2}=\Sigma{\alpha^{2}}{\beta}-6\alpha\beta\gamma$
9. Prove that: $(\beta^{2}\gamma+\beta\gamma^{2}+\gamma^{2}\alpha+\gamma\alpha^{2}+\alpha^{2}\beta+\alpha\beta^{2})(\alpha+\beta+\gamma)=\Sigma{\alpha^{2}}\beta+2\Sigma{\alpha^{2}}{\beta^{2}}+2\alpha\beta\gamma\Sigma{\alpha}$
10. Prove that : $a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+abc(a+b+c)=\Sigma{a^{2}}.\Sigma{ab}$.
11. Prove that: $(a+b-c)(a^{2}+b^{2}-c^{2})+(b+c-a)(b^{2}+c^{2}-a^{2})+(c+a-b)(c^{2}+a^{2}-b^{2})=3\Sigma{a^{3}}-\Sigma{a^{2}{b}}$
12. Prove that: $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=(ax+by+cz)^{2}+(bz-cy)^{2}+(cx-az)^{2}+(ay-bz)^{2}$
13. Prove that: $(b^{2}-ac)(c^{2}-ab)+(c^{2}-ab)(a^{2}-bc)+(a^{2}-bc)(b^{2}-ac)=-(bc+ca+ab)(a^{2}+b^{2}+c^{2}-bc-ca-ab)$
14. Prove that: $(a^{2}-bc)(b^{2}-ac)(c^{2}-ab)=abc(a^{2}+b^{2}+c^{2})-(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2})$
15. If one of the numbers a, b, and c is the geometric mean of the other two, use the previous problem to prove the following: $abc(a^{2}+b^{2}+c^{2})=b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}$
16. If the numbers x, y, z taken in some order or other form an AP, use problem 3 to prove that $2(x+y+z)^{2}+27xyz=9(x+y+z)(yz+zx+xy)$
17. Express $2(a-b)(a-c)+2(b-c)(b-a)+2(c-a)(c-b)$ as the sum of three squares. Hence, show that $(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)$ is negative for all real values of a, b, c except when $a=b=c$. Hint: Put $b-c=x$, $c-a=y$, $a-b=z$, and notice that $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=(x+y+z)^{2}=0$.
18. If $x+y+z=0$, show that (i) $2yz=x^{2}-y^{2}-z^{2}$; (ii) $(y^{2}+z^{2}-x^{2})(z^{2}+z^{2}-y^{2})(x^{2}+y^{2}-z^{2})+8x^{2}y^{2}z^{2}=0$ (iii) $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy$ can be expressed in the form $px^{2}+qy^{2}+rz^{2}$; and, find p, q, r in terms of a, b, c, f, g, h.

Cheers,

Nalin Pithwa.

Number theory: let’s learn it the Nash way !

Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}

Cheers,

Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1

https://www.amazon.in/INSTRUMENTS-TI-Nspire-CX-II-CAS/dp/B07XCM6SZ3/ref=sr_1_1?crid=3RNR2QRV1PEPH&keywords=ti+nspire+cx+cas&qid=1585782633&s=electronics&sprefix=TI+n%2Caps%2C253&sr=8-1