Indian Statistical Institute B.Math & B.Stat : Number Theory

Let $S$ be the set of all integers $k$, \( 1 \leq k \leq n\), such that $g.c.d(k,n)=1$. What is the arithmetic mean of the integers in $S$?. $$$$ First note that \( |S| = \phi(n) \). Now let $k \in S$ then there exists intergers $u,v$ such that $ku+nv=1\dots (A)$. $$$$ The integer \( n-k \in \{1,2,\dots,n-1\} \) because $k$ can never be equals $n$, for $g.c.d(n,n)=n$ and $k \in S$. We will now show that $g.c.d(n-k,n)=1.$ Adding $-nu$ to both sides of $(A)$ we get \( -nu+ku+nv=-nu+1 \implies -u(n-k)+(v+u)n = 1 \implies g.c.d(n-k,n)=1 \) $$$$ So, for \( k \in \{1,2,\dots,n-1\}\) if \( k \in S \implies n-k \in S \) Thus $S$ can be written in the form, \( S=\{k_1,k_2,k,.....,k_r,n-k_r,......,n-k_2,n-k_1\} \) where $|S| = \phi(n)$ $$$$ Clearly the sum of the elements of $S$ is \( (k_1+n-k_1)+(k_2+n-k_2)+\dots+(k_r+n-k_r) = \frac{n\phi(n)}{2} \) ( Pairing reduces the terms to half the original ($\phi(n)))$. $$$$ arithmetic mean \[ =\frac{\frac{n\phi(n)}{2}}{\phi(n)} = \frac{n}{2} \]

Indian Statistical Institute B.Math & B.Stat : Inequality

If \(a,b,c \in (0,1) \) satisfy $a+b+c=2$, prove that \( \frac{abc}{(1-a)(1-b)(1-c)} \geq 8. \) $$$$ Let \( p = 1-a, q = 1-b, r = 1-c \). $p+q+r= 3-(a+b+c)=1$.Clearly $p,q,r$ are positive. Substituting in the given inequality, it transforms to $$$$ \( \frac{(1-p)(1-q)(1-r)}{pqr} \geq 8 \iff {(1-p)(1-q)(1-r)} \geq 8{pqr} \) $$$$ \( \iff 1-(p+q+r)+qr+rp+pq-pqr \geq 8{pqr} \iff qr+rp+pq-pqr \geq 8{pqr} \) $$$$ \( \iff qr+rp+pq \geq 9{pqr} \iff \frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9 \) $$$$ Thus proving the above inequality reduces to proving \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9\) subjected to $p+q+r=1$ $$$$ Since $p,q,r$ are positive, apllying $A.M \geq H.M$ we have \( \frac{p+q+r}{3} \geq \frac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}\) $$$$ Noting $p+q+r=1$, we have \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \geq 9 \)

Indian Statistical Institute B.Math & B.Stat : Combinatorics

Problem 1: Points on a Curve

Question: A point \( P \) with coordinates \( (x,y) \) is said to be good if both \( x \) and \( y \) are positive integers. Find the number of good points on the curve \( xy=27027 \).

Solution:

First, note the prime factorization of the number: \( 27027 = 3^3 \times 13^1 \times 11^1 \times 7^1 \).

Since \( x \) and \( y \) must be positive integers, for every chosen positive integer \( x \) that divides 27027, there is exactly one corresponding integer \( y \) (where \( y = \frac{27027}{x} \)). Therefore, the number of good points is simply equal to the total number of positive divisors of 27027.

Using the divisor function formula \( \tau(n) \), if \( n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k} \), the number of divisors is \( (a_1 + 1)(a_2 + 1) \dots (a_k + 1) \).

Applying this to our factorization:

\[ \tau(27027) = (3+1)(1+1)(1+1)(1+1) = 4 \times 2 \times 2 \times 2 = 32 \]

There are 32 good points on the curve.

Problem 2: Ordered Triplets

Question: What is the number of ordered triplets \( (a, b, c) \), where \( a, b, c \) are positive integers (not necessarily distinct), such that \( abc = 1000 \)?

Solution:

First, find the prime factorization of 1000, which is \( 2^3 \times 5^3 \).

Any ordered triplet \( (a, b, c) \) satisfying the condition must be of the form:

• \( a = 2^{l_1} \times 5^{m_1} \)
• \( b = 2^{l_2} \times 5^{m_2} \)
• \( c = 2^{l_3} \times 5^{m_3} \)

For \( abc = 2^3 \times 5^3 \) to hold true, the powers of each prime base must add up to 3. This gives us two independent equations:

1. \( l_1 + l_2 + l_3 = 3 \)
2. \( m_1 + m_2 + m_3 = 3 \)

We need to find the number of non-negative integer solutions for each equation. Using the "Stars and Bars" combinatorics method, the number of solutions to \( x_1 + x_2 + x_3 = n \) is given by \( \binom{n+r-1}{r-1} \), where \( r \) is the number of variables.

For the powers of 2: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

For the powers of 5: \( \binom{3+3-1}{3-1} = \binom{5}{2} = 10 \) solutions.

Since the distributions of the powers of 2 and 5 are independent, we multiply the possibilities:

\[ 10 \times 10 = 100 \]

Thus, in total 100 ordered triplets are possible.

Problem 3: Derangements and Boxes

Question: There are 8 balls numbered \( 1,2,\dots,8 \) and 8 boxes numbered \( 1,2,\dots,8 \). Find the number of ways one can put these balls in the boxes so that each box gets one ball and exactly 4 balls go into their corresponding numbered boxes.

Solution:

First, we must choose which 4 balls will go into their correct boxes. This can be done in \( \binom{8}{4} \) ways. For any such selection—say we choose \( \{Ball_2, Ball_5, Ball_3, Ball_7\} \)—there is exactly 1 way for them to go into their corresponding boxes.

Now, the remaining 4 balls—in this example, \( \{Ball_1, Ball_4, Ball_6, Ball_8\} \)—must be placed into the remaining 4 boxes such that none of them end up in a box matching their own number. This is the classic definition of a derangement.

The number of ways to derange \( n \) objects is denoted as \( D_n \). The formula for a derangement is:

\[ D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n\frac{1}{n!} \right) \]

For our remaining 4 balls, we calculate \( D_4 \):

\[ D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9 \]

The total number of valid configurations is the product of our choices:

\[ \text{Total Ways} = \binom{8}{4} \times D_4 = 70 \times 9 = 630 \]

```

Indian Statistical Institute B.Math & B.Stat : Combinatorics

Using only the digits $2$, $3$ and $9$, how many six digit numbers can be formed which are divisible by $6$? $$$$ First note that a number is divisible by $6$, $iff$ it is divisible by $2$ and by $3.$ ($2$ and $3$ being co-prime)$$$$ The above argument clearly shows that the number $222222$ is divisible by $6$. Use divisibility test of $2$ and $3$. First note that a number will be divisible by $2$, $iff$ the digit at the unit place is $2.$ Now observe that, if the number which is divisible by $2$ has to be divisible by $3$, the sum of the digits must be divisible by $3.$ So the 5 places of the number (except the digit at the unit place, which is $2$) is a combination of the digits $2$, $3$ and $9$ such that the sum including the digit at the unit place is divided by $3$.$$$$ Now observe that among the $5$ places to be filled, exactly $two$ places can be the digit $2$. (Convince yourself why?! thin if not...) And the remaining $three$ places can be filled by either $3$ or $9$. Thus giving \( \binom {5} {2} \times 2\times2\times2 \) possibilities with desired condition. $$$$ therefore, total number of numbers \(= \binom {5} {2} \times 2\times2\times2+1 = 80+1=81 \)

Indian Statistical Institute B.Math & B.Stat : Integration

Evaluate: \[\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx\] $$$$ Let \( I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx\), Put $x=\frac{1}{t}$ \( \implies I = -\int_{2014}^\frac{1}{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt =\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{t}}{t} dt = \int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx \) $$$$ \(\mathbb{Therefore,}\) $$$$ \[2I=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x}{x} dx+\int_\frac{1}{2014}^{2014} \frac{tan^{-1}\frac{1}{x}}{x} dx\] $$$$ \[=\int_\frac{1}{2014}^{2014} \frac{tan^{-1}x+tan^{-1}\frac{1}{x}}{x} dx \] $$$$ \[=\int_\frac{1}{2014}^{2014} \frac{\frac{\pi}{2}}{x} dx \] $$$$ \[=\frac{\pi}{2}\int_\frac{1}{2014}^{2014} \frac{dx}{x} \] $$$$ \[=\frac{\pi}{2}[\log x]_{\frac{1}{2014}}^{2104} \] $$$$ \[=\frac{\pi}{2}\big[\log 2014-\log \frac{1}{2014}\big]\] $$$$ \[=\frac{\pi}{2}\log 2014^2\] $$$$ \[=\pi\log 2014\] $$$$

Indian Statistical Institute B.Math & B.Stat : Continuity

Let \( P: \mathbb{R} \rightarrow \mathbb{R}\) be a continuous function such that $P(x)=x$ has no real solution. Prove that $P(P(x))=x$ has no real solution. $$$$ If possible let, $P(P(x))=x$ has a real solution for $x=x_0$. Then $P(P(x_0))=x_0\dots(1)$ $$$$ Now let, $P(x_0)=y_0$ \( \implies P(y_0)=x_0\) using $(1)$ $$$$ Note that $x_0 \neq y_0$, otherwise we will have a solution to the equation $P(x)=x$! A contradiction to the hypothesis. Without loss of generality assume that \(x_0 < y_0\) $$$$ Construct a function \( Q:[x_0,y_0]\rightarrow \mathbb{R}\) where \(Q(x)=P(x)-x\),since $P$ is given to be continuous on $\mathbb{R}$, $Q$ is continuous on \([x_0,y_0].\) $$$$ Observe that, \( Q(x_0)=P(x_0)-x_0=y_0-x_0 > 0\) and \( Q(y_0)=P(y_0)-y_0=x_0-y_0 < 0\) $$$$ \(\implies Q(x_0)Q(y_0) < 0 \implies \) there exists a point $c\in$ \([x_0,y_0]\) such that $Q(c)=0$ using $ Intermediate-Value-Theorem$ $$$$ \(\implies P(c)-c=0 \implies P(c)=c\) for a real value, which contradicts the hypothesis, thus the assumption $P(P(x))=x$ has a real solution is not tenable.

Combinatorics :Indian Statistical Institute B.Math & B.Stat

How many $three-digits$ numbers of distinct digits can be formed by using the digits $1,2,3,4,5,9$ such that the sum of digits is at least 12? $$$$ First note that since the digits must be distinct there must be no repetition and further note that the maximum sum of such three digits is $9+5+4=18.$ $$$$ We will find all such possible by considering their sum of the digits. $$$$ Case I: Sum of the digits is 12. In this case the selection of digits can be \(\{1,2,9\},\{3,4,5\}\) $$$$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$ Case II: Sum of the digits is 13. In this case the selection of digits can be \(\{1,3,9\}\) $$$$ So, a total of $3!=6$ $three-digits$ numbers possible. $$$$ Case III: Sum of the digits is 14. In this case the selection of digits can be \(\{1,4,9\},\{2,3,9\}\) $$$$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$ case IV: Sum of the digits is 15. In this case the selection of digits can be \(\{1,5,9\},\{2,4,9\}\) $$$$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$ Case V: Sum of the digits is 16. In this case the selection of digits can be \(\{2,5,9\},\{2,5,9\}\) $$$$ So, a total of $3!+3!=12$ $three-digits$ numbers possible. $$$$ Case VI: Sum of the digits is 17. In this case the selection of digits can be \(\{3,5,9\}\) $$$$ So, a total of $3!=6$ $three-digits$ numbers possible. $$$$ Case VII: Sum of the digits is 18. In this case the selection of digits can be \(\{4,5,9\}\) $$$$ So, a total of $3!=6$ $three-digits$ numbers possible. $$$$ Adding all the cases we have $12+6+12+12+12+6+6=66$, $three-digits$ numbers of distinct digits can be formed.

Complex Numbers :Indian Statistical Institute B.Math & B.Stat

Let \(\omega\) be the complex cube root of unity. Find the cardinality of the set \(S\) where \(S = \{(1+\omega+\omega^2+\dots+\omega^n)^m \mid m,n = 1,2,3,\dots\}\)

Note that the sum \(1+\omega+\omega^2+\dots+\omega^n\) contains \(n+1\) terms. The power \(n\) must be of the form \(3k\), \(3k+1\) or \(3k+2\) where \(k \in \mathbb{N} \cup \{0\}\).

When \(n\) is of the form \(3k+2\), there are \(3k+3\) terms (a multiple of 3). Because \(1+\omega+\omega^2 = 0\), the sum evaluates to \(0\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 0^m = 0\) \(\forall m \in \mathbb{N}\).

When \(n\) is a multiple of 3 (i.e., \(3k\)), there are \(3k+1\) terms. The sum evaluates to \(1\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = 1^m = 1\) \(\forall m \in \mathbb{N}\).

When \(n\) is of the form \(3k+1\), there are \(3k+2\) terms. The sum evaluates to \(1+\omega = -\omega^2\). Whence \((1+\omega+\omega^2+\dots+\omega^n)^m = (-\omega^2)^m = (-1)^m \omega^{2m}\) \(\forall m \in \mathbb{N}\).

In this final case, as \(m\) varies over the natural numbers, the expression \((-1)^m \omega^{2m}\) generates 6 distinct values: \(-\omega^2, \omega, -1, \omega^2, -\omega, 1\).

Combining the three cases, we see that \( S = \{0, -1, 1, \omega, -\omega, \omega^2, -\omega^2\} \), therefore \(|S|=7\).

Common terms of two A.P Series : Indian Statistical Institute B.Math & B.Stat

Consider the two arithmetic progressions \( 3,7,11,\dots,407\) and \(2,9,16,\dots,709\). Find the number of common terms of these two progressions. $$$$ Let $a_n$ and $a_m$ be the last terms of the progressions respectively \( \Rightarrow 407 = 3+(n-1)4\) and \( 709 = 2+(m-1)7 \) $$$$ Solving we get, \( n,m = 102\). To find the common terms, assume that the $n^{th}$ term of the first progression is equal to the $m^{th}$ term of the second progression. $$$$ \(\Rightarrow 3+(n-1)4=2+(m-1)7 \Rightarrow 3+4n-4-2+7=7m \Rightarrow 4(n+1)=7m,\) where \( n,m \in \{1,2,3,\dots,102\}\) $$$$ R.H.S is a multiple of 7, while L.H.S is 4(n+1). Since $g.c.d(4,7)=1$ L.H.S will be multiple of $7$, $iff$ $n+1$ is a multiple of $7$. $$$$ \( \Rightarrow n = 6,13,20,\dots \) Again since $n$ is bounded by $102$. The largest possible value of $n$ is $97$. $$$$ So, \( n \in \{6,13,20,\dots\,97}\) Which has $14$ terms. Thus the number of common terms of the progression is $14$

Application of Rolle's Theorem

Have you ever looked at a polynomial and wondered if it has a root in a specific interval, without actually having to solve it?

Calculus gives us an incredibly elegant workaround for this exact scenario: Rolle's Theorem. Let's walk through a classic problem that demonstrates how we can find hidden roots by stepping up one level of abstraction.

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The Problem

Suppose we have four real numbers \(a_0, a_1, a_2,\) and \(a_3\) that satisfy the following condition:

$$a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0$$

The Goal: Prove that the polynomial \(f(x) = a_0 + a_1x + a_2x^2 + a_3x^3\) has at least one root in the interval \((0, 1)\).

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The Intuition: Where do we start?

When asked to prove that a function \(f(x)\) equals zero at some point, a calculus student's first instinct should be to look for an antiderivative.

Why? Because Rolle's Theorem tells us that if a continuous, differentiable function starts and ends at the exact same height on a graph (for instance, if it equals \(0\) at both ends of an interval), it must flatten out somewhere in between. At that flat point, its derivative is exactly zero.

So, if we can find a parent function \(g(x)\) such that \(g'(x) = f(x)\), and we can show that \(g(x)\) is tied down at \(x=0\) and \(x=1\), the theorem will do the heavy lifting for us.

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The Proof

Step 1: Construct the parent function, \(g(x)\).
To find our function \(g(x)\), we simply integrate \(f(x)\) with respect to \(x\):

$$g(x) = \int (a_0 + a_1x + a_2x^2 + a_3x^3) \,dx$$
$$g(x) = a_0x + \frac{a_1}{2}x^2 + \frac{a_2}{3}x^3 + \frac{a_3}{4}x^4$$

(Note: We can ignore the constant of integration \(+C\) here by simply setting it to \(0\), as we just need one valid parent function).

Step 2: Evaluate the endpoints of our interval \([0, 1]\).
Let's plug our bounds into \(g(x)\) to see how the function behaves at the edges.

At \(x = 0\):

$$g(0) = a_0(0) + \frac{a_1}{2}(0)^2 + \frac{a_2}{3}(0)^3 + \frac{a_3}{4}(0)^4 = 0$$

At \(x = 1\):

$$g(1) = a_0(1) + \frac{a_1}{2}(1)^2 + \frac{a_2}{3}(1)^3 + \frac{a_3}{4}(1)^4$$
$$g(1) = a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4}$$

Notice something familiar? The problem explicitly states that \(a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \frac{a_3}{4} = 0\). Therefore:

$$g(1) = 0$$

Step 3: Apply Rolle's Theorem.
We now have a function \(g(x)\) where:

1. \(g(x)\) is a polynomial, meaning it is perfectly continuous on the closed interval \([0,1]\) and differentiable on the open interval \((0,1)\).
2. \(g(0) = g(1) = 0\).

Because both endpoints are at the same \(y\)-value, the graph of \(g(x)\) must turn around at least once between \(x=0\) and \(x=1\).

By Rolle's Theorem, there must exist at least one point \(c\) in the interval \((0, 1)\) where the derivative is zero:

$$g'(c) = 0$$

Since \(g'(x)\) is just our original polynomial \(f(x)\), this means:

$$f(c) = a_0 + a_1c + a_2c^2 + a_3c^3 = 0$$

Conclusion:

We have successfully proven that \(f(x)\) has at least one root exactly where we wanted it: inside the interval \((0, 1)\).

Problem from Indian Statistical Institute: B.Stat. (Hons.)

Let $P(x)$ be a polynomial of degree $11$ such that \( P(x) = \frac{1}{x+1}\), for \( x = 0,1,2,\dots,11.\) $$$$ Find the value of $P(12)$ $$$$ Solution: Let \( f(x) = (x+1)P(x)-1\), clearly $f(x)$ is a polynomial of degree 12. $$$$ Now for \( x \in \{0,1,2,\dots,11\}\), \(f(x) = (x+1)P(x)-1=\frac{x+1}{x+1}-1=1-1=0\) $$$$ This shows that $f(x)$ vanishes at the points \( x = 0,1,2,\dots,11.\) $$$$ $f(x)$ being of degree 12, the above statement assures that \( x = 0,1,2,\dots,11.\) are the possible roots of $f(x)$ $$$$ Therefore \(f(x)=(x+1)P(x)-1=a_{0}(x-0)(x-1)(x-2)\dots(x-11)\) $$$$ Letting $x=-1$ in the above equality, we have \(-1=a_{0}(-1)(-2)(-3)\dots(-12) \Rightarrow a_{0} = \frac{-1}{12!}\) $$$$ Therefore \(f(x)=(x+1)P(x)-1=\frac{-1}{12!}(x-0)(x-1)(x-2)\dots(x-11)\) $$$$ Now, letting $x=12$ we have \(13P(12)-1=\frac{-1}{12!}(12)(11)(10)\dots(1)=\frac{-12!}{12!}=-1\) $$$$ \(\Rightarrow 13P(12)-1=-1 \Rightarrow P(12) = 0 \)

CBSE Mathematics Solved Paper 2015 : XII

CBSE Mathematics Solved Paper 2015 : XII  

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Remainder Theorem

The term containing the highest power of $x$ in the polynomial $f(x)$ is $2x^4$. $$ $$ Two of the roots of the equation \(f(x)=0\) are -1 and 2. Given that $x^2-3x+1$ is a quadratic $$ $$ factor of $f(x)$, find the remainder when $f(x)$ is divided by $2x-1.$ $$$$ Since degree of $f(x)$ is $4$ and $x^2-3x+1$ is a factor of $f(x)$, it can be written as product of two quadratics $$ $$ Therefore, \(f(x) = ( x^2-3x+1 ) (ax^2+bx+c ) \). Again since, $2x^4$ is the leading term $a$, must be equal to $2$ $$$$ So, \(f(x) = ( x^2-3x+1 ) (2x^2+bx+c ) \). Given $-1$ and $2$ are the roots of $f(x)$ $\Rightarrow$ $f(-1)=0$ & $f(2)=0 $ $$$$ Note that $x^2-3x+1$ does not vanishes at \(x = -1 , 2 \) $\Rightarrow$ $2x^2+bx+c$ must vanishes at this two points. $$$$ \( \Rightarrow 2-b+c = 0 , 8+2b+c=0\) Solving the equations we get \(b=-2, c=-4\) $$$$ Thus \(f(x) = ( x^2-3x+1 ) (2x^2-2x-4 ) \). Required remaninder is \(f(\frac{1}{2}) = \frac{9}{8} \)