1. a and b are two numbers having the same no. of digits and same sum of digits (=28). Can one be a multiple of the other? a is not equal to b. (courtesy Abhra Abir Kundu)
2. Is $latex e^x-sinx $ a polynomial ? (courtesy Tias Kundu)
3. Find the number of onto function from set A containing n elements to set B containing m elements (m<n) (courtesy Tias Kundu)
4. If a+b+c=30,  how many (a,b,c) tuples possible (a,b,c all non-negative). (courtesy Saikat Palit)
5. Can sin(x) be expressed as a polynomial in x? (courtesy Soumik Bhattacharyya)
6. Integers 1-64 are placed in a 8X8 chessboard. How many ways are there to place them such that all numbers in the 1st row and column are in AP? (courtesy Soumik Bhattacharyya)

(more…)

This post contains an interesting problem from ISI BMath 2014 based on the jump of a frog and the lotus. Solve and enjoy this problem.

Problem: Jump of a frog Problem

n (> 1) lotus leafs are arranged in a circle. A frog jumps from a particular leaf by the following rule: It always moves clockwise. From staring point it skips 1 leaf and jumps to the next. Then it skips 2 leaves and jumps to the following. That is in the 3rd jump it skips 3 leaves and 4th jump it skips 4 leaves and so on. In this manner it keeps moving round and round the circle of leaves. It may go to one leaf more than once. If it reaches each leaf at least once then n (the number of leaves) cannot be odd.

Discussion:
Suppose we number the leaves as 1, 2, 3, upto n. Leaf 1 being the starting point of the frog.
Step 1 - Leaf 1
Step 2 - Leaf 3 (because the frog skips 1 leaf)
Step 3 - Leaf 6 (1+2+3) ( because it skips 2 leaves in second jump)
In this manner the frog will be in the
$\mathbf{ \frac{m(m+1)}{2} } $ (modulo n) leaf in the mth step.

If n = 2k+1 (that is if we have odd number of leaves) then in the 2k step and 2k+1 step it will be on Leaf 2k+1.
This is because $\mathbf{\frac{(2k+1)(2k)}{2} = k(2k+1) \equiv 0 \text{mod} 2k+1 }$ and $\mathbf{\frac{(2k+1)(2k + 2)}{2} = (k+1)(2k+1) \equiv 0 \text{mod} 2k+1 }$
Also notice that after the first 2K+1 steps the position of the frog repeats.
This is because for even value (2m) $\mathbf{\frac{2m(2m+1)}{2} \equiv m(2m+1) \equiv \frac{(2m + 2k + 1) (2m+ 2k + 2)}{2} \equiv (2m+2k+1)(m+k+1) mod 2k+1 }$
Since $\mathbf{ (2m+2k+1)(m+k+1) \equiv (2m)(m+k+1) + (2k+1)(m+k+l) \equiv 2m^2 + 2mk + 2m }$
But $\mathbf{ 2m^2 + 2mk + 2m \equiv 2m^2 + m + 2mk + m \equiv m(2m+1) + m(2k+1) \equiv m(2m+1) mod 2k+1 }$
Similarly for odd values (2m+1)
$\mathbf{\frac{(2m+2)(2m+1)}{2} \equiv (m+1)(2m+1) \equiv \frac{(2m + 1 + 2k + 1) (2m + 1 + 2k + 2)}{2} \equiv (m+k+1)(2m+2k+3) mod 2k+1 }$
Since $\mathbf{(m+k+1)(2m+2k+3) \equiv (2m+2)(m+k+1) + (2k+1)(m+k+l) \equiv 2m^2 + 2mk + 2m + 2m + 2k +2 }$
But $\mathbf{ (2k+1)m + (2k+1) + 1 + 3m + 2m^2 \equiv (m+1)(2m+1) mod 2k+1 }$

As every 2k+1 values (modulo 2K+1) need to be distinct if the frog visits every leaf, it visits leaf number 2k+1 in the last two steps, hence some leaf must be missing.

Special Note

Extension: The frog may visit each leaf if and only if number of leaves is a power of 2.

This is a problem from CMI Entrance 2014 based on Map from a power set to n-set.

Problem: Map from a power set to n-set

(1) Let A = {1, ... , k} and B = {1, ... , n}. Find the number of maps from A to B .
(2) Define $latex \mathbf{ P_k } $ be the set of subsets of A. Let f be a map from $latex \mathbf{P_k to B }$ such that if $latex \mathbf{ U , V \in P_k }$ then $latex \mathbf{ f(U \cup V) }$= $latex \mathbf{\text{max} { f(U) , f(V) } }$ . Find the number of such functions. (For example if k = 3 and n =4 then answer is 100)

Discussion:

(1) For each member x of set A we have n choices for f(x) in B. Hence the number of functions is $latex mathbf{ n^k }$

(2) Claim (i): $latex \mathbf{ f(\phi) }$ is minimum for any such function f from $latex \mathbf{P_k to B }$ . This is because $latex \mathbf{ f(A_1) = f(\phi \cup A_1 ) = \text{max} { f(A_1), f(\phi) } }$ hence $latex \mathbf{f(A_1)}$ must be larger than $latex \mathbf{ f(\phi) }$ for any member $latex \mathbf{A_1}$ of $latex \mathbf{ P_k }$

Claim (ii) If we fix the values of the singleton sets then the entire function is fixed. That is if we fix the values of f({1}) , f({2}) , ... , f({k}). Since for any member $latex \mathbf{A_1}$ of $latex \mathbf{P_k , {A_1} }$ is union of several singleton sets. Hence it's value is the maximum of the functional values of those singleton sets. For example let $latex \mathbf{ A_1 = (1, 2) }$ then $latex \mathbf{f(A_1) = f({1}\cup{2}) = \text {max} {f({1}) , f({2}) } }$

Claim (iii) f({1}) , ... , f({k}) are individually independent of each other.

Now we fix $latex \mathbf{ f(\phi) = i}$. Since it is the smallest, the singleton sets map to i to n.
Hence if $latex \mathbf{ f(\phi) = 1}$ each of the singleton sets have n choices from 1 to n; hence there are $latex \mathbf{ n^k }$ functions. Similarly $latex \mathbf{ f(\phi) = 2}$ each of the singleton sets have n-1 choices from 2 to n; hence there are $latex \mathbf{ (n-1)^k }$ functions.

Thus the total number of functions = $latex \mathbf{ \sum_{i=1}^n i^k }$

This is a problem from Chennai Mathematical Institute, CMI Entrance 2014 based on area of a region. Try to solve it.

Problem: Area of a region

$latex \mathbf{ A= {(x, y), x^2 + y^2 \le 144 , \sin(2x+3y) le 0 } } $ . Find the area of A.

Discussion:

$latex \mathbf{ x^2 + y^2 \le 144 }$ is a disc of radius 12 with center (0, 0).

Now notice that $latex \mathbf{ \sin(2x + 3y) \le 0 \implies \sin ( 2(-x) + 3(-y) ) = \sin(-(2x+3y)) = -\sin(2x+3y) \ge 0 }$

Hence if a point (x, y) is in A then (-x, -y) is not in A.
Similarly if there is a point (x, y) not in A then we get a corresponding point (-x, -y) in A.

Therefore we have a bijection between points in A and not in A.

Thus area of A is exactly half the area of the disc = $latex \mathbf{ \frac{\pi (12)^2 }{2} = 72 \pi }$

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Limit of square roots - Video

Let $latex \mathbf{ x \in \mathbb{R} , x^{2014} - x^{2004} , x^{2009} - x^{2004} in \mathbb{Z} }$ . Then show that x is an integer. (Hint: First show that x is a rational number)

Discussion:

$latex \mathbf{ x^{2014} - x^{2004} - x^{2009} + x^{2004} = x^{2014} - x^{2009} = x^{2009}(x^{5} - 1 ) }$ is an integer
$latex \mathbf{x^{2009} - x^{2004} = x^{2004}(x^5 - 1) }$ is also an integer.
Hence ratio of those two are $latex \mathbf{ \frac{ x^{2009}(x^5 -1)}{x^{2004}(x^5 - 1)} = x^5 }$ is a rational.
Again $latex \mathbf{(x^5)^{400} = x^{2000}, x^5 - 1 }$ are rational. Also it is given that $latex \mathbf{ x^{2000}\cdot x^4 cdot (x^5 - 1) }$ is integer. Hence $latex \mathbf{ x^4 }$ is rational.
Therefore $latex \mathbf{ \frac{x^5}{x^4} =x }$ is rational. Since ratio of rationals is rational.

Suppose x = p/q, then $latex \mathbf{ \frac{p^{2014}}{q^{2014}} - \frac{p^{2004}}{q^{2004}} = k }$ where gcd(p, q) = 1 and k is an integer. $latex \mathbf{ p^{2014} - p^{2004}q^{10} = kq^{2014} }$
But this implies q divides p which means q = 1.

Hence x is an integer. (Proved)

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Limit of square roots – Video

This post contains problem from Chennai Mathematics Institute, CMI 2014 B.Sc. Entrance Paper. Try to solve them out.

Help us to add and rectify problems and solutions to this paper. We are collecting problems from student feed back.

4 Point Problems

1. Find the minimum value for x for which $latex \mathbf{ 50!/ (24)^n }$ is not an integer.
2. Find the slope of the line L which satisfies the following conditions: (i) L is tangent to the graph of $latex \mathbf{y = x^3}$ (ii) L passes through the point (0, 2000)
3. If $latex \mathbf{f(x) = (x-a)(x-b)^3(x-c)^5(x-d)^7}$ is a polynomial with 16 real roots such that 4 are distinct. How many real roots will its derivative have? And how many are distinct?
4. Find the area of the 12 sided regular polygon inscribed in the unit circle. Find the greatest integer lesser than or equal to the area of the polygon with 2014 sides.
5. Given that the sum of the lengths of the 12 sides of a cuboid is 60, find the range of possible volumes. If the total surface area is 56 square units, find the length of the longest diagonal and the volume, if possible.
6. There is a regular 100-gon. Choose any three vertices of it. What is the probability that this triangle is a right triangle
7. Consider $latex \mathbf{ e^{i \theta } , e^{2i \theta } , ... , e^{13 i \theta } }$. Let $latex \mathbf{ \sum_{t=1}^{13} e^{t i \theta} }$ . Then find the maximum and minimum of |A|.
8. How many triangles are possible in each of these cases?
   (a) $latex \mathbf{ \angle A = 95^o , \angle B = 55^{o} , \angle C = 30^{0} }$
   (b) a=95, b=55, c=30
   (c) $latex \mathbf{ \angle A = 95^o , \angle B = 55^{0} , c= 30 }$
   (d) $latex \mathbf{ b = 95 , b = 55 , \angle C = 30^{0} }$

3 Point Problems

1. If $latex \mathbf{f(x) = x^2e^x}$ when $latex \mathbf{x ge 0}$, and $latex \mathbf{f(x) = x e^{(-x)}}$ when x < -1, answer the following with true or false:
   f(x) is continuous at all points
   f(x) is differentiable at all points
   f(x) is one-one
   f(x) takes all possible real values

10 point for problem 1 and rest are 15 point Problems

1. $latex \mathbf{ A= {(x, y), x^2 + y^2 \le 144 , sin(2x+3y) \le 0 } } $ . Find the area of A.
   Solution
2. Let $latex \mathbf{ x \in \mathbb{R} , x^{2014} - x^{2004} , x^{2009} - x^{2004} \in \mathbb{Z} }$ . Then show that x is an integer. (Hint: First show that x is a rational number)
   Solution
3. Let A = {1, ... , k} and B = {1, ... , n}. Find the number of maps from A to B .
   Define $latex \mathbf{ P_k } $ be the set of subsets of A. Let f be a map from $latex \mathbf{P_k to B }$ such that if $latex \mathbf{ U , V \in P_k }$ then $latex \mathbf{ f(U \cup V) }$= $latex \mathbf{\text{max} { f(U) , f(V) } }$ . Find the number of such functions. (For example if k = 3 and n =4 then answer is 100).
   Solution
4. (1) Let S, T be two circles intersecting at X, Y. Let AB and CD be two chords of circle S such that AX and DY meet on the circumference of circle T at M and BY, CX meet on the circumference of T at N.
   (2) There is a triangle EFI in which EF|| GH.  GF , EH  are joined to meet at L.  Then a cevian is drawn from I to EF passing through L which cut GH at J(say) & EF at K(say).
   Then prove that j is midpoint of GH & K is midpoint of EF.
   Hint: use ceva’s theorem our assume GHFE as cyclic quadrilateral.
   (3) Using part (1) and (2) and an unmarked straight edge find the center of a given circle
5. Suppose f is a function continuous over [-1,1] and differentiable at 0. Also, define $latex \mathbf{ g(x)=\frac{f(x)-f(0)}{x} }$ for $latex \mathbf{ x \in [-1,0) \cup (0,1] }$.
   i) If g is to be continuous over [-1,1], what should the value of g(0) be?
   ii) Prove that the $latex \mathbf{ \lim_{r to 0+} \int_{-1}^{-r} \frac{f(x)}{x} dx + \int_r^1 \frac{f(x)}{x} dx }$ exists.
   iii) Give an example to show that (ii) need not hold when f is not differentiable at 0.
6. i) For a polynomial F(x), define the discrete derivative of F at x as F(x)-F(x-1). Determine the leading term of F’(x) in terms of leading term of F(x).
   ii) Define $latex \mathbf{p_0(x)=1, p_1(x)=x, p_2(x)=\frac{x(x-1)}{2!} }$ and so on. Show that any n degree polynomial can be written as $latex \mathbf{ \sum_{r=0}^n b_r \cdot p_r(x)}$ for real values of $latex \mathbf{b_r}$.
   iii) Let G(x) be a polynomial which achieves integer values for all integral x. Using (i) and (ii), show that if G is written in the summation form as mentioned in (ii), $latex \mathbf{b_i}$ is an integer for all values of i.

This post contains problem from Chennai Mathematics Institute, CMI BSc Math Entrance 2014 Model Problem set.

In each problem you have to fill in 4 blanks as directed. Points will be given based only on the filled answer, so you need not explain your answer. Each correct answer gets 1 point and having all 4 answers correct will get 1 extra point for a total of 5 points per problem.

But each wrong/illegible/unclear answer will get minus 1 point. Negative points from any problem will be counted in your total score, so it is better not to guess! If you are unsure about a part, you may leave it blank without any penalty. If you write something and then want it not to count, cross it out and clearly write no attempt" next to the relevant part.

1. Let f(x) be a function such that f(x+y) = f(x) + f(y). For each statement below write TRUE or FALSE
   1. If domain and codomain of f(x) be positive integers then the only  function that satisfies the above relation is f(x) = cx
   2. If domain and codomain of f(x) be integers then the only  function that satisfies the above relation is f(x) = cx
   3. If domain and codomain of f(x) be rational numbers then the only  function that satisfies the above relation is f(x) = cx
   4. If domain and codomain of f(x) be real numbers then the only  function that satisfies the above relation is f(x) = cx
2. Let f(x) be a function that is defined for rationals and irrationals separately in the interval [0, 1]. For each statement below write TRUE or FALSE
   1. f(x) = x if x is rational, f(x) = (1-x) otherwise; then f(x) is discontinuous at all points
   2. f(x) = 1/q if x = p/q (a rational) otherwise f(x) = 0 and f(0) = 0; then f(x) is continuous at x=0
   3. f(x) = x if x is irrational and f(x) = (1-x) otherwise; then f(x) is continuous at all rational points
   4. f(x) = 1/q if x = p/q (a rational) otherwise f(x) = 0 and f(0) = 0; then f(x) is continuous at all rational points
3. Suppose ABC is any triangle, P is any point inside it and RS be any line segment inside it (where R, S may or may not lie on the sides). For each statement below write TRUE or FALSE
   1. PX + PY +PZ (X, Y, Z are feet of perpendiculars dropped on sides of the triangle from P) is smaller than AT (where AT is altitude dropped from A)
   2. RS is always smaller than at least one side of the triangle.
   3. RS is always smaller than PX + PY +PZ
   4. If P is now taken to be any point on the plane of the triangle ABC then area of triangle XYZ ( (X, Y, Z are feet of perpendiculars dropped on sides of the triangle from P) is always larger than a constant c.
4. Suppose A is a set of n elements. B, C are subsets of this set. For each of the cases specified below, write an expression for the number of such elements. Do NOT try to simplify your answers.
   1. Ordered pair of subsets (A, B)
   2. Ordered pair of disjoint subsets (A, B)
   3. Ordered pair of disjoint non empty subsets (A, B)
   4. Unordered pair of disjoint non empty subsets {A, B}
5. Calculate the following limits (or write if they do not exist). Note that [x] denotes the greatest integer smaller than x
   1. $latex \mathbf{ \lim_{n to \infty} \sqrt { n^2 + 2n} - [\sqrt {n^2 + 2n} ] }$
   2. $latex \mathbf{ \lim_{x to 0 } \sin(x) \sin( \frac {1}{x}) }$
   3. $latex \mathbf{ \lim_{n to \infty} prod_{i=1}^n \frac{n^3 -1}{n^3 +1} }$
   4. $latex \mathbf{ \lim_{n to \infty} (1 - \frac{1}{n^2} )^n }$
6. Let $latex \mathbf {A_1,A_2,...., A_n}$ be a regular n gon inscribed in a unit circle and P be a point inside it. Find the values of the given expressions:
   1.  $latex \mathbf{ \overline{A_1A_2}\cdot\overline{A_1A_3}\cdots\overline{A_1A_n} }$
   2. $latex \mathbf{ \sin\frac{\pi}{n}\sin\frac{2\pi}{n}\cdots \sin \frac{(n-1)\pi}{n}}$
   3. $latex \mathbf{ \sin\frac{\pi}{2n}\sin\frac{3\pi}{2n}\cdots \sin \frac{(2n-1)\pi}{2n}}$
   4. $latex \mathbf{ \overline{PB_1}+\overline{PB_2}+\cdots+\overline{PB_n}}$ where $latex \mathbf{B_1,B_2,...., B_n}$ are feet of perpendiculars of from P on the n sides
7. Let $latex \mathbf{ x^2 + bx + c , b, c \in \mathbb{Z}}$ be a quadratic equation. Then find the values of the given expressions
   1. Probability of getting a rational root between (0,1) given b, c are from the set {-100, -99, ... , 99, 100}
   2. Probability of getting a rational root between [0,1] given b, c are from the set {-100, -99, ... , 99, 100}
   3. Probability of getting a rational root between (0, 1) given b, c are any integer
   4. Probability of getting a rational root between (1, 10) given b, c are any integer

(For solutions to this problem set please click on 'follow' button in the right bottom corner of this blog. You will get a confirmation e mail in which you have to click a confirmation link. We will send solutions to the followers)

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Limit of square roots – Video

This is a problem from ISI BMath 2014 Subjective Solution based on Mulitple roots or Real root. Try to solve this problem.

Problem: Multiple roots or real root 

Let $latex \mathbf { y = x^4 + ax^3 + bx^2 + cx +d , a,b,c,d,e \in \mathbb{R}}$. it is given that the functions cuts the x axis at least 3 distinct points. Then show that it either cuts the x axis at 4 distinct point or 3 distinct point and at any one of these three points we have a maxima or minima.

Discussion:

Since all the coefficients are real, complex roots occur as conjugates. Hence the fourth root (it is a four degree polynomial hence has a fourth root), must be real (if it is complex then we must have at least one more complex root, but all the other three roots are given to be real).

Let l, m, n be the three roots. Then the fourth root is either distinct from l, m, n or it is equivalent to exactly one of them say 'n'.

If it is equal to n then we may rewrite the polynomial as $latex \mathbf { y = (x-n)^2 (x-l)(x-m) }$
We take first and second derivative of the y with respect to x.
$latex \mathbf { y' = (x-n)^2 (x-l) + (x-n)^2 (x-m) + 2(x-n) (x-l)(x-m) }$. At x=n the first derivative vanishes. Hence x=n is a critical point. We want to show that this is also a point of maxima or minima. For that we must show that the second derivative at x=n is positive or negative (not zero).
$latex \mathbf { y'' = 2(x-n)(x-l) + (x-n)^2 + (x-n)^2+ 2(x-n) (x-m) + 2(x-n) (x-m) + 2(x-l)(x-m) + 2(x-n) (x-l) }$
Hence at x=n $latex \mathbf { y'' = 2(n-l)(n-m) }$ . Since n is distinct from m or l, hence the second derivative is either positive or negative and not zero. Hence we have maxima or minima at that point.
Proved.

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Limit of square roots – Video

Let PQR be a triangle. Take a point A on or inside the triangle. Let f(x, y) = ax + by + c. Show that $latex \mathbf { f(A) \le \max { f(P), f(Q) , f(R)} }$

Discussion:

Basic idea is this: First we take A on a side, say PQ. We show $latex \mathbf { f(A) \le \max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }$
Next we take A in the interior. Lets join RA and produce it to meet PQ at T. Using the previous argument we show $latex \mathbf { f(A) \le \max { f(T), f(R) } \text{but} f(T) \le max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }$

Hence it is sufficient to show for a point A on the line PQ (and the rest will follow):

Let $latex \mathbf { A = (x, y), P = (x_p , y_p) , Q = (x_q, y_q) }$ Since A is on PQ, it is possible to write $latex \mathbf { x = \frac{m x_p + n x_q }{ m+n} , y = \frac{ m y_p + n y_q} {m+n}, m,n in \mathbb{R}^{+} }$

Suppose f(A) is larger than both f(P) and f(Q). Then f(A) - f(P) and f(A) - f(Q) are both positive.

Let $latex \mathbf { \lambda_1 = \frac{m}{m+n} , \lambda_2 = \frac{n}{m+n} implies \lambda_1 + \lambda_2 = 1 }$

$latex \mathbf { f(A) - f(P) = ax+by+c - ax_p - by_p -c = a(x-x_p) + b(y - y_p) = \newline a(\lambda_1 x_p + \lambda_2 x_q - x_p) + b (\lambda_1 y_p + \lambda_2 y_q - y_p) }$
$latex \mathbf {\implies f(A) - f(P) = a(\lambda_2 x_q - (1-\lambda_1) x_p) + b(\lambda_2 y_q - (1-\lambda_1) y_p) = a \lambda_2 (x_q - x_p) + b \lambda_ 2 (y_q - y_p) }$
Similarly we can show $latex \mathbf { f(A) - f(Q) = a \lambda_1 (x_p - x_q) + b \lambda_1 (y_p - y_q) }$
$latex \mathbf { f(A) - f(Q) = \frac{-\lambda_2}{\lambda_1} (f(A) - f(P) ) }$
Hence f(A) - f(Q) and f(A) - f(P) are opposite signs. The conclusion follows.

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ISI CMI Entrance Program

Limit of square roots – Video