ISL 2018 C7

by randomusername, Jan 7, 2024, 8:31 PM

Problem. There are $2018$ circles on the plane, configured such that any pair meets twice. We go along each circle and record a label $+$ or $-$ at each intersection point with another circle. Thus, every point where two circles meet is labelled with $++$ , $--$ , or $\pm$ . Show that if one of the circles has at least $2061$ labels of $\pm$ , then at least one region determined by the circles has only $\pm$ labels along its perimeter. $\square$

First of all, observe that each circle arc proceeds from $+$ to $-$ in some order. It follows that any closed path along the circle arcs has an equal number of $+$ and $-$ labels.

A natural question to ask is whether the desired region occurs, provided that all the points along a circle are labelled $\pm$ . Observe that in this case, the labels of all the points are determined, up to interchanging $+$ and $-$ everywhere. By induction, I was able to prove that actually, all the points will have $\pm$ .

I began by drawing an arrow along each circle arc from $+$ to $-$ . Then, I noticed that in a region with only $\pm$ labels along its perimeter, the arrows along the arcs that form its perimeter agree in their orientation, all clockwise or all counterclockwise. Hence, if a new circle is introduced where all intersections were $\pm$ , the labels of each point between its two meeting points must change, say, in the exterior of the novel circle. Regions that it doesn't intersect continue to be all $\pm$ . Regions that the new circle does cross are split into a clockwise and a counterclockwise region, interior and exterior to it. This naturally determines arrows on each arc of the new circle, oriented as antidirected, so that $+$ and $-$ labels may be written alternately along it.

Now that we know of a standard labelling of all the circles where each meeting point has $+$ and $-$ , we consider that changing the labels on $k$ of the circles other than the given one with $2061$ labels $\pm$ causes $2k$ points on the given circle to be $++$ or $--$ . It follows that $2k\le 2\cdot 2018-2061=1975$ , so $k\le 987$ .

Which points have $++$ or $--$ after $k$ circles are changed up? Exactly those whose pair of circles has been reversed only once. This means that $2k(2018-k)$ points are such. Furthermore, each region has an equal number of $++$ and $--$ points, meaning $0$ or more than one. Thus, as each such point lies in $4$ of the regions, at most $4k(2018-k)$ regions have $++$ or/and $--$ labels.

If the number of regions determined by $1+n$ pairwise intersecting circles is $a_n$ , then $a_n=2n+a_{n-1}$ , because removing a circle causes $2n$ arcs to be removed that all bound disjoint pairs of regions. As $a_0=2$ and $n+(n-1)+\dots+1=\frac{n(n+1)}{2}$ is well-known, we see that $a_n=n^2+n+2$ .

The number of regions with only $\pm$ is at least $(2017^2+2017+2)-4\cdot 987\cdot 1031=-80$ . This is inconclusive, although I could obtain the positive bound of $100$ unless $k=987$ . If only one could verify that more than $80$ regions have two $++$ points, the answer would be clear. This occurs when the reversed circles do not form contiguous blocks.

Graphic degree sequences

by randomusername, Dec 4, 2023, 8:15 PM

In my past life as a high school Olympiad student, it seemed important that I keep track of my progress, partly as a way to hold on to good ideas while trying to make an impact, but also as part of looking for new approaches to learning and practicing mathematics. At university, however, the state-level elite and international events that motivated me no longer existed, and I was entrusted with continuing a tradition of mathematical exploration much older than the fast-growing and youthful scene of the IMO. Arriving with unusually mature experience in self-improvement and teaching, I knew that any thought pattern that gets repeated the same way for over a year is just exasperating, and so a curriculum that hardly ever changes and cannot adapt each year to the passion of students bored me dearly.

Currently, as a PhD student in combinatorics with a view towards graph theory, it has occurred to me that even within combinatorics, basic definitions and systems of knowledge have to be reintroduced with each new presentation. Therefore, I do not find it surprising that the following question has not been discussed in any academic environment I have yet encountered. Let $G$ be a graph on $n$ vertices, and let $d_1\ge d_2\ge \dots \ge d_n$ be a sequence of non-negative integers. For which values of the $d_i$ does there exist a graph $G$ whose vertices $v_1,v_2,\dots,v_n$ have degree $d_1,d_2,\dots,d_n$ , respectively? There are two known answers to this: an explicit one by Erdős and Gallai 1960, and an implicit one by Havel 1955.

Fact. If the maximal degree value $\Delta$ is at most $n-1$ , the degree sequence may be written as $\Delta\ge D_1\ge \dots\ge D_\Delta\ge d_1\ge \dots \ge d_m$ . Consider the sequence $D_1-1,D_2-1,\dots,D_\Delta-1,d_1,\dots,d_m$ rearranged in non-increasing order. Clearly, if these are the degrees of a graph on $n-1$ vertices, then a new vertex joined to the $\Delta$ former vertices yields a graph on $n$ vertices of the specific degrees. The converse is true, as well!

Proof. If the vertex $v$ of degree $\Delta$ is joined to vertices of degrees $D_1,\dots,D_\Delta$ , then removing it gives rise to the graph required. If it is joined to a vertex $t$ of degree $d_j$ instead of a vertex $u$ of degree $D_i$ , consider that as $d_j\le D_i$ , there must be a vertex $w$ such that $wt$ is not an edge but $wu$ is. Then, swapping the edges $vt$ and $wu$ for the edges $vu$ and $wt$ changes no degree, yet brings us closer to the case when the neighbours of $v$ have the largest possible degrees, which is resolved. Thus, we have an inductive description of so-called graphic sequences $d_1,\dots,d_n$ . $\blacksquare$

Onwards, the $l$ largest degree vertices have degree sum counting internal edges twice and outgoing edges once. Therefore, as there are at most $d_i$ and at most $l$ edges ending at vertex $v_i$ for $i>l$ , we obtain
$d_1+d_2+\dots+d_l\le 2\binom{l}{2}+\sum_{i>l}\min\{d_i,l\}.$ This inequality for all possible values of $l$ is not only necessary but also sufficient for the existence of a graph $G$ admitting degrees $d_1,\dots,d_n$ with even sum. To verify this, recall the fact above, and show that if the length $n$ sequence obeys these inequalities, then so does the length $n-1$ version. Treating the case $l\le \Delta$ distinctly in view of how $\Delta\ge D_1$ , $D_1\ge D_2$ , and so on, this is not so hard to check.

Problem N.1 from Kömal

by randomusername, Oct 11, 2018, 11:21 AM

I'd like to transcribe a solution, along with the path that led to it. Do you think this writing technique improves the quality of the solution?

Problem: The expression $x^{10}+ \, ? \, x^9 + \, ? \,x^8+\ldots+ \, ? \, x+1$ is written on the board. Player $A$ , then $B$ , then $A$ , etc. makes a move: this means writing a real number in place of one of the question marks. Does the polynomial at the end have a real root?: a) if yes, $A$ wins, b) if yes, $B$ wins. Who has a winning strategy.

Solution stream of consciousness: Clearly $\to \infty$ as $x\to \infty$ , hence to get root suffices to make $f(1)\le 0$ , for example (IVT). $A$ goes last, so if $A$ wants to make $f(1)=0$ she can do it, so in a), $A$ wins.

For b)? Maybe $A$ can prevent $B$ from winning, but $B$ could win so easily if only he could make some value of $f$ non-positive. Making e.g. $f(1)\le 0$ means controlling nine coefficients: that's beyond $B$ 's power. But idea: $\frac12(f(1)+f(-1))=1+a_8+a_6+a_4+a_2+1$ .

Can $B$ make $f(1)+f(-1)$ negative? No: $A$ can arrange to choose the last remaining free value of $a_8,a_6,a_4,a_2$ (though $A$ must do this, else $B$ wins), e.g. by always taking one of them after $B$ takes one of them. Still, this is a bit promising, and there's a lot to milk in this idea.

Have $(x^{10}+Ax^k+1)$ initially. Then $B$ , $A$ alternate: $+cx^\ell$ . Our idea was looking at $g(x)=f(ax)+a^\ell f(-x)$ instead of $f(x)$ (for $\ell$ odd), which knocks out $\ell$ -th powers. Observation: have $f\le 0$ if have $g\le 0$ . Using steps like this, we can get to some $f^*$ from $f$ which has all the even powers, of course, but only one odd power, say: if once the even power terms are fixed $B$ can choose that odd power, $B$ can guarantee $f^*\le 0$ somewhere $\leadsto$ $f\le 0$ , win.

So $B$ does have a winning strategy: I'll write it out. First, he takes even terms until none remain. If he gets to take all the even terms, $f(1)+f(-1)\le 0$ doable, win. Else all even taken and $B$ takes odd term on his last move: the $m$ -th term. Consider the map $\phi_\ell:f(x)\mapsto f(2x)+2^\ell f(-x)=\sum (2^i-2^\ell)a_ix^i$ on polynomials. Do $\phi_\ell$ for all $\ell\in \{1,3,5,7,9\}\setminus \{m\}$ , call the result $f^*$ , which just looks like $\sum \lambda_i a_i x^i$ for known constants $\lambda_i$ which are zero iff $i$ is one of the chosen $\ell$ . So on his last move $B$ can make $f^*(1)\le 0$ so done.

This post has been edited 1 time. Last edited by randomusername, Oct 11, 2018, 11:22 AM

"Trivial" problems, bare hands

by randomusername, Aug 1, 2017, 4:06 PM

1. Consider a square and a circle. Prove that there exists a continuous transformation between the two of them.

2. We say a curve from $A$ to $B$ is continuous if it is defined by a continuous function $[0;1]\to \mathbb{R}^n$ . (Continuous means $f(x)$ and $f(y)$ are close enough if $x,y$ are close enough. We can replace $\mathbb{R}^n$ by any metric space.) We say that objects $U$ and $V$ in $\mathbb{R}^n$ are homeomorphic if there exists a homeomorphism between them, i.e. a bijection between their points which is continuous in both directions. Prove that any two continuous curves are homeomorphic.

3. [I'm don't know how to do this] Prove that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic for $m\neq n$ . Are $RP_2$ (projective plane) and $\mathbb{R}^2$ homeomorphic? Is the surface of a torus and the surface of a sphere homeomorphic?

4. [This also might be a bit hard] Consider the unit disk $D$ : $x^2+y^2\le 1$ , and the unit circle $K$ : $x^2+y^2=1$ . Let $a,b,c,d$ be four points on $K$ forming a convex quadrilateral. Suppose that $\gamma_1$ and $\gamma_2$ are continuous curves from $a$ to $c$ and from $b$ to $d$ , respectively, and that they lie within $D$ (and only meet $K$ at $a,c$ or at $b,d$ ). Prove that $\gamma_1$ and $\gamma_2$ intersect each other.

5. [This is definitely very hard; Jordan's theorem] Take a continuous non-self-intersecting curve $\gamma$ from $A$ to $A$ in $\mathbb{R}^2$ . Prove that $\mathbb{R}^2\setminus \gamma$ contains exactly two connected components (i.e. inside which any two points can be connected by a continuous curve), one of which is bounded.

6. Show that the Möbius strip is non-orientable.

This post has been edited 2 times. Last edited by randomusername, May 19, 2018, 9:17 AM

More intrinsic connection between ord_p and quadr-recipr

by randomusername, May 28, 2017, 12:46 PM

There are two problems which come to mind at the moment: PEN A17, RMM 2013 P1, and I've just thought of $n|2^{n-1}+1\implies n=1$ as well.

Solution to the PEN problem:

The Binomial theorem begets $3m|m^n+3^n+1$ from $3m|(m+3)^n+1$ . Next, note that for even $n$ , $3$ cannot divide the sum as perfect squares are $0,1\pmod 3$ , $\boxed{n\text{ is odd}}$ . Hence, $3m|m^n-m=(m^{n-1}-1)m$ , so we get $3m|m+3^n+1$ , which in turn is equivalent to [ $3|m+1$ and $m|3^n+1$ ].

Let $m=2^\alpha m'$ with $m'$ odd, then we want to prove that $v_2(3m)=v_2(x)$ where
$x=(m+3)^n+1=(2^\alpha m'+3)^n+1=2^{2\alpha}y+2^\alpha(3nm')+(3^n+1).$ For odd $n$ , $v_2(3^n+1)=2\ge \alpha$ , and interestingly, we're done for $\alpha=0$ and $\alpha=1$ , while not for $\alpha=2$ . There's more deep stuff involved!

We cannot have $\alpha=2$ , i.e. every divisor of $\frac{3^n+1}{4}$ is of the form $3k+1$ . This is because for any odd prime $p|3^n+1$ , $\left(\frac{p}{3}\right)=1$ . Indeed, define $d$ as follows

$d$ is the minimal odd number with $p|3^d+1$ , OR
$d=\frac{\text{ord}_p(3)}{2}$

(either definition works; it's easy to check they're equivalent). Then $d|p-1$ , by Fermat's Little Theorem. In fact, $d$ is odd, so $d|\frac{p-1}{2}$ . The interesting idea is that if $\frac{p-1}{2}$ is odd, then $3^{\frac{p-1}{2}}\equiv -1\pmod{p}$ , but if it's even, then $3^{\frac{p-1}{2}}\equiv +1\pmod{p}$ . But by $\left(\frac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\pmod{p}$ , this precisely means $\left(\frac{3}{p}\right)=(-1)^{p-1}$ . So by Quadratic Reciprocity, $\left(\frac{p}{3}\right)=(-1)^{\frac{3-1}{2}}=-1$ . $\blacksquare$

This post has been edited 1 time. Last edited by randomusername, May 28, 2017, 12:47 PM

Some set theory

by randomusername, May 25, 2017, 8:57 PM

OK I need to record this before I forget.

Problem. Infinitely people in a room are wearing red or blue hats such that they can see everyone else's hat but not their own. Simultaneously they all guess the color of their hat. Prove that it's possible for the participants to devise a strategy so that at most finitely many people get their color wrong.

Solution. Well, there is a set of possible strategies, call it $\Sigma$ . What I mean by this is that $\Sigma$ contains all possible combinations of guesses our set $S$ of people can make, i.e. $\Sigma$ is in bijection with its power set $2^S$ . The current hat assignment $\kappa$ determines a subset $\Sigma_\kappa$ of $\Sigma$ containing the set of strategies for which only finitely many people get their color wrong.

It is a huge insight to define these, but that's just rephrasing the problem, getting the most general form of what we're dealing with out there (a good thing to do in general, mind you). The key idea is that for any $\sigma\in \Sigma$ , each and every player can decide whether applying $\sigma$ will lead to infinitely many mistakes or not. If you aren't smiling yet, that's because you need to be told that each player thus knows $\Sigma_\kappa$ . By the Axiom of Choice, the players can agree beforehand on a fixed representative $\sigma_\kappa\in \Sigma_\kappa$ ; thus, the players will apply $\sigma_\kappa$ , ensuring only finitely many mistakes, as needed. $\blacksquare$

New problem solution 2

by randomusername, May 24, 2017, 12:03 PM

Problem. Consider the proposition
$F=f(x_1,x_2,x_3)+f(x_2,x_3,x_4)+\dots+f(x_{n-1},x_n,x_1)+f(x_n,x_1,x_2)=x_1+x_2+\dots+x_n$ for any real numbers $x_1,\dots,x_n$ . Prove that if this proposition holds for $n=5$ , then it holds for all $n\ge 3$ .

Solution. Replace the original $f$ by $f(x,y,z)-x$ , then the RHS is replaced by $0$ .

Lemma: using the new $f$ , the quantity $f(a,b,x)+f(b,x,d)+f(x,d,e)$ is independent of the choice of $x$ . Proof: this is a trivial corollary of the proposition for $n=5$ with $(x_1,\dots,x_5)=(a,b,x,d,e)$ .

Corollary: the value of $F(x_1,x_2,\dots,x_n)$ is independent of the choice of any of the variables. Since $5f(0,0,0)=0$ , plugging in all zeros gives the desired result $F=0$ . $\blacksquare$

New problem solution

by randomusername, May 15, 2017, 9:17 AM

Problem: Given the constraints
$\begin{align*} x_1\ge x_2\ge \dots \ge x_n&>0 \\ y_1\ge y_2\ge \dots\ge y_n&>0 \\ x_1x_2\dots x_k &\ge y_1y_2\dots y_k \quad \forall k=1,2,\dots,n \end{align*}$ prove that $\sum x_i\ge \sum y_i$ .

Solution:

We induct on $n$ , from the base case $n=1$ . Start by increasing $y_1$ until $x_1x_2\dots x_k=y_1y_2\dots y_k$ for some $k$ . If this occurs for $1\le k<n$ , then by the induction hypothesis,
$\begin{align*} x_1+\dots+x_k&\ge y_1+\dots+ y_k \\ x_{k+1}+\dots+x_n&\ge y_{k+1}+\dots+y_n, \end{align*}$ done. Else we have $x_1x_2\dots x_n=y_1y_2\dots y_n$ , and our constraints can be rearranged as
$\begin{align*} x_1\ge x_2\ge \dots \ge x_n&>0 \\ y_1\ge y_2\ge \dots\ge y_n&>0 \\ y_n&\ge x_n \\ y_ny_{n-1}&\ge x_nx_{n-1} \\ \dots \\ y_ny_{n-1}\dots y_2&\ge x_nx_{n-1}\dots x_2. \end{align*}$ We will create a sequence of $X=(x_1,x_2,\dots,x_n)$ satisfying the constraints with decreasing sums $\sum x_i$ , until reaching a sequence for which $x_n=y_n$ . Then we will be done, as the claim holds for $x_n=y_n$ , by induction. In particular, our sequences will be $X_1,X_2,\dots$ where the last $j$ terms of $X_j$ are the same: $x_n=x_{n-1}=\dots=x_{n-j+1}$ .

Take a sequence $X_j$ where $x_n=x_{n-1}=\dots=x_{n-j+1}=w$ . We will start increasing $w$ and decreasing $x_{n-j}$ towards each other in a way that none of the constraints will be violated. The relevant constraints are $w\le x_{n-j}(\le x_{n-j-1})$ , $w\le y_n$ and $y_ny_{n-1}\dots y_{n-j}\ge w^jx_{n-j}$ (the rest are automatically satisfied because $y_n\le y_{n-1}\le \dots$ ): if these are satisfied, the new sequence also satisfies the system of constraints.

Move $w\le x_{n-j}$ towards each other such that $P:=w^jx_{n-j}$ remains fixed. I claim this decreases $\sum x_i$ , i.e. that this decreases $jw+x_{n-j}$ . Trivially so: the sum is $jw+\frac{P}{w^j}$ , which as a function in $w$ has derivative $j-j\frac{P}{w^{j+1}}<0$ as long as $w<x_{n-j}$ , i.e. $w^{j+1}<P$ .

While moving $w$ and $x_{n-j}$ toward each other, we either violate the condition $w\le y_n$ at some point, in which case we'll get a sequence with $x_n=y_n$ , or we don't, and we end up with $w=x_{n-j}$ , thus creating $X_{j+1}$ from $X_j$ .

We're done: either we get to $x_n=y_n$ and we're done by induction, or we get to $X_n$ , in which case $y_n>x_n$ would imply the contradiction $y_1y_2\dots y_n>x_1x_2\dots x_n=x_n^n$ , so $x_n=y_n$ arises again.

More cool stuff for storage

by randomusername, Dec 2, 2016, 1:23 PM

First of all, note that $\log(xy)=\log x+\log y$ because
$\int_1^{xy} \frac{dt}{t}=\int_1^x \frac{dt}{t}+\int_1^y\frac{dt}{t}$ follows after subtracting $\int_1^x\frac{dt}{t}$ from both sides and applying a substitution.

Second of all, note that
$\sum_{n=1}^\infty\left(e-\left(1+\frac 1n\right)^n\right)$ is divergent, because
$\left(1+\frac1n\right)^n=1+\binom n1\frac 1n+\sum_{k=2}^n \left(1-\frac1n\right)\dots\left(1-\frac{k-1}{n}\right)\cdot \frac1{k!}\le 1+1+\left(1-\frac1n\right)\sum_{k=2}^n\frac1{k!},$ hence
$e-\left(1+\frac 1n\right)^n\ge \frac1n(e-2),$ and so the series diverges like the harmonic series.

When does \sum 1/f(n) converge?

by randomusername, Nov 22, 2016, 3:19 PM

[If you don't know analysis, this stuff will be awesome for you, because you'll get to Google enough things I talk about to have some idea about how it works. So stay with me.]

I have known for ages that $\sum_{n=1}^\infty \frac 1n$ , the harmonic series, diverges, and that $\sum_{n=2}^\infty \frac1{n\log n}$ also diverges (fun fact: this is more or less the reciprocal sum of the primes by the PNT).

BTW elementary proof that $\sum \frac1{n\log n}$ diverges: WLOG convert to $\log_2$ and notice that
$\sum_{n=2^k}^{2^{k+1}-1} \frac1{n\log_2 n}\ge 2^k\cdot \frac1{2^{k+1}\log_2 2^{k+1}}=\frac12 \frac1{k+1},$ and summing this over $k$ gives something which diverges.

We also have
$\sum_{n\le x}\frac1{n}\sim \log x,\qquad \sum_{n\le x}\frac1{n\log n}\sim \log\log x$ which basically comes from integrals. (Here, $f\sim g$ means that $\lim_{x\to\infty}(f/g)=1$ .) Then there must be a series which diverges like $\log\log\log x$ , too! Right? Yes, in fact, integrals tell us
$\underbrace{\log\log\dots\log}_{\text{k times}}x\sim \sum_{n\le x}(n\log n\log\log n\cdot\ldots\cdot \underbrace{\log\log\dots\log}_{\text{(k-1) times}} n)^{-1}.$
Nice. But what about the other side, i.e. $\sum f(n)^{-1}$ for $f(x)=x^s$ , for $f(x)=x(\log x)^s$ etc. with $s>1$ ? Again, integrals trivialize this, and in fact show that just like the Riemann $\zeta$ has a single pole at $1$ with residue $1$ , or expressed in the world of real analysis,
$\zeta(s):=\sum_{n=1}^\infty n^{-s}\sim \frac1{s-1}\qquad \text{as }s\to 1^+,$ the same property is true of $s\mapsto \sum_n (n(\log n)^s)^{-1}$ as $s\to 1^+$ , i.e. that it is $\sim \frac1{s-1}$ .

Of course, again, there is elementary treatment which may lead us to conjecture this, namely:
$\sum_{n=2^k}^{2^{k+1}-1}(n(\log n)^s)^{-1}\le 2^k\cdot (2^k\cdot k^s)^{-1}=k^{-s},$ which, when summed over $k$ , gives something like $\zeta(s)$ . Surprisingly enough,
$\sum_{n=2^k}^{2^{k+1}-1}(n(\log n)^s)^{-1}\ge 2^k\cdot (2^{k+1}\cdot (k+1)^s)^{-1}=\frac 12 (k+1)^{-s},$ so we get something like
$\frac12(\zeta(s)-1)\le \sum_{n=2}^\infty \frac1{n(\log n)^s}\le \zeta(s),$ which at first is extremely surprising, but at a second glance, well, it still is! The resolve here is that $\zeta(s)$ starts out with somewhat smaller summands (for example, $3^{1.1}\approx 3.348$ but $3(\log3)^{1.1}\approx 3.327$ ), and $\zeta(s)$ has enough small summands to overtake $\sum \frac1{n(\log n)^s}$ , however, these small summands don't amount to a substantial enough difference.

Our result therefore is that the border between $\sum \frac 1{f(n)}$ converging and not is somewhere at
$f(x)=x\log x\log\log x\cdot \dots.$ [Of course, the composite of logs only makes sense for large enough $x$ , so an infinite product like this would die.]

Challenge problem: decide whether $\sum_n \frac1{f(n)}$ converges or diverges, if $l(t):=\log((t-1)+e)$ and $l_2(t)=l(l(t))$ etc., and we define $f$ as the infinite product
$f(x)=x\prod_{k=1}^\infty l_k(x).$

This post has been edited 2 times. Last edited by randomusername, Feb 14, 2017, 4:31 PM

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