# Tag Info

0

For $n = 1$ any $p$ is divisible by $1$ so $(1, p)$ is a solution for any prime $p$. Let $n > 1$. Since for $p = 2$ the only solution is $(2, 2)$, assume $p$ odd. Assume, also, $p \not | n$. Denote $n = pk + r$ for $0 \leqslant k \leqslant 1$ and $0 < r < p$. We have $$n^{p - 1} = (pk + r)^{p - 1} \equiv r^{p - 1} \pmod {p}.$$ Since $(r, p) = 1$, ...

1

The trick is the following. Let $ON \cap AB = P'$. It suffices to show that $P'$ lie on the circumcircle of $BOM$. After that, we will have $P'=P$, and then we will have $P$ lie on $ON$. Let $\angle BAM = a$. We begin angle chasing. We have $$\angle BP'O + \angle BMO = (a+ \angle ANO) + (\angle BMA - \angle OMA) = a+90+(180- \angle B - a) -(90 - \angle ... 0 For even n=2k, take a shape formed by a column of k+1 squares next to one of k-1 squares; for odd n=2k+1, k+1 next to k. Assemble two copies to form a rectangle of size 2\times n. Assemble n\times2 such pairs to form a square of size 2n\times2n. Then arrange n squares to form the original shape, for a total of 4n^3 copies. 4 You can always add up to 100 if 8 are removed. To add up to 100 you can use 4 pairs of numbers that add to 25. There are 12 such pairs. If you remove 8 numbers you can only eliminate at most 8 such pairs. Thus at least 4 such pairs will always be left. Thus 4 pairs will always add to 100. 2 For the first this is the solution 0 [I found this solution collaboratively with someone else offline.] \def\nn{\mathbb{N}} \def\rr{\mathbb{R}} Let T(n) = ( \text{The theorem is true for any length-n sequence from \rr} ), for any n \in \nn. If T(n) is false for some n \in \nn: Let m \in \nn be the minimum such that T(m) is false [by well-ordering]. Let ... 3 Here's another approach. First write out what h(h(x)) is$$h(h(x))=h\left(\frac{ax+b}{bx+c}\right)=\frac{a\left(\tfrac{ax+b}{bx+c}\right)+b}{b\left(\frac{ax+b}{bx+c}\right)+c}=\frac{a(ax+b)+b(bx+c)}{b(ax+b)+c(bx+c)}.$$Next set this equal to x and clear out the numerator to get$$a(ax+b)+b(bx+c)=x(b(ax+b)+c(bx+c))=b(a+c)x^2+(b^2+c^2)x,$$and moving ... 1 HINT:$$h(h(x))={1 \left( {\frac {a \left( ax+b \right) }{bx+c}}+b \right) \left( { \frac {b \left( ax+b \right) }{bx+c}}+c \right) ^{-1}} $$1 Observe that k f (2k) < f (4k) \leqslant 2k f (2k) for k \geqslant 4 a power of 2. We use induction on n to prove the inequality. If n = 3, then it is obvious that$$2^{3^2/4} < f(8) = 10 < 2^{3^2/2}.$$Suppose the inequality holds for n = m, we now prove it for n = m + 1. Since f (2^m) < 2^{m^2/2} and using f (4k) \leqslant 2k ... 1 Starting from one of such points on the plane draw a rectangle whose sides are m and n. Let R_1 be the total area of black squares and R_2 be that of white squares, in the rectangle. Since by drawing a diagonal we can divide the rectangle into two identical right-angled triangles as described, we have R_1 = 2S_1 and R_2 = 2S_2. Then, |R_1 - R_2| ... 0 Let x_i - x_{i - 1} = p occur k times and x_i - x_{i - 1} = -q occur n - k times. Since$$0 = x_n - x_0 = \sum_{i = 1}^{n} (x_i - x_{i - 1}) = kp - (n - k)q,$$we have k (p + q) = nq. Since p + q < n, we have nq = k (p + q) < nk, which means that q < k. Since -q > -k, it is obvious that kp - k (n - k) < 0, which in turn ... 0 For p = 2, we can choose q = 3 or q = 7. Assume p is odd. By Fermat's Little Theorem, we have n^p \equiv n \pmod {p}. Assume contrary: let n^p - p \equiv 0 \pmod {q} for all primes q. Since (p, q) = 1 and by Chinese Remainder Theorem, we have n^p \equiv p^2 \pmod {pq}. But since n^p \equiv nq \pmod {pq}, we deduce that p^2 \equiv nq ... 3 We have three cases: a = b, a > b, a < b. If a = b, we have a ^ {a ^ 2} = a ^ a, which gives a ^ 2 = a for a \geqslant 1, that is, a = 1 so b = 1: (a, b) = (1, 1). Observe now the lemma below: Lemma 1. a and b are made up of the same primes. Proof. Let$$a = p_1 ^ {r_1} p_2 ^ {r_2} \cdots p_n ^ {r_n} = \prod_{i = 1}^{n} p_i ^ ...

0

Hint: RELATIVE VELOCITY .See the $total time = sum of both trains(distance)/speed$ . Therefore $y=\frac{w+x}{r_1+r_2}$. So $w=y(r_1+r_2)-x$. They are going in opposite direction so $V_{AB}=V_{A}+V_{B}$.$w$ for more info you can go through the physics concept of relative velocity. is the length of trainB. Hope its clear.

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Let's say that you guess $a=c$, e.g. by observing equations (2) and (3) as you have them above. Then from equations (5) and (6) you may see that $x=y$ works. For $a=c$ and $x=y$ equation (2) becomes equal to equation (3) and (5) equal to (6), and so you get the system of equations: \begin{align*}2a^2+b^2+2x^2&=1\\ ax^2+bx^2+a\lambda&=0\\ ...

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An easier variant of RMM 2009 P4. I will prove a stronger statement - the RMM problem itself. It generalizes the two integers $x,y$ to a set of positive integers. This solution is similar to the solution presented in the link. First, for convenience, we define $f(x)=\arctan x$. Also, for a set $X$, we define $g(X)=\sum_{x \in X} f(\frac{1}{x})$. Now, ...

1

Since $\gcd{(3n + 2, 3)} = 1$ we conclude that $b^2=3n+2$ and $a^2−ab=3$. There is another possibility: $b^2=-(3n+2)$ and $a^2−ab=-3$. And these conditions can be satisfied, for example, with $a=1$, $b=4$ and $n=-6$. So this proof is incorrect.

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Most probably you might have miscomputed $Pr(T^2)$ Since we are computing probability (a ratio), we can as well use the $\%$ figures directly, if you aren't getting it, try

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$P(D)=0.0001$ $P(T_n\mid D) = .99, P(T_n\mid D')=0.05$ Find $P(D\mid T_1, T_2) = \dfrac{\overline{\underline{|\qquad\qquad|}}}{\overline{\underline{|\qquad\qquad|}}+\overline{\underline{|\qquad\qquad|}}} \approx 0.037{\small 7}$

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You get a tractable recurrence if, instead of considering where the largest number is, you look at the smallest number. The indexing will seem strange here, but bear with me. Let $f(n,l,r)$ denote the number of permutations of $\{0,\ldots,n\}$ where $l+1$ elements are visible from the left and $r+1$ are visible from the right. Clearly $f(n,l,r)=0$ if ...

1

Let $A(0,10),B(0,0),C(15,0),P(p,q)$ where $0\lt q$. Then, solving $$p^2+q^2=12^2\quad \text{and}\quad (p-15)^2+q^2=9^2$$ gives $$p=\frac{48}{5},\quad q=\frac{36}{5}.$$ So, $$PA=\sqrt{\left(\frac{48}{5}\right)^2+\left(10-\frac{36}{5}\right)^2}=10.$$

1

There are some great solutions here, but there is a far simpler way of approaching this problem. Address this problem starting with the variables of highest coefficient. Thus, we clearly start with $w$. Notice that if you plug in $2$ for $w$, you have $54$ oz. Plugging in $3$ for the remaining variables, your total weight comes up short at only $93$ oz. ...

1

If you set $x' = x - 1$, $y' = y - 1$, $z' = z - 1$, and $w' = w - 1$, then $x', y', z', w' \in \{0, 1, 2\},$ and $x+3y+9z+27w=95$ implies that $$55 = 3^3 w' + 3^2 z' + 3y' + x'.$$ So basically you are being asked to convert the decimal number $55$ to base three. The answer is $$55_{10} = 2001_3,$$ which implies that $(w', z', y', x') = (2, 0, 0, 1)$ ...

2

Hint: The $13$ coins from bags $1,2,3$ weigh from $13$ through to $39$ ounces so the $27$ coins from bag $4$ weigh from ... through to ... and so must weigh ... each and ... in total, leaving ... for the $13$ coins. The $4$ coins from bags $1,2$ weigh from $4$ through to $12$ ounces so the $9$ coins from bag $3$ weigh from ... through to ... and so must ...

1

Here is some R code that solves your problem: #Matrix for saving results mat <- matrix(NA,ncol = 1,nrow = 4) colnames(mat) <- c("x","y","z","w") #Loop through all possible values for (x in c(1,2,3)){ for (y in c(1,2,3)){ for (z in c(1,2,3)){ for (w in c(1,2,3)){ if(x+3*y+9*z+27*w == 95){ mat <- cbind(mat,c(x,y,z,w)) ...

3

Hint: If you consider your equation $x+3y+9z+27w=95$ modulo $3$. You will get $$x\equiv 2 \pmod{3}.$$ Based on the fact that $x \in \{1,2,3\}$, you get $x=2$. For $y$ try modulo $9$ and so on. Hopefully you can handle the rest.

5

Being successful at Olympiad mathematics is certainly correlated with being successful in later studies and research, but there is no implication in either direction. This is what you would expect a priori: coming up with creative ideas is a part of the work of a research mathematician, but by no means the only (and arguably not the most important) part. ...

8

I am not one thousandth the mathematician that Terry Tao is, but my own feeling is rather different. I had a college classmate who was far better at competition mathematics than I was, and when we went to grad school (together), he seemed good at following a prescribed path, but not so good at striking out on his own. In later professional life, he made no ...

9

Find Terry Tao's blog. He talks about his experience of learning at different levels of mathematics education. Among other insights, he writes how patterns from competition problems he later discovered to be examples of more general, deep and beautiful results. What I took away from all that was that while solving competition problems isn't directly ...

3

If two functions have the same output and the same domain, they are the same function. You know that $g(x)=kx+n$, and that $g^{-1}(x) = g(x)=kx+n$. Now, using the properties of inverse functions, you also know that for each $x$, you have $x = \mathrm{id}(x) = (g\circ g^{-1})(x) = (g\circ g)(x)=g(g(x))$, and this equation should give you a lot of ...

3

Indeed, no polynomial with degree $3$ can satisfy this property because it has to have a negative value at some point. Also, it is clear that (by taking very large $x$), the degree of this polynomial must be $2,3$ or $4$. So this polynomial has to have a degree of $2$ or $4$. Case 1. deg$p$ is $2$. Let $p(x)=ax^2+bx+c$. Since $0 \le p(0)=c \le 1$, we ...

2

Hint : No polynomial with degree $3$ can have this property because it has negative values. It is easy to find out the polynomials with degree $2$ and $4$ satisfying the given bounds. Note that $x^2>x^4$ for $0<x<1$

1

Here is a more inspired solution. I will prove a stronger claim - that $$\frac{m+1}{n}+\frac{n+1}{m}=3$$ has infinitely many solutions in positive integers. Setting $$m=2d(d+a), n=2d(d-a)$$ We have $$\frac{m+1}{n}+\frac{n+1}{m}=\frac{2d^2+2a^2+1}{d^2-a^2}$$ So for $\frac{m+1}{n}+\frac{n+1}{m}=3$ to hold, we only need $d^2-5a^2=1$. There are infinitely ...

1

I told my friend Ruby how to check the numbers by Hagen, zhoraster and two test numbers by me (n=4212345, n=0). She did the check and got: n digits: [0:198, 1:202, 2:202, 3:201, 4:202, 5:202, 6:201, 7:202, 8:202, 9:201] OK - congrats! n digits: [0:227, 1:231, 2:231, 3:231, 4:231, 5:231, 6:231, 7:231, 8:231, 9:230] OK - congrats! n digits: [0:0, 1:1, 2:2, ...

10

Suppose $N>2000$ is an integer such that the period length of the (eventually) repeating $\frac1N$ equals $N$. Then in computing the decimal expansion all remainders $1,\ldots,N-1$ occur at some place. Then the fractions $\frac1N,\frac2N,\ldots, \frac{2000}N$ turn out to lead to the very same period, merely shifted. In this situation, we have ...

8

The answer is positive. Moreover, there exists a number $n$ such that additionally all numbers $2n,3n,\dots,2000n$ are obtained from $n$ by cyclic permutation of digits (naturally, we need to add some zeroes before $n$ for that). Here is this number: ...

0

With a little bit of cheating, you don't need the whole hexagon... Let $O$ be the centre of the unit circle with equilateral $\triangle ABC$ inscribed in it. Extend $\vec {AO}$ to meet the circle at D. As $BC$ and $OD$ are perpendicular bisectors of each other, $\triangle OBD$ is isosceles, and hence $|BD|=1$. But this must be smaller than the minor ...

1

I am not sure if this is redundant, but: If an equilateral triangle is inscribed in a unit circle, and if, on each side of the inscribed triangle, an isosceles triangle is further inscribed in the circle, then an equilateral hexagon with each side of length $=1$ results; but then $6 < 2\pi$ implies $3 < \pi$. So is this something you are after?

8

The inscribed hexagon in the unit circle has perimeter $6$. The perimeter of the circle is $2\pi$, hence $\pi > 3$.

4

The inscribed $12$-gon in the unit circle has area $\frac{12}{2}\sin (2\pi/12)=3$. The area of the unit circle is $\pi$. Hence $\pi\ge 3$.

1

It is clear that $$(a_1,b_1,c_1) | (b_1,c_1)$$ and if $m|n$ then $[m,n] = n$ so $$(b_3,c_3) = [(a_1,b_1,c_1),(b_1,c_1)] = (b_1,c_1) = a_2$$

0

Let's consider your example of selecting three numbers from the set $\{1, 2, 3, \ldots, 50\}$ that are in numerical order. There are $50$ ways to select the first number, $49$ ways to select the second number, and $48$ ways to select the third number. However, the three numbers we have selected may not be in numerical order. Let's say we have selected your ...

2

For $2^i\le j< 2^{i+1}$, we have $\frac{1}{j^2}\le\frac{1}{2^{2i}}$. As such, if we take $N=2^k-1$ for some $k$, then $$\sum\limits_{n=1}^{N}{\frac{1}{n^2}} = \sum\limits_{i=1}^{k-1}{\sum\limits_{j = 2^i}^{2^{i+1}-1}{\frac{1}{j^2}}}\le\sum\limits_{i=1}^{k-1}{\sum\limits_{j=2^i}^{2^{i+1}-1}{\frac{1}{2^{2i}}}} = ... 9 Since x \mapsto \dfrac1{x^2} is a monotone decreasing function over [1,\infty), then we have$$ \sum_{n=1}^N\frac1{n^2}\leq1+\int_1^N\frac1{x^2}dx=1+\left[-\frac1x \right]_1^N=2-\frac1N, \quad N\geq1. $$From which you deduce easily that$$ \sum_{d|N} \frac1{d^2}\leq\sum_{n=1}^N\frac1{n^2}\leq 2 $$as announced. Alternatively, you may use a ... 1 First we know x^2+y^2+z^2 is even so either two odd one even or three even. In the case of two odd one even we have x^2+y^2+z^2\equiv2\pmod{4} while 2xyz\equiv0\pmod{4} so this is impossible so all three are even. Let x=2a,y=2b,z=2c we have a^2+b^2+c^2=4abc. Now a,b,c must be all even again. Let a=2a_2,b=2b_2,c=2c_2 we have a_2,b_2,c_2 ... 4 From z^2 - 2 (xy) z + (x^2 + y^2) = 0, we get$$z = xy \pm \sqrt{x^2 y^2 - x^2 - y^2}$$so we basically need to search for x, y such that x^2 y^2 - x^2 - y^2 is a square say$$x^2 y^2 - x^2 - y^2 = t^2$$or$$(x^2 - 1) (y^2 - 1) = t^2 + 1$$Now a^2 \equiv 0, 1 \mod 4 hence a^2 - 1 \equiv -1, 0 \mod 4 corresponding to a even, odd respectively so ... 3 Subtracting the equation will give you a^2+ab+b^2=3(-a-b) and similar for the other two. Subtracting again will give you (a+b+c)(b-c)=-3(b-c)\implies a+b+c=-3. Now add the three equations up we get 2a^2+2b^2+2c^2+ab+bc+ca=3(-2a-2b-2c) so 2(a+b+c)^2-3ab-3bc-3ca=-6(a+b+c)\implies ab+bc+ca=0 Hence a^2+b^2+c^2=9 Now add the original three equations ... 7 Hint: AM-GM. Note that the product of our fractions is 1. 0 While the final answer posted above is true, the logic is flawed. Maximizing a sum is not equivalent to maximizing each term. I am sure one can give plenty of counter examples to that claim. To prove the estimate 28 more rigorously I give the following hints: First note that a_k + |a_k-a_{k-1}| = a_{k-1} + b_k where b_k = 0 if a_k < a_{k-1} and ... 0 By using the triangle inequality$$|x+y|\leq|x|+|y| $\left|a_1-a_2\right|+|a_2-a_3|+|a_3-a_4|+...+|a_6-a_7|+a_7$ $\leq |a_1|+|a_2|+|a_2|+...+|a_7|+|a_7|$ $=a_1+2(a_2+a_3+a_4+...+a_7)$ $=2(a_1+a_2+a_3+...+a_7)-a_1$ $=2(1+2+3+4+5+6+7)-a_1$ $\because$ it fully depends on $a_1$ Choose $a_1=1$, $\therefore$Maximum$\, =2(28)-1=55$

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