Challenge Math Challenge - June 2019

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7. Five vessels contain ##100## balls each. Some vessels contain only balls of ##10\, g## mass, while the other vessels contain only balls of ##11\, g## mass. How can we determine with a single weighing which results in a mass, which vessels contain balls of ##10\, g## and which contain balls of ##11\, g##? (It is allowed to remove balls from the vessels.)
From 1 to 31, all natural numbers can be written in one unique way in the following form $$ a\cdot 1 + b\cdot 2 + c\cdot 2^2 +d\cdot 2^3 + e \cdot 2^4 = a\cdot 1 +b\cdot 2 +c\cdot 4 +d\cdot 8+ e\cdot 16$$ wherein a, b, c, d, e are each 0 or 1.


In the first vessel, we leave one weight, in the second vessel 2, 4 in the third, 8 in the fourth and 16 in the fifth. We weigh all of these weights. Since each weight is either 10 or 11 g, if we subtract 310 g, then we get exactly the weight that the 11 g weights contributed to the total weight, with their weight above 10 g. The difference is expressed in the above form, and a, b, c, d, e show which vessels contributed to the total weight by weight over 10 g.

Let's look at an example: First vessel: 10 g; second 11 g; third 11 g; fourth 10 g; fifth 11g. Using the above method, the total weight ## 1 \cdot 10 + 2 \cdot 11 +4 \cdot 11 +8 \cdot 10 +16 \cdot 11 = 332##. Furthermore ##332-310 =22 =2+4+16.##
 

scottdave

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Remove 1 ball from the first vessel, two balls from the second vessel, three from the third, four from the fourth and five from the fifth. Weigh all fifteen balls. If they weigh 150 grams, all vessels contain 10 g balls. If they weigh 152 grams, all vessels but #2 weigh 10 grams and vessel #2 weighs 11 grams. If they weigh 155 g, only #2 and #3 have 11 gram balls.
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Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.
 
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Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.
I also thought of prime numbers for the first time, but here we are talking about sums. Then I thought of binary writing of the numbers. It was confusing that 32 and 64 were not needed, only 16.
 
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This way you cannot tell which vessel has 10g balls and which 11g balls, only how many of each.
I WILL JUST PUT SOME LABELS ON EACH MEASURED BALL :)
 
Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{2\sinh(x)-2x}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)-\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}
 
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fresh_42

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Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{(2\sinh(x)-2x)'}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)+\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}
Modulo a typo or two this is correct. The idea is that de L'Hôpital can be applied several times.
 
Solve the following equation: $$(3x^2y^2+x^2)dx+(2x^3y+y^2)dy=0$$ (1).

Solution:
We need to determine if (1) is an exact equation or not.
Let $$M(x,y)=(3x^2y^2+x^2)$$ and $$N(x,y)=(2x^3y+y^2)$$. For testing for exactness, we need to take the partial derivatives with respect to y and x for M and N, respectively.
So, $$M_y(x,y)=6x^2y$$ and $$N_x(x,y)=6x^2y$$.
Since $$M_y(x,y)=N_x(x,y)$$, we know that (1) is an exact equation. So
\begin{align*}
F_x=(3x^2y^2+x^2)\\
&\int F_x \, dx =\int (3x^2y^2+x^2) \, dx\\
&F(x,y)= x^3y^2+\frac{x^3}{3}+g(y) \, & (\text{we need to determine what g(y), so taking the partial derivative of F with respect to y and set equal to N(x,y) and integrate both side by y})\\
&F_y(x,y)=2x^3y+g'(y)\\
2x^3y+y^2=2x^3y+g'(y)\\
g(y)=\frac{y^3}{3}
\end{align*}
The final solution of (1) is $$F(x,y)=x^3y^2+\frac{x^3}{3}+\frac{y^3}{3}=C$$.
 
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Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.
 

fresh_42

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Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.
Probably not, as we always have that the entire sum of points equals ten, which should be an equivalent information, but I haven't checked. I just sounds better in the question to mention all players.
 

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