The3Mathketeers

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UsernameThe3Mathketeers
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 #2
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My method of solving this is to consider all the possible ways in which N could have a sum of digits no more than 4 and adding those together.

 

\(2 \times 10^7 \leq N < 10^8 \\ 20\;000\;000 \leq N < 100\;000\;000\)

 

Considering all the values of N is a tad overwhelming, so I started with limited my search to the range \(20\;000\;000 \leq N < 30\;000\;000\). How can the values of N within this range have a sum of digits no more than 4?

 

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

2) One way is to ensure that one 1 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen.

3) One way is to ensure that 2 1's appear in any digit except the leading digit. There are 7 digits available for the first 1 and 6 remaining choices for the second 1. However, order is immaterial, so we must exclude the cases where different orders create the same number. \(\frac{7 * 6}{2} = 21 \text{ ways}\).

4) One way is to ensure that one 2 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen.

 

I then considered the range \(30\;000\;000 \leq N < 40\;000\;000\). Luckily, the cases are simpler than the previous range because the leading digit of 3 restricts some of the possibilities.

 

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

2) One way is to ensure that one 1 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen. 

 

I then considered the range \(40\;000\;000 \leq N < 50\;000\;000\). This case is by far the simplest case.

 

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

 

I then realized that there are no more options. For the range \(50\;000\;000 \leq N < 100\;000\;000\), there are no combinations of digits that meet the criteria since the leading digit exceeds the maximum sum.

 

Now, just add all the possible ways. \(1 + 7 + 21 + 7 + 1 + 7 + 1 = 45\) positive integers N that meet both criteria.

Aug 21, 2023
 #3
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This problem just requires some clever algebraic manipulation. I know others have already posted the answer to this system of equations, but I will show you some work as to how one might find one of the variables because it may not be too straightforward of a process. If I can avoid it, I try to resist having fractions in a problem involving system of equations, so I decided to eliminate the fractions from the original equation. There is probably a faster way, but it is typically hard to find.

 

\(\frac{xy}{x + y} = 1 \\ xy = x + y\) \(\frac{xz}{x + z} = 2 \\ xz = 2x + 2z\) \(\frac{yz}{y + z} = 4 \\ yz = 4y + 4z\)

 

Afterwards, I solved for one of the variables in both equations. I chose to solve for y, but any choice should suffice. I used the technique of grouping all the terms with y and then factoring.

 

\(xy = x + y \\ xy - y = x \\ y(x - 1) = x \\ y = \frac{x}{x - 1}\) \(yz = 4y + 4z \\ yz - 4y = 4z \\ y(z - 4) = 4z \\ y = \frac{4z}{z - 4}\)

 

We have two equations solved for y, so we can find a relationship between x and z variables.

 

\(y = \frac{x}{x - 1}, y = \frac{4z}{z - 4} \\ \frac{x}{x - 1} = \frac{4z}{z - 4} \\ x(z - 4) = 4z(x - 1) \\ xz - 4x = 4xz - 4z \\ -3xz = 4x - 4z\)

 

Now, we can conveniently use the \(xz = 2x + 2z\) equation to help solve this system of equations.

 

\(xz = 2x + 2z \\ 3xz = 6x + 6z\)

 

Now, we can use technique of elimination to find a relationship between x and z.

 

\(\begin{align*} 3xz &= &6x &+ 6z \\ -3xz &= &4x &- 4z \\ 0 &= &10x &+ 2z \end{align*} \\ 5x + z = 0 \\ z = -5x\)

 

Now, we can substitute this relationship for z into an equation with x and z variables and solve for x.

 

\(xz = 2x + 2z \\ x * -5x = 2x + 2 * -5x \\ -5x^2 = -8x \\ -5x^2 + 8x = 0 \\ -x(5x - 8) = 0 \\ x = 0 \text{ or } 5x - 8 = 0 \\ x = 0 \text{ or } x = \frac{8}{5}\)

 

Now, you might notice that we obtained x = 0 as a potential solution, but we can reject \(x = 0\) immediately because if we were to substitute this x-value into the original equation \(\frac{xy}{x + y} = 1\), there would be no solution, so \(x = 0\) is not a candidate. At this point, I will not go any further because the process should be similar to find the other variables. Happy solving!

Aug 21, 2023
 #1
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It is best for these problems to use a picture, and the picture should make it clearer how to proceed. With the given information, I was able to create the following diagram below:

 

Note that G is the center of the circle. Our goal is to find the length of \(\widehat{DE}\), which is essentially asking for the arc length. The typical arc length formula for degrees is \(S = 2 \pi r * \frac{\theta}{360^{\circ}}\) where S is the arc length, r is the radius of the circle, and \(\theta\) is the angle of the central angle of the arc, which is \(\angle DGE\) in this case.

 

We can modify this formula somewhat because we already know the circumference. Since the given information states that \(C = 2 \pi r = 24\), we can substitute that into the arc length formula, which turns it into \(S = 24 * \frac{\theta}{360^{\circ}}\).

 

Now, we must find the central angle, which we can use the Inscribed Angle Theorem to find the measure of the central angle. The Inscribed Angle Theorem states that the measure of the central angle is twice that of the corresponding inscribed angle. Applying that theorem to this particular case, \(2m \angle DFE = m \angle DGE\).

 

\(2 m \angle DFE = m\angle DGE \\ 2 * 60^\circ = m \angle DGE \\ m \angle DGE = 120^\circ\)

 

Now that we know the measure of the central angle, we can find the arc length finally.

 

\(S = 24 * \frac{\theta}{360^{\circ}} \\ S = 24 * \frac{120^{\circ}}{360^{\circ}} \\ S = 24 * \frac{1}{3} \\ S = 8 \text{ units}\)

 

The question asks to represent the answer as a common fraction, so I suppose \(S = \frac{8}{1} \text{ units}\) would be the right answer in this case.

Aug 20, 2023