Input:
**User**:
Two runners, $A$ and $B,$ start at a point $O$ on a linear track, and start running in the same direction.  Runner $B$ runs three times as fast as runner $A.$  An observer stands at point $P$ so that $\overline{OP}$ is perpendicular to the track.  Find the maximum of $\angle APB,$ in degrees.

[asy]
unitsize(2 cm);

pair A, B, O, P;

A = (0.4,0);
B = (1.2,0);
O = (0,0);
P = (0,1);

draw((-0.5,0)--(2,0));
draw(O--P);
draw(P--A);
draw(P--B);

label("$A$", A, S);
label("$B$", B, S);
label("$O$", O, S);
label("$P$", P, N);
[/asy]
Please reason step by step, and put your final answer within \boxed{}.

<think>
Okay, so there's this problem about two runners, A and B, starting from the same point O on a linear track. They run in the same direction, and B is three times as fast as A. There's an observer at point P, where OP is perpendicular to the track. We need to find the maximum angle APB, in degrees.

First, let me visualize the setup. The track is a straight line, and OP is perpendicular to it. So, if I imagine the track as the x-axis, then point O is at the origin (0,0), and point P is somewhere along the y-axis, say at (0, h) for some height h. The Asymptote code shows P at (0,1), but maybe the actual coordinates aren't important, just the relative positions.

Runners A and B start at O and run in the same direction. Let me assume the track is horizontal, so the direction they're running is along the positive x-axis. Since B is three times as fast as A, if A's speed is v, then B's speed is 3v.

The question is about the angle APB. So, we need to consider the positions of A and B at some time t after they start running, and then find the angle at point P between points A and B. The goal is to find the maximum value of this angle.

Let me think about how to model this. Let's denote the position of runner A at time t as (x_A(t), 0), and runner B as (x_B(t), 0). Since they start at O, which is (0,0), and run in the same direction (let's say positive x-direction), then:

x_A(t) = v * t

x_B(t) = 3v * t

But since they are moving along the track, their positions are along the x-axis. The observer is at point P, which is at (0, h). Wait, in the Asymptote code, P is at (0,1), but maybe h is arbitrary? Or maybe we can assume h is 1 for simplicity? Or maybe it's better to keep h as a variable? Hmm.

But since the problem doesn't specify the distance from O to P, maybe the maximum angle is independent of h? Or maybe not. Wait, but the angle APB might depend on h. However, since the problem says "an observer stands at point P so that OP is perpendicular to the track," but doesn't specify where P is. Wait, but maybe the maximum angle is achieved regardless of where P is? Or maybe there's a specific position? Wait, no, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, OP is perpendicular, meaning P is somewhere off the track, but the exact position isn't given. Wait, but maybe the problem is assuming that P is fixed? Wait, the Asymptote code shows P at (0,1), but maybe that's just an example.

Wait, but the problem says "Find the maximum of angle APB." So, perhaps we need to consider all possible positions of P? Or is P fixed? Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, P is fixed, but the problem doesn't specify where. Wait, but maybe the maximum angle is independent of the position of P? Or maybe there's a specific position where the maximum occurs? Wait, but the problem says "Find the maximum of angle APB," so maybe we need to find the maximum possible angle that can be achieved as the runners move, given that P is fixed at some position where OP is perpendicular. Wait, but the problem doesn't specify where P is. Wait, maybe the answer is the same regardless of where P is? Or maybe there's a specific position?

Wait, maybe the problem is similar to the classic problem where you have a point P off the track, and you want to find the maximum angle subtended by two moving points. In such cases, the maximum angle is achieved when the circle through A and B is tangent to the track. But I need to think more carefully.

Alternatively, maybe we can parametrize the positions of A and B as functions of time, then express the angle APB in terms of time, and then find its maximum. Let me try that approach.

Let me set up coordinates. Let me assume the track is the x-axis, O is at (0,0), and P is at (0, h), where h is some positive number. Let me assume that h is fixed. Then, as time t increases, runner A is at (vt, 0) and runner B is at (3vt, 0).

We need to find angle APB. So, angle at point P between points A and B. To find this angle, we can use coordinates. Let me recall that the angle between two lines can be found using the tangent formula or using vectors.

Alternatively, we can use coordinates to find vectors PA and PB, then compute the angle between them.

Let me denote coordinates:

Point P: (0, h)

Point A: (vt, 0)

Point B: (3vt, 0)

So vector PA is A - P = (vt - 0, 0 - h) = (vt, -h)

Similarly, vector PB is B - P = (3vt - 0, 0 - h) = (3vt, -h)

Then, the angle between vectors PA and PB is angle APB.

The formula for the angle θ between two vectors u and v is:

cosθ = (u · v)/(|u| |v|)

So let's compute the dot product of PA and PB:

PA · PB = (vt)(3vt) + (-h)(-h) = 3v²t² + h²

The magnitude of PA is sqrt( (vt)^2 + h² ) = sqrt(v²t² + h² )

Similarly, magnitude of PB is sqrt( (3vt)^2 + h² ) = sqrt(9v²t² + h² )

Therefore,

cosθ = [3v²t² + h²] / [ sqrt(v²t² + h² ) * sqrt(9v²t² + h² ) ]

So θ = arccos [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ]

Our goal is to find the maximum value of θ. Since arccos is a decreasing function, maximizing θ is equivalent to minimizing the expression [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ].

Alternatively, maybe it's easier to work with tanθ or something else. Let me think.

Alternatively, since angle APB is formed by lines PA and PB, maybe we can use coordinates to find tanθ. Let me try that.

Alternatively, since we have coordinates, maybe we can compute the slopes of PA and PB, then compute the angle between them.

But maybe using coordinates is more complicated. Let me think again.

Alternatively, since we have coordinates, maybe we can use coordinates to find tanθ. Let me recall that if we have two lines with slopes m1 and m2, then tanθ = |(m2 - m1)/(1 + m1m2)|. But since angle APB is the angle between PA and PB, which are two lines from P to A and P to B.

So, let me compute the slopes of PA and PB.

Slope of PA: (0 - h)/(vt - 0) = -h / (vt)

Slope of PB: (0 - h)/(3vt - 0) = -h / (3vt)

So, m1 = -h/(vt), m2 = -h/(3vt)

Then tanθ = |(m2 - m1)/(1 + m1m2)|

Let me compute that:

Numerator: m2 - m1 = [ -h/(3vt) - (-h/(vt)) ] = -h/(3vt) + h/(vt) = ( -1/3 + 1 ) h/(vt) = (2/3) h/(vt)

Denominator: 1 + m1m2 = 1 + [ (-h/(vt)) * (-h/(3vt)) ] = 1 + [ h² / (3v²t²) ]

Therefore tanθ = | (2/3 h/(vt)) / (1 + h²/(3v²t²)) | = (2/3 h/(vt)) / (1 + h²/(3v²t²))

Simplify denominator:

Multiply numerator and denominator by 3v²t²:

Numerator becomes: (2/3 h/(vt)) * 3v²t² = 2 h v t

Denominator becomes: 3v²t² + h²

Therefore tanθ = (2 h v t) / (3v²t² + h² )

So tanθ = (2 h v t) / (3v²t² + h² )

So now, we have tanθ expressed in terms of t. Our goal is to find the maximum θ, which corresponds to maximizing tanθ, since tanθ is increasing in [0, π/2).

Therefore, we can find the maximum of tanθ as a function of t, then take arctangent to find θ.

So let me denote f(t) = (2 h v t) / (3v²t² + h² )

We need to find the maximum of f(t). Let me set u = t, so f(u) = (2 h v u) / (3v²u² + h² )

To find maximum, take derivative with respect to u, set to zero.

Let me compute f'(u):

Using quotient rule:

f'(u) = [ (2 h v)(3v²u² + h² ) - (2 h v u)(6v²u) ] / (3v²u² + h² )²

Simplify numerator:

First term: 2 h v (3v²u² + h² )

Second term: - 2 h v u * 6v²u = -12 h v³ u²

So numerator:

2 h v * 3v²u² + 2 h v * h² - 12 h v³ u²

= 6 h v³ u² + 2 h² v - 12 h v³ u²

= (6 h v³ u² - 12 h v³ u² ) + 2 h² v

= (-6 h v³ u² ) + 2 h² v

Set numerator equal to zero:

-6 h v³ u² + 2 h² v = 0

Divide both sides by 2 h v (assuming h ≠ 0, v ≠ 0):

-3 v² u² + h = 0

So:

-3 v² u² + h = 0

=> 3 v² u² = h

=> u² = h / (3 v² )

=> u = sqrt( h / (3 v² )) = (sqrt(h)/v ) / sqrt(3 )

Therefore, critical point at t = sqrt( h ) / ( v sqrt(3) )

Now, check if this is a maximum. Since f(t) approaches 0 as t approaches 0 and as t approaches infinity, this critical point must be a maximum.

Therefore, maximum tanθ is f(t) evaluated at t = sqrt( h ) / ( v sqrt(3) )

Let me compute that:

First, compute numerator: 2 h v t = 2 h v * [ sqrt(h) / (v sqrt(3)) ] = 2 h v * sqrt(h) / (v sqrt(3)) ) = 2 h sqrt(h) / sqrt(3) = 2 h^(3/2)/sqrt(3)

Denominator: 3v² t² + h²

Compute t²: h / (3 v² )

So 3v² t² = 3v² * [ h / (3 v² ) ] = h

Therefore denominator is h + h² ?

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Wait denominator is 3v² t² + h² = h + h² ?

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Solution:
Without loss of generality, assume that $OP = 1.$  Let $OA = x$ and $OB = 3x.$  Let $\alpha = \angle OPA$ and $\beta = \angle OPB,$ so $\tan \alpha = x$ and $\tan \beta = 3x,$ so from the angle subtraction formula,
\begin{align*}
\tan \angle APB &= \tan (\angle OPB - \angle OPA) \\
&= \tan (\beta - \alpha) \\
&= \frac{\tan \beta - \tan \alpha}{1 + \tan \alpha \tan \beta} \\
&= \frac{2x}{1 + 3x^2}.
\end{align*}We want to maximize this expression.  Maximizing this expression is equivalent to minimizing $\frac{1 + 3x^2}{2x}.$  By AM-GM,
\[\frac{1 + 3x^2}{2x} \ge \frac{2 \sqrt{1 \cdot 3x^2}}{2x} = \frac{2x \sqrt{3}}{2x} = \sqrt{3},\]so
\[\tan \angle APB \le \frac{1}{\sqrt{3}},\]which means $\angle APB \le 30^\circ.$  Equality occurs when $x = \frac{1}{\sqrt{3}},$ so the maximum of $\angle APB$ is $\boxed{30^\circ}.$


Input:
**User**:
How many $y$-intercepts does the graph of the parabola $x = y^2 - 4y - 1$ have?
Please reason step by step, and put your final answer within \boxed{}.

To determine how many **y-intercepts** the graph of the parabola
$$
x = y^2 - 4y - 1
$$
has, we need to understand what a **y-intercept** is. A y-intercept is a point where the graph crosses the **y-axis**, which occurs when the **x-coordinate is 0**. So, to find the y-intercepts, we set $ x = 0 $ and solve for $ y $.

---

### Step 1: Set $ x = 0 $

Substitute $ x = 0 $ into the equation:

$$
0 = y^2 - 4y - 1
$$

This is a **quadratic equation** in terms of $ y $:

$$
y^2 - 4y - 1 = 0
$$

---

### Step 2: Solve the Quadratic Equation

We can solve this using the **quadratic formula**:

$$
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

For the equation $ y^2 - 4y - 1 = 0 $, the coefficients are:

- $ a = 1 $
- $ b = -4 $
- $ c = -1 $

Now compute the **discriminant**:

$$
D = b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20
$$

Since the discriminant is **positive**, the equation has **two distinct real solutions**.

---

### Step 3: Find the Solutions

Using the quadratic formula:

$$
y = \frac{4 \pm \sqrt{20}}{2}
$$

Simplify $ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} $, so:

$$
y = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}
$$

Thus, the two y-values are:

$$
y = 2 + \sqrt{5} \quad \text{and} \quad y = 2 - \sqrt{5}
$$

These correspond to the **y-intercepts** of the graph, which are the points:

$$
(0, 2 + \sqrt{5}) \quad \text{and} \quad (0, 2 - \sqrt{5})
$$

---

### Step 4: Interpret the Result

Since the equation is a **horizontal parabola** (solved for $ x $ in terms of $ y $), it opens to the right or left. The vertex form of the equation is:

$$
x = (y - 2)^2 - 5
$$

This confirms the parabola opens to the right and has a vertex at $ (-5, 2) $. Since the vertex is to the left of the y-axis and the parabola opens to the right, it **must cross the y-axis at two points**, which matches our earlier result.

---

### Final Answer

$$
\boxed{2}
$$

Solution:
A $y$-intercept is a point on the graph that lies on the $y$-axis, so $x = 0$.  Hence, the number $y$-intercepts corresponds to the number of real solutions of the quadratic equation $y^2 - 4y - 1 = 0$.  The discriminant of this quadratic equation is $(-4)^2 + 4 \cdot 1 \cdot (-1) = 20$, which is positive, so the quadratic has two distinct real roots.  Therefore, the number of $y$-intercepts is $\boxed{2}$.

[asy]
size(150);
real ticklen=3;
real tickspace=2;

real ticklength=0.1cm;
real axisarrowsize=0.14cm;
pen axispen=black+1.3bp;
real vectorarrowsize=0.2cm;
real tickdown=-0.5;
real tickdownlength=-0.15inch;
real tickdownbase=0.3;
real wholetickdown=tickdown;
void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool

useticks=false, bool complexplane=false, bool usegrid=true) {

import graph;

real i;

if(complexplane) {

label("$\textnormal{Re}$",(xright,0),SE);

label("$\textnormal{Im}$",(0,ytop),NW);

} else {

label("$x$",(xright+0.4,-0.5));

label("$y$",(-0.5,ytop+0.2));

}

ylimits(ybottom,ytop);

xlimits( xleft, xright);

real[] TicksArrx,TicksArry;

for(i=xleft+xstep; i<xright; i+=xstep) {

if(abs(i) >0.1) {

TicksArrx.push(i);

}

}

for(i=ybottom+ystep; i<ytop; i+=ystep) {

if(abs(i) >0.1) {

TicksArry.push(i);

}

}

if(usegrid) {

xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray

(0.22),extend=true),p=invisible);//,above=true);

yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.22),extend=true),

p=invisible);//,Arrows);

}

if(useticks) {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx ,

pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));

} else {

xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));

yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));

}
};
real lowerx, upperx, lowery, uppery;
real f(real x) {return x^2 - 4*x - 1;}
lowery = -1;
uppery = 5;
rr_cartesian_axes(-6,5,lowery,uppery);
draw(reflect((0,0),(1,1))*(graph(f,lowery,uppery,operator ..)), red);
dot((0,2 + sqrt(5)));
dot((0,2 - sqrt(5)));
[/asy]


Input:
**User**:
Convert $\frac{21}{2^2 \cdot 5^7}$ to a terminating decimal.
Please reason step by step, and put your final answer within \boxed{}.

To convert the fraction
$$
\frac{21}{2^2 \cdot 5^7}
$$
into a **terminating decimal**, we begin by analyzing the structure of the denominator.

---

### Step 1: Understand the Denominator

The denominator is:
$$
2^2 \cdot 5^7
$$

This is a product of powers of 2 and 5, which are the prime factors of 10. Since the denominator only contains the prime factors 2 and 5, the fraction will indeed convert to a **terminating decimal**.

---

### Step 2: Convert the Denominator to a Power of 10

To make the denominator a power of 10, we need to balance the number of 2s and 5s. The current denominator has:
- $2^2$
- $5^7$

To make it $10^n$, we need equal powers of 2 and 5. So we multiply both numerator and denominator by $2^5$ to make the total number of 2s equal to 7:
$$
\frac{21}{2^2 \cdot 5^7} \cdot \frac{2^5}{2^5} = \frac{21 \cdot 2^5}{2^7 \cdot 5^7} = \frac{21 \cdot 32}{10^7}
$$

---

### Step 3: Simplify the Numerator

Now compute the numerator:
$$
21 \cdot 32 = 672
$$

So the fraction becomes:
$$
\frac{672}{10^7}
$$

---

### Step 4: Convert to Decimal

Since the denominator is $10^7 = 10,000,000$, we can write the fraction as:
$$
\frac{672}{10,000,000} = 0.0000672
$$

This is a **terminating decimal** with 7 decimal places.

---

### Final Answer

$$
\boxed{0.0000672}
$$

Solution: A terminating decimal can be written in the form $\frac{a}{10^b}$, where $a$ and $b$ are integers. So we try to get a denominator of the form $10^b$: $$\frac{21}{2^2\cdot5^7}\cdot\frac{2^5}{2^5}=\frac{21\cdot32}{10^7}=\frac{672}{10^7}=\boxed{.0000672}.$$


Input:
**User**:
In the diagram, four circles of radius 1 with centres $P$, $Q$, $R$, and $S$ are tangent to one another and to the sides of $\triangle ABC$, as shown. [asy]
size(200);
pair A, B, C, P, Q, R, S;
R=(0,0);
Q=(-2,0);
S=(2,0);
P=(1,1.732);
B=(-5.73,-1);
C=(3.732,-1);
A=(1.366,3.098);
draw(A--B--C--A);
draw(circle(P, 1));
draw(circle(Q, 1));
draw(circle(R, 1));
draw(circle(S, 1));
label("A", A, N);
label("B", B, SW);
label("C", C, SE);
dot(P);
dot(Q);
dot(R);
dot(S);
label("P", P, N);
label("Q", Q, SW);
label("R", R, SW);
label("S", S, SE);
[/asy]

What is the degree measure of the smallest angle in triangle $PQS$?
Please reason step by step, and put your final answer within \boxed{}.

To determine the **smallest angle** in triangle $ PQS $, we begin by analyzing the given configuration and the coordinates of the centers of the four circles:

- $ P = (1, \sqrt{3}) $
- $ Q = (-2, 0) $
- $ S = (2, 0) $

These coordinates are provided in the Asymptote code and are consistent with the diagram. We are told that all four circles have radius 1 and are tangent to one another and to the sides of triangle $ ABC $. While the exact configuration of triangle $ ABC $ is not necessary for solving the problem, the coordinates of $ P, Q, S $ allow us to compute the angles of triangle $ PQS $ directly.

---

### Step 1: Compute the Side Lengths of Triangle $ PQS $

Using the distance formula:

- $ PQ = \sqrt{(1 - (-2))^2 + (\sqrt{3} - 0)^2} = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} $
- $ QS = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4 $
- $ PS = \sqrt{(2 - 1)^2 + (0 - \sqrt{3})^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 $

So the side lengths of triangle $ PQS $ are:
- $ PQ = 2\sqrt{3} $
- $ QS = 4 $
- $ PS = 2 $

---

### Step 2: Identify the Smallest Angle

In any triangle, the **smallest angle** is opposite the **shortest side**. Here, the shortest side is $ PS = 2 $, which is opposite angle $ Q $.

Thus, the **smallest angle** is $ \angle Q $.

---

### Step 3: Use the Law of Cosines to Find $ \angle Q $

We apply the Law of Cosines to triangle $ PQS $:

$$
\cos(\angle Q) = \frac{PQ^2 + QS^2 - PS^2}{2 \cdot PQ \cdot QS}
$$

Substitute the known values:

$$
\cos(\angle Q) = \frac{(2\sqrt{3})^2 + 4^2 - 2^2}{2 \cdot 2\sqrt{3} \cdot 4}
= \frac{12 + 16 - 4}{16\sqrt{3}}
= \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}
$$

$$
\angle Q = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ
$$

---

### Step 4: Confirm the Triangle Type

We can also verify that triangle $ PQS $ is a **right triangle** by checking the Pythagorean Theorem:

$$
PS^2 + PQ^2 = 2^2 + (2\sqrt{3})^2 = 4 + 12 = 16 = QS^2
$$

So, triangle $ PQS $ is a **right triangle** with a right angle at $ P $, and the angles are:

- $ \angle Q = 30^\circ $
- $ \angle S = 60^\circ $
- $ \angle P = 90^\circ $

---

### Final Answer

The **smallest angle** in triangle $ PQS $ is:

$$
\boxed{30}
$$

Solution:
Join $PQ$, $PR$, $PS$, $RQ$, and $RS$. Since the circles with center $Q$, $R$ and $S$ are all tangent to $BC$, then $QR$ and $RS$ are each parallel to $BC$ (as the centres $Q$, $R$ and $S$ are each 1 unit above $BC$). This tells us that $QS$ passes through $R$. When the centers of tangent circles are joined, the line segments formed pass through the associated point of tangency, and so have lengths equal to the sum of the radii of those circles. Therefore, $QR=RS=PR=PS=1+1=2$.

[asy]
size(200);
pair P, Q, R, S;
Q=(0,0);
R=(2,0);
S=(4,0);
P=(3,1.732);
label("Q", Q, SW);
label("R", R, dir(270));
label("S", S, SE);
label("P", P, N);
draw(circle(Q,1), dashed);
draw(circle(P,1), dashed);
draw(circle(R,1), dashed);
draw(circle(S,1), dashed);
draw(P--Q--S--P--R);
[/asy]

Since $PR=PS=RS$, we know $\triangle PRS$ is equilateral, so $\angle PSR=\angle PRS=60^\circ$.  Since $\angle PRS=60^\circ$ and $QRS$ is a straight line, we have $\angle QRP=180^\circ-60^\circ=120^\circ$. Since $QR=RP$, we know $\triangle QRP$ is isosceles, so $$\angle PQR = \frac{1}{2}(180^\circ-120^\circ)= 30^\circ.$$Since $\angle PQS=30^\circ$ and $\angle PSQ=60^\circ$, we have $\angle QPS = 180^\circ - 30^\circ - 60^\circ = 90^\circ$, so $\triangle PQS$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Thus, the answer is $\boxed{30^\circ}$.


Input:
**User**:
How many positive two-digit integers are factors of both 100 and 150?
Please reason step by step, and put your final answer within \boxed{}.

To determine how many **positive two-digit integers** are factors of **both 100 and 150**, we follow a structured approach based on the concept of **common factors** and the **greatest common divisor (GCD)**.

---

### Step 1: Understand the Problem

We are looking for **two-digit integers** that divide **both 100 and 150** without leaving a remainder. That is, we want the **common factors** of 100 and 150 that are **between 10 and 99 inclusive**.

---

### Step 2: Find the GCD of 100 and 150

The **common factors** of two numbers are exactly the **factors of their GCD**. So, we first compute the GCD of 100 and 150.

Using the **Euclidean algorithm**:

- $ 150 \div 100 = 1 $ remainder $ 50 $
- $ 100 \div 50 = 2 $ remainder $ 0 $

So, the GCD is **50**.

---

### Step 3: Find All Factors of the GCD (50)

Now, we find all the **factors of 50**:

$$
\text{Factors of } 50 = \{1, 2, 5, 10, 25, 50\}
$$

---

### Step 4: Identify the Two-Digit Factors

From the list of factors of 50, we select those that are **two-digit numbers**:

- 10
- 25
- 50

These are the only two-digit numbers that divide both 100 and 150.

---

### Step 5: Final Verification

Let’s verify that each of these numbers divides both 100 and 150:

- $ 100 \div 10 = 10 $, $ 150 \div 10 = 15 $ → ✅
- $ 100 \div 25 = 4 $, $ 150 \div 25 = 6 $ → ✅
- $ 100 \div 50 = 2 $, $ 150 \div 50 = 3 $ → ✅

All three numbers are valid.

---

### Final Answer

$$
\boxed{3}
$$

Solution: The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100.  Of these, the $\boxed{3}$ factors 10, 25 and 50 have two digits and are factors of 150 as well.