https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Math4life2020&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T18:49:29ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_12&diff=1467622019 AIME I Problems/Problem 122021-02-15T06:35:16Z<p>Math4life2020: Added a proof.</p>
<hr />
<div>==Problem 12==<br />
Given <math>f(z) = z^2-19z</math>, there are complex numbers <math>z</math> with the property that <math>z</math>, <math>f(z)</math>, and <math>f(f(z))</math> are the vertices of a right triangle in the complex plane with a right angle at <math>f(z)</math>. There are positive integers <math>m</math> and <math>n</math> such that one such value of <math>z</math> is <math>m+\sqrt{n}+11i</math>. Find <math>m+n</math>.<br />
<br />
==Solution 1==<br />
Notice that we must have <cmath>\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .</cmath>However, <math>f(t)-t=t(t-20)</math>, so<br />
<cmath><br />
\begin{align*}<br />
\frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\<br />
&=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\<br />
&=(z-19)(z+1)\\<br />
&=(z-9)^2-100.<br />
\end{align*}<br />
</cmath><br />
Then, the real part of <math>(z-9)^2</math> is <math>100</math>. Since <math>\text{Im}(z-9)=\text{Im}(z)=11</math>, let <math>z-9=a+11i</math>. Then, <cmath>100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.</cmath>It follows that <math>z=9+\sqrt{221}+11i</math>, and the requested sum is <math>9+221=\boxed{230}</math>.<br />
<br />
(Solution by TheUltimate123)<br />
<br />
==Solution 2==<br />
<br />
We will use the fact that segments <math>AB</math> and <math>BC</math> are perpendicular in the complex plane if and only if <math>\frac{a-b}{b-c}\in i\mathbb{R}</math>. To prove this, when dividing two complex numbers you subtract the angle of one from the other, and if the two are perpendicular, subtracting these angles will yield an imaginary number with no real part. <br />
<br />
Now to apply this: <br />
<cmath>\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}</cmath><br />
<cmath>\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}</cmath><br />
<cmath>\frac{z^2-20z}{z^4-38z^3+341z^2+380z}</cmath><br />
<cmath>\frac{z(z-20)}{z(z+1)(z-19)(z-20)}</cmath><br />
<cmath>\frac{1}{(z+1)(z-19)}\in i\mathbb{R}</cmath><br />
<br />
The factorization of the nasty denominator above is made easier with the intuition that <math>(z-20)</math> must be a divisor for the problem to lead anywhere. Now we know <math>(z+1)(z-19)\in i\mathbb{R}</math> so using the fact that the imaginary part of <math>z</math> is <math>11i</math> and calling the real part r, <br />
<br />
<cmath>(r+1+11i)(r-19+11i)\in i\mathbb{R}</cmath><br />
<cmath>r^2-18r-140=0</cmath><br />
<br />
solving the above quadratic yields <math>r=9+\sqrt{221}</math> so our answer is <math>9+221=\boxed{230}</math><br />
<br />
==Solution 3==<br />
I would like to use a famous method, namely the coni method. <br />
<br />
Statement .If we consider there complex number <math>A,B,C</math> in argand plane then <math>\angle ABC =\arg{\frac{B-A}{B-C}}</math>.<br />
<br />
According to the question given, we can assume ,<math>A= f(f(z)),B=f(z),C= x</math> respectively.<br />
<br />
WLOG,<math>Z_1= \frac{f(f(z))-f(z)}{z-f(z)}</math>.<br />
According to the question <math>\arg{Z_1}=\frac{\pi}{2}</math>.<br />
<br />
So,<math>\Re (Z_1)=0</math>.<br />
<br />
Now, <math>Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}</math>.<br />
<br />
<math>\implies Z_1= -(z^2-18z-19)</math>.<br />
WLOG,<math>z=a+11i</math> .where <math>a=m+\sqrt{n}</math>.<br />
<br />
So,<math>\Re (Z_1)= -(a^2-18a-140)</math>.<br />
Solving,<math>a^2-18a-140=0</math> .get ,<br />
<br />
<math>a=9</math>±<math>\sqrt{221}</math>.<br />
So, possible value of <math>a=9+\sqrt{221}</math>.<br />
<br />
<math>m+n=\boxed{230}</math>.<br />
~ftheftics.<br />
<br />
==Proof of this method==<br />
Note that if we translate a triangle, the measures of all of its angles stay the same. So we can translate <math>ABC</math> on the complex plane so that <math>B=0</math>. Let the images of <math>A, B, C</math> be <math>A', B', C'</math> respectively. Then, we can use the formula:<br />
<br />
<math>re^{i\theta}=rcos(\theta)+i \cdot rsin(\theta)</math>. (This is known as Euler's Theorem.)<br />
<br />
Using Euler's theorem (represent each complex number in polar form, then use exponent identities), we can show that <math>arg(\frac{A'}{C'})=arg(A')-arg(C')=\angle(A'B'C')=\angle(ABC)</math>. So this method is valid.<br />
~Math4Life2020<br />
<br />
==Solution 4==<br />
It is well known that <math>AB</math> is perpendicular to <math>CD</math> iff <math>\frac{d-c}{b-a}</math> is a pure imaginary number. Here, we have that <math>A=z</math>, <math>B,C=f(z)</math>, and <math>D=f(f(z))</math>. This means that this is equivalent to <math>\frac{f(f(z))-f(z)}{f(z)-z}</math> being a pure imaginary number. Plugging in <math>f(z)=z^2-19z</math>, we have that <math>\frac{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}</math> being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to <math>(z-19)(z+1)</math> being pure imaginary. We let <math>z=a+bi</math>, so this is equivalent to <math>(a+bi-19)(a+bi+1)</math> being pure imaginary. Expanding the product, this is equivalent to <math>a^2+abi+a+abi-b^2+bi-19a-19bi-19</math> being pure imaginary. Taking the real part of this, and setting this equal to <math>0</math>, we have that <math>a^2-18a-b^2-19=0</math>. Since <math>b=11</math>, we have that <math>a^2-18a-140=0</math>. By the quadratic formula, <math>a=9 \pm \sqrt{221}</math>, and taking the positive root gives that <math>a=9+ \sqrt{221}</math>, so the answer is <math>9+221=230</math><br />
<br />
~smartninja2000<br />
<br />
==See also==<br />
{{AIME box|year=2019|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Math4life2020https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_23&diff=1414452019 AMC 10B Problems/Problem 232021-01-03T22:27:04Z<p>Math4life2020: Added another solution.</p>
<hr />
<div>==Problem==<br />
<br />
Points <math>A=(6,13)</math> and <math>B=(12,11)</math> lie on circle <math>\omega</math> in the plane. Suppose that the tangent lines to <math>\omega</math> at <math>A</math> and <math>B</math> intersect at a point on the <math>x</math>-axis. What is the area of <math>\omega</math>?<br />
<br />
<math>\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) }<br />
\frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}</math><br />
<br />
==Solution 1==<br />
First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is <math>(x, 0)</math>, the Pythagorean Theorem gives <math>\sqrt{(x-6)^2 + 13^2} = \sqrt{(x-12)^2 + 11^2}</math>. This simplifies to <math>x = 5</math>.<br />
<br />
Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) defined by the circle's center, <math>A</math>, <math>B</math>, and <math>(5, 0)</math> is cyclic. Therefore, we can apply Ptolemy's Theorem to give<br />
<math>2\sqrt{170}x = d \sqrt{40}</math>, where <math>x</math> is the radius of the circle and <math>d</math> is the distance between the circle's center and <math>(5, 0)</math>. Therefore, <math>d = \sqrt{17}x</math>. Using the Pythagorean Theorem on the triangle formed by the point <math>(5, 0)</math>, either one of <math>A</math> or <math>B</math>, and the circle's center, we find that <math>170 + x^2 = 17x^2</math>, so <math>x^2 = \frac{85}{8}</math>, and thus the answer is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 2 (coordinate bash)==<br />
We firstly obtain <math>x=5</math> as in Solution 1. Label the point <math>(5,0)</math> as <math>C</math>. The midpoint <math>M</math> of segment <math>AB</math> is <math>(9, 12)</math>. Notice that the center of the circle must lie on the line passing through the points <math>C</math> and <math>M</math>. Thus, the center of the circle lies on the line <math>y=3x-15</math>. <br />
<br />
Line <math>AC</math> is <math>y=13x-65</math>. Therefore, the slope of the line perpendicular to <math>AC</math> is <math>-\frac{1}{13}</math>, so its equation is <math>y=-\frac{x}{13}+\frac{175}{13}</math>. <br />
<br />
But notice that this line must pass through <math>A(6, 13)</math> and <math>(x, 3x-15)</math>. Hence <math>3x-15=-\frac{x}{13}+\frac{175}{13} \Rightarrow x=\frac{37}{4}</math>. So the center of the circle is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. <br />
<br />
Finally, the distance between the center, <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>, and point <math>A</math> is <math>\frac{\sqrt{170}}{4}</math>. Thus the area of the circle is <math>\boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 3==<br />
The midpoint of <math>AB</math> is <math>D(9,12)</math>. Let the tangent lines at <math>A</math> and <math>B</math> intersect at <math>C(a,0)</math> on the <math>x</math>-axis. Then <math>CD</math> is the perpendicular bisector of <math>AB</math>. Let the center of the circle be <math>O</math>. Then <math>\triangle AOC</math> is similar to <math>\triangle DAC</math>, so <math>\frac{OA}{AC} = \frac{AD}{DC}</math>.<br />
The slope of <math>AB</math> is <math>\frac{13-11}{6-12}=\frac{-1}{3}</math>, so the slope of <math>CD</math> is <math>3</math>. Hence, the equation of <math>CD</math> is <math>y-12=3(x-9) \Rightarrow y=3x-15</math>. Letting <math>y=0</math>, we have <math>x=5</math>, so <math>C = (5,0)</math>.<br />
<br />
Now, we compute <math>AC=\sqrt{(6-5)^2+(13-0)^2}=\sqrt{170}</math>,<br />
<math>AD=\sqrt{(6-9)^2+(13-12)^2}=\sqrt{10}</math>, and<br />
<math>DC=\sqrt{(9-5)^2+(12-0)^2}=\sqrt{160}</math>.<br />
<br />
Therefore <math>OA = \frac{AC\cdot AD}{DC}=\sqrt{\frac{85}{8}}</math>,<br />
and consequently, the area of the circle is <math>\pi\cdot OA^2 = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
<br />
==Solution 4 (how fast can you multiply two-digit numbers?)==<br />
Let <math>(x,0)</math> be the intersection on the x-axis. By Power of a Point Theorem, <math>(x-6)^2+13^2=(x-12)^2+11^2\implies x=5</math>. Then the equations are <math>13(x-6)+13=y</math> and <math>\frac{11}{7}(x-12)+11=y</math> for the tangent lines passing <math>A</math> and <math>B</math> respectively. Then the lines normal to them are <math>-\frac{1}{13}(x-6)+13=y</math> and <math>-\frac{7}{11}(x-12)+11=y</math>. Thus, <br />
<br />
<br />
<br />
<cmath>-\frac{7}{11}(x-12)+11=-\frac{1}{13}(x-6)+13</cmath><br />
<cmath>\frac{13\cdot7x-11x}{13\cdot11}=\frac{84\cdot13-6\cdot11-2\cdot11\cdot13}{11\cdot13}</cmath><br />
<cmath>13\cdot7x-11x=84\cdot13-6\cdot11-2\cdot11\cdot13</cmath><br />
<br />
After condensing, <math>x=\frac{37}{4}</math>. Then, the center of <math>\omega</math> is <math>\left(\frac{37}{4}, \frac{51}{4}\right)</math>. Apply distance formula. WLOG, assume you use <math>A</math>. Then, the area of <math>\omega</math> is <cmath>\sqrt{\frac{1^2}{4^2}+\frac{13^2}{4^2}}^2\pi=\frac{170\pi}{16} \implies \boxed{\textbf{(C) }\frac{85}{8}\pi}.</cmath><br />
<br />
==Solution 5 (power of a point)==<br />
<br />
Firstly, the point of intersection of the two tangent lines has an equal distance to points <math>A</math> and <math>B</math> due to power of a point theorem. This means we can easily find the point, which is <math>(5, 0)</math>. Label this point <math>X</math>. <math>\triangle{XAB}</math> is an isosceles triangle with lengths, <math>\sqrt{170}</math>, <math>\sqrt{170}</math>, and <math>2\sqrt{10}</math>. Label the midpoint of segment <math>AB</math> as <math>M</math>. The height of this triangle, or <math>\overline{XM}</math>, is <math>4\sqrt{10}</math>. Since <math>\overline{XM}</math> bisects <math>\overline{AB}</math>, <math>\overleftrightarrow{XM}</math> contains the diameter of circle <math>\omega</math>. Let the two points on circle <math>\omega</math> where <math>\overleftrightarrow{XM}</math> intersects be <math>P</math> and <math>Q</math> with <math>\overline{XP}</math> being the shorter of the two. Now let <math>\overline{MP}</math> be <math>x</math> and <math>\overline{MQ}</math> be <math>y</math>. By Power of a Point on <math>\overline{PQ}</math> and <math>\overline{AB}</math>, <math>xy = (\sqrt{10})^2 = 10</math>. Applying Power of a Point again on <math>\overline{XQ}</math> and <math>\overline{XA}</math>, <math>(4\sqrt{10}-x)(4\sqrt{10}+y)=(\sqrt{170})^2=170</math>. Expanding while using the fact that <math>xy = 10</math>, <math>y=x+\frac{\sqrt{10}}{2}</math>. Plugging this into <math>xy=10</math>, <math>2x^2+\sqrt{10}x-20=0</math>. Using the quadratic formula, <math>x = \frac{\sqrt{170}-\sqrt{10}}{4}</math>, and since <math>x+y=2x+\frac{\sqrt{10}}{2}</math>, <math>x+y=\frac{\sqrt{170}}{2}</math>. Since this is the diameter, the radius of circle <math>\omega</math> is <math>\frac{\sqrt{170}}{4}</math>, and so the area of circle <math>\omega</math> is <math>\frac{170}{16}\pi = \boxed{\textbf{(C) }\frac{85}{8}\pi}</math>.<br />
<br />
==Solution 6 (Similar to #3)==<br />
Let the tangent lines from A and B intersect at X. Let the center of <math>\omega</math> be C. Let the intersection of AB and CX be M. Using the techniques above, we get that the coordinate of X is <math>(5, 0)</math>. However, notice that CMX is the perpendicular bisector of AB. Thus, AM is the altitude from A to CX. Using the distance formula on AX, we get that the length of <math>AX=\sqrt{170}=\sqrt{17}\sqrt{10}</math>. Using the distance formula on AM, we get that <math>AM=\sqrt{10}</math>. Using the distance formula on MX, we get that <math>MX=4\sqrt{10}</math>. To get AC (the radius of <math>\omega</math>), we use either of these methods:<br />
<br />
Method 1: Since CAX is a right angle, the altitude AM is the geometric mean of XM and MC. We get that <math>MC=\frac{\sqrt{17}}{4}</math>. Thus, XC has length <math>XC=\frac{17\sqrt{10}}{4}</math>. Using the Pythagorean Theorem on CAX yields <math>CA=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Method 2: Note that CAX and AMX are similar. Thus, <math>\frac{AM}{MX}=\frac{AC}{AX}</math>. Solving for AC yields <math>\frac{AX \cdot AM}{MX}=\frac{\sqrt{170}}{4}</math>. <br />
<br />
Using the area formula for a circle yields that the area is <math>\frac{85\pi}{8} \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
==Video Solution==<br />
For those who want a video solution: (Is similar to Solution 1)<br />
https://youtu.be/WI2NVuIp1Ik<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{AMC12 box|year=2019|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Math4life2020https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_25&diff=1408802019 AMC 10B Problems/Problem 252020-12-29T02:43:10Z<p>Math4life2020: Added new solution</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}<br />
<br />
==Problem==<br />
<br />
How many sequences of <math>0</math>s and <math>1</math>s of length <math>19</math> are there that begin with a <math>0</math>, end with a <math>0</math>, contain no two consecutive <math>0</math>s, and contain no three consecutive <math>1</math>s?<br />
<br />
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math><br />
<br />
==Solution 1 (recursion)==<br />
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.<br />
Thus we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, where <math>f(n)</math> is the number of valid sequences of length <math>n</math>.<br />
<br />
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.<br />
<br />
It is easy to find <math>f(5) = 1</math> with the only possible sequence being <math>01010</math> and <math>f(6) = 2</math> with the only two possible sequences being <math>011010</math> and <math>010110</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 2 (casework)==<br />
After any particular <math>0</math>, the next <math>0</math> in the sequence must appear exactly <math>2</math> or <math>3</math> positions down the line. In this case, we start at position <math>1</math> and end at position <math>19</math>, i.e. we move a total of <math>18</math> positions down the line. Therefore, we must add a series of <math>2</math>s and <math>3</math>s to get <math>18</math>. There are a number of ways to do this:<br />
<br />
'''Case 1''': nine <math>2</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.<br />
<br />
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.<br />
<br />
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.<br />
<br />
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.<br />
<br />
==Solution 3 (casework and blocks)==<br />
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework: <br />
<br />
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>. <br />
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks. <br />
<br />
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways. <br />
<br />
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks. <br />
<br />
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>. <br />
<br />
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer. <br />
<br />
~Equinox8<br />
<br />
==Solution 4 (similar to #3)==<br />
Any valid sequence must start with a 0. We can then think of constructing a sequence as adding groups of terms to this 0, each ending in 0. (This is always possible, because every valid string ends in 0.) For example, we can represent the string 01011010110110 as: 0 - 10 - 110 - 10 - 110 - 110.<br />
To not have any consecutive 0s, we must have at least one 1 before the next 0. However, we cannot have 3 or more 1s before the next 0 because we cannot have 3 consecutive 1s. Consequently, we can only have one or two 1s. So we can have the groups: <br />
10, 110.<br />
<br />
After the initial 0, we have 18 digits left to fill in the string. Let the number of "10" blocks be x, and "110" be y. Then x and y must satisfy <math>2x+3y=18</math>. We recognize this as a Diophantine equation. Taking <math>(mod 2)</math> yields <math>y=0 (mod 2)</math>. Since x and y must both be nonnegative, we get the solutions (9, 0); (6, 2); (3, 4); (0, 6). We now handle each of these cases separately.<br />
<br />
(9, 0): Only one arrangement, namely all "01".<br />
<br />
(6, 2): Of the 8, we choose 2 to be "001". This has 8C2=28 cases.<br />
<br />
(3, 4): Of the 7, we choose 4 to be "001". This has 7C3=35 cases.<br />
<br />
(0, 6): Only one arrangement, namely all "001".<br />
<br />
Adding these, we have <math>1+28+35+1=65 \longrightarrow \boxed{(C)}</math>.<br />
~Math4Life2020<br />
<br />
==Video Solution==<br />
For those who want a video solution: https://youtu.be/VamT49PjmdI<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Math4life2020https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_19&diff=1406542017 AMC 8 Problems/Problem 192020-12-26T18:37:16Z<p>Math4life2020: Added another solution.</p>
<hr />
<div>==Problem 19==<br />
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ?<br />
<br />
<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math> <br />
<br />
==Solution 1==<br />
Factoring out <math>98!+99!+100!</math>, we have <math>98!(1+99+99*100)</math> which is <math>98!(10000)</math> Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.<br />
<br />
==Solution 2==<br />
The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math> to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.<br />
<br />
==Solution 3==<br />
We can rewrite the expression as <math>98!+99!+100!=98!(1+99+99\cdot100)=98!(100+99\cdot100)=98!\cdot10,000</math>.<br />
The exponent of <math>5</math> in <math>10,000</math> is <math>4</math>. Onto the <math>98!</math> part.<br />
<br />
Remember that <math>98!</math> is the product of the integers from 1 to 98. Among these, there are multiples of <math>5</math> and multiples of <math>25</math>.<br />
The number of multiples of <math>5</math> below or equal to <math>98</math> is <math>\left\lfloor\frac{98}{5}\right\rfloor</math> (try to see why), which is <math>19</math>. Every such number has a factor of <math>5</math>, so they contribute <math>1</math> to the total each. So these numbers contribute <math>19</math> to the exponent of <math>5</math> in <math>98!</math>.<br />
<br />
However, we forgot about multiples of <math>25</math>. Using similar logic, we have <math>\left\lfloor\frac{98}{25}\right\rfloor=3</math>, so there are 3 multiples of 25 in this range. Multiples of 25 contribute 2 to the total each. We already counted 1 of 2 contributions while counting multiples of 5. So we need to add another 1 to the exponent of 5 for every multiple of 25. So these numbers contribute 3 to the exponent of 5 in <math>98!</math>.<br />
<br />
<br />
We thus have that the exponent of <math>5</math> in <math>98!</math> is <math>19 + 3 = 22</math>, and so our answer is <math>22 + 4 = 26 \longrightarrow \boxed{(D)26}</math>.<br />
<br />
~Math4Life2020<br />
==See Also==<br />
{{AMC8 box|year=2017|num-b=18|num-a=20}}<br />
<br />
{{MAA Notice}}</div>Math4life2020https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=1404302007 AMC 12B Problems/Problem 242020-12-24T05:40:51Z<p>Math4life2020: Added another solution.</p>
<hr />
<div>== Problem ==<br />
How many pairs of positive integers <math>(a,b)</math> are there such that <math>\text{gcd}(a,b)=1</math> and <math>\frac{a}{b} + \frac{14b}{9a}</math> is an integer?<br />
<br />
<math>\mathrm {(A)}\ 4\quad\mathrm {(B)}\ 6\quad\mathrm {(C)}\ 9\quad\mathrm {(D)}\ 12\quad\mathrm {(E)}\ \text{infinitely many}</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer.<br />
<br />
Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math><br />
<br />
Since <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction(divide by <math>9</math> on both top and bottom) as <math>\frac{a^2 + 14n^2}{3an}</math><br />
<br />
Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>.<br />
<br />
But since <math>1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)</math>, we must have <math>n=1</math>, and thus <math>b=3</math>.<br />
<br />
For <math>b=3</math> the original fraction simplifies to <math>\frac{a^2 + 14}{3a}</math>.<br />
<br />
For that to be an integer, <math>a</math> must be a factor of <math>14</math>, and therefore we must have <math>a\in\{1,2,7,14\}</math>. Each of these values does indeed yield an integer.<br />
<br />
Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm{(A)}</math><br />
<br />
=== Solution 2 ===<br />
Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m</math> We get<br />
<br />
<math>9a^2 - 9mab + 14b^2 = 0</math><br />
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Factoring this, we get <math>4</math> equations-<br />
<br />
<math>(3a-2b)(3a-7b) = 0</math><br />
<br />
<math>(3a-b)(3a-14b) = 0</math><br />
<br />
<math>(a-2b)(9a-7b) = 0</math><br />
<br />
<math>(a-b)(9a-14b) = 0</math><br />
<br />
(It's all negative, because if we had positive signs, <math>a</math> would be the opposite sign of <math>b</math>)<br />
<br />
Now we look at these, and see that-<br />
<br />
<math>3a=2b</math> <br />
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<math>3a=b</math><br />
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<math>3a=7b</math> <br />
<br />
<math>3a=14b</math><br />
<br />
<math>a=2b</math> <br />
<br />
<math>9a=7b</math><br />
<br />
<math>a=b</math> <br />
<br />
<math>9a=14b</math><br />
<br />
This gives us <math>8</math> solutions, but we note that the middle term needs to give you back <math>9m</math>.<br />
<br />
For example, in the case <br />
<br />
<math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal by <math>-9m</math> for any integer <math>m</math>.<br />
<br />
Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math><br />
<br />
=== Solution 3 ===<br />
Let <math>u = \frac{a}{b}</math>. Then the given equation becomes <math>u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}</math>.<br />
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Let's set this equal to some value, <math>k \Rightarrow \frac{9u^2 + 14}{9u} = k</math>.<br />
<br />
Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>:<br />
<br />
<math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math><br />
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Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9k)^2 - 504}</math> is an integer.<br />
<br />
Let <math>\sqrt{(9k)^2 - 504} = x</math>. Squaring and rearranging yields:<br />
<br />
<math>(9k)^2 - x^2 = 504</math><br />
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<math>(9k+x)(9k-x) = 504</math>.<br />
<br />
In order for both <math>x</math> and <math>a</math> to be an integer, <math>9k + x</math> and <math>9k - x</math> must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let <math>9k + x = 2m</math> and <math>9k - x = 2n</math>.<br />
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Then:<br />
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<math>2m \cdot 2n = 504</math><br />
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<math>mn = 126</math>.<br />
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Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18),</math> and <math>(9,14)</math>.<br />
<br />
Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42)</math> and <math>(6,21)</math>. This means that there are <math>2</math> values of <math>k</math> such that <math>u</math> is an integer. But looking back at <math>u</math> in terms of <math>k</math>, we have <math>\pm</math>, meaning that there are <math>2</math> values of <math>u</math> for every <math>k</math>. Thus, the answer is <math>2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}</math>.<br />
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=== Solution 4 ===<br />
Rewriting the expression over a common denominator yields <math>\frac{9a^2 + 14b^2}{9ab}</math>. This expression must be equal to some integer <math>m</math>.<br />
<br />
Thus, <math>\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm</math>. Taking this <math>\pmod{a}</math> yields <math>14b^2 \equiv 0\pmod{a}</math>. Since <math>\gcd(a,b)=1</math>, <math>14 \equiv 0\pmod{a}</math>. This implies that <math>a|14</math> so <math>a = 1, 2, 7, 14</math>.<br />
<br />
We can then take <math>9a^2 + 14b^2 = 9abm \pmod{b}</math> to get that <math>9 \equiv 0 \pmod{b}</math>. Thus <math>b = 1, 3, 9</math>.<br />
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However, taking <math>9a^2 + 14b^2 = 9abm \pmod{3}</math>, <math>b^2 \equiv 0\pmod{3}</math> so <math>b</math> cannot equal 1.<br />
<br />
Also, note that if <math>b = 9</math>, <math>\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}</math>. Since <math>a|14</math>, <math>\frac{14}{a}</math> will be an integer, but <math>\frac{a}{9}</math> will not be an integer since none of the possible values of <math>a</math> are multiples of 9. Thus, <math>b</math> cannot equal 9.<br />
<br />
Thus, the only possible values of <math>b</math> is 3, and <math>a</math> can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is <math>\mathrm{(A)}</math>.<br />
<br />
=== Solution 5 (Similar to Solution 1) ===<br />
Rewriting <math>\frac{a}{b} + \frac{14b}{9a}</math> over a common denominator gives <math>\frac{9a^2 + 14b^2}{9ab}.</math><br />
<br />
Thus, we have <math>9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.</math><br />
<br />
Next, we have <math>ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.</math><br />
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Thus, <math>a \in (1,2,7,4).</math> <br />
<br />
Next, we have <math>b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.</math><br />
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Thus, <math>b \in (1,3,9).</math><br />
<br />
Now, we simply do casework on <math>b.</math> <br />
<br />
Plugging in <math>b = 1,3</math> and <math>9</math> gives that there are <math>4</math> total solutions for <math>(a,b).</math> <br />
<br />
~coolmath2017<br />
=== Solution 6 (Similar to solution 3) ===<br />
Let a/b = r. So <math>r + \frac{14}{9r}=I</math>, where I is an integer. Algebraic manipulations yield: <math>r^2-Ir+\frac{14}{9}=0</math>. The discriminant of this must be the square of a rational number, call this R. So <math>I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}</math>. I is <math>\frac{1}{2}</math> the sum of <math>I-R</math> and <math>I+R</math>. To have an integer sum, <math>I-R</math> and <math>I+R</math> must have the same denominator, namely 3. We proceed with casework.<br />
<br />
Case 1. <br />
<math>I+R=56/3</math>, <math>I-R=1/3</math>. This yields <math>I=19/2</math>, which is not an integer. This case produces 0 solutions.<br />
<br />
Case 2.<br />
<math>I+R=28/3</math>, <math>I-R=2/3</math>. This yields <math>I=5</math>. Substituting into our original equation yields: <math>r^2-5r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{1}{3}</math>, <math>r=\frac{14}{3}</math>. This case produces 2 solutions, namely (1,3) and (14,3).<br />
<br />
Case 3. <br />
<math>I+R=14/3</math>, <math>I-R=4/3</math>. This yields <math>I=3</math>. Substituting into our original equation yields: <math>r^2-3r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{2}{3}</math>, <math>r=\frac{7}{3}</math>. This case produces 2 solutions, namely (2,3) and (7,3).<br />
<br />
Case 4. <br />
<math>I+R=8/3</math>, <math>I-R=7/3</math>. This yields <math>I=5/2</math>, which is not an integer. This case produces 0 solutions.<br />
<br />
Altogether, we have 4 solutions, so our answer is <math>\boxed{(A)}</math>.<br />
<br />
~Math4Life2020<br />
== See Also ==<br />
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Math4life2020