https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathwizard07&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T05:25:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_9&diff=1048962010 AIME II Problems/Problem 92019-03-24T04:41:01Z<p>Mathwizard07: </p>
<hr />
<div>== Problem ==<br />
Let <math>ABCDEF</math> be a [[regular polygon|regular]] [[hexagon]]. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the [[midpoint]]s of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The [[segment]]s <math>\overline{AH}</math>, <math>\overline{BI}</math>, <math>\overline{CJ}</math>, <math>\overline{DK}</math>, <math>\overline{EL}</math>, and <math>\overline{FG}</math> bound a smaller regular hexagon. Let the [[ratio]] of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.<br />
<br />
__TOC__<br />
<br />
==Solution==<br />
<center><asy><br />
defaultpen(0.8pt+fontsize(12pt));<br />
pair A,B,C,D,E,F;<br />
pair G,H,I,J,K,L;<br />
A=dir(0);<br />
B=dir(60);<br />
C=dir(120);<br />
D=dir(180);<br />
E=dir(240);<br />
F=dir(300);<br />
draw(A--B--C--D--E--F--cycle,blue);<br />
<br />
G=(A+B)/2;<br />
H=(B+C)/2;<br />
I=(C+D)/2;<br />
J=(D+E)/2;<br />
K=(E+F)/2;<br />
L=(F+A)/2;<br />
<br />
int i;<br />
for (i=0; i<6; i+=1) {<br />
draw(rotate(60*i)*(A--H),dotted);<br />
}<br />
<br />
<br />
pair M,N,O,P,Q,R;<br />
M=extension(A,H,B,I);<br />
N=extension(B,I,C,J);<br />
O=extension(C,J,D,K);<br />
P=extension(D,K,E,L);<br />
Q=extension(E,L,F,G);<br />
R=extension(F,G,A,H);<br />
draw(M--N--O--P--Q--R--cycle,red);<br />
<br />
<br />
label('$A$',A,(1,0));<br />
label('$B$',B,NE);<br />
label('$C$',C,NW);<br />
label('$D$',D, W);<br />
label('$E$',E,SW);<br />
label('$F$',F,SE);<br />
label('$G$',G,NE);<br />
label('$H$',H, (0,1));<br />
label('$I$',I,NW);<br />
label('$J$',J,SW);<br />
label('$K$',K, S);<br />
label('$L$',L,SE);<br />
label('$M$',M);<br />
label('$N$',N);<br />
label('$O$',(0,0),NE); dot((0,0));<br />
</asy></center><br />
<br />
Let <math>M</math> be the intersection of <math>\overline{AH}</math> and <math>\overline{BI}</math><br />
<br />
and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>.<br />
<br />
Let <math>O</math> be the center.<br />
<br />
===Solution 1===<br />
Let <math>BC=2</math> (without loss of generality).<br />
<br />
Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, and so has degree <math>120^\circ</math>.<br />
<br />
Because <math>\triangle ABH</math> and <math>\triangle BCI</math> are rotational images of one another, we get that <math>\angle{MBH}=\angle{HAB}</math> and hence <math>\triangle ABH \sim \triangle BMH \sim \triangle BCI</math>.<br />
<br />
Using a similar argument, <math>NI=MH</math>, and<br />
<br />
<cmath>MN=BI-NI-BM=BI-(BM+MH).</cmath><br />
<br />
Applying the [[Law of cosines]] on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}</math><br />
<br />
<cmath>\begin{align*}\frac{BC+CI}{BI}&=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH} \\<br />
BM+MH&=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}} \\<br />
MN&=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}} \\<br />
\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}&=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}</cmath><br />
<br />
Thus, the answer is 4 + 7 = <math>\boxed{011}</math>.<br />
<br />
===Solution 2===<br />
We can use coordinates. Let <math>O</math> be at <math>(0,0)</math> with <math>A</math> at <math>(1,0)</math>, <br />
<br />
then <math>B</math> is at <math>(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>, <br />
<br />
<math>C</math> is at <math>(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>, <br />
<br />
<math>D</math> is at <math>(\cos(180^\circ),\sin(180^\circ))=(-1,0)</math>, <br />
<br />
<cmath>\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\<br />
&I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}</cmath><br />
<br />
<br/><br />
<br />
Line <math>AH</math> has the slope of <math>-\frac{\sqrt{3}}{2}</math><br />
and the equation of <math>y=-\frac{\sqrt{3}}{2}(x-1)</math><br />
<br />
<br/><br />
<br />
Line <math>BI</math> has the slope of <math>\frac{\sqrt{3}}{5}</math><br />
and the equation <math>y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)</math><br />
<br />
<br/> <br />
<br />
Let's solve the system of equation to find <math>M</math><br />
<br />
<cmath>\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\<br />
-5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\<br />
x&=\frac{1}{7} \\<br />
y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}</cmath><br />
<br />
Finally,<br />
<br />
<cmath>\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\<br />
&\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}</cmath><br />
<br />
Thus, the answer is <math>\boxed{011}</math>.<br />
<br />
==Solution 3==<br />
Use the diagram. Now notice that all of the "overlapping triangles" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=8|num-a=10|n=II}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_1&diff=1047902019 AIME II Problems/Problem 12019-03-22T22:09:48Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two different points, <math>C</math> and <math>D</math>, lie on the same side of line <math>AB</math> so that <math>\triangle ABC</math> and <math>\triangle BAD</math> are congruent with <math>AB = 9</math>, <math>BC=AD=10</math>, and <math>CA=DB=17</math>. The intersection of these two triangular regions has area <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
<asy><br />
unitsize(10);<br />
pair A = (0,0);<br />
pair B = (9,0);<br />
pair C = (15,8);<br />
pair D = (-6,8);<br />
draw(A--B--C--cycle);<br />
draw(B--D--A);<br />
label("$A$",A,dir(-120));<br />
label("$B$",B,dir(-60));<br />
label("$C$",C,dir(60));<br />
label("$D$",D,dir(120));<br />
label("$9$",(A+B)/2,dir(-90));<br />
label("$10$",(D+A)/2,dir(-150));<br />
label("$10$",(C+B)/2,dir(-30));<br />
label("$17$",(D+B)/2,dir(60));<br />
label("$17$",(A+C)/2,dir(120));<br />
<br />
draw(D--(-6,0)--A,dotted);<br />
label("$8$",(D+(-6,0))/2,dir(180));<br />
label("$6$",(A+(-6,0))/2,dir(-90));<br />
</asy><br />
- Diagram by Brendanb4321<br />
<br />
Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so <br />
that you have a rectangle. The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABE</math> and <math>\triangle DCE</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>E</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABE</math>.<br />
<cmath>\frac{7}{3}=\frac{y}{x}</cmath><br />
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath><br />
<cmath>7x=24-3x</cmath><br />
<cmath>10x=24</cmath><br />
<cmath>x=\frac{12}{5}</cmath><br />
<br />
This means that the area is <math>A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}</math>. This gets us <math>54+5=\boxed{059}.</math><br />
<br />
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_1&diff=1047882019 AIME II Problems/Problem 12019-03-22T22:08:25Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two different points, <math>C</math> and <math>D</math>, lie on the same side of line <math>AB</math> so that <math>\triangle ABC</math> and <math>\triangle BAD</math> are congruent with <math>AB = 9</math>, <math>BC=AD=10</math>, and <math>CA=DB=17</math>. The intersection of these two triangular regions has area <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
==Solution==<br />
<asy><br />
unitsize(10);<br />
pair A = (0,0);<br />
pair B = (9,0);<br />
pair C = (15,8);<br />
pair D = (-6,8);<br />
draw(A--B--C--cycle);<br />
draw(B--D--A);<br />
label("$A$",A,dir(-120));<br />
label("$B$",B,dir(-60));<br />
label("$C$",C,dir(60));<br />
label("$D$",D,dir(120));<br />
label("$9$",(A+B)/2,dir(-90));<br />
label("$10$",(D+A)/2,dir(-150));<br />
label("$10$",(C+B)/2,dir(-30));<br />
label("$17$",(D+B)/2,dir(60));<br />
label("$17$",(A+C)/2,dir(120));<br />
<br />
draw(D--(-6,0)--A,dotted);<br />
label("$8$",(D+(-6,0))/2,dir(180));<br />
label("$6$",(A+(-6,0))/2,dir(-90));<br />
</asy><br />
- Diagram by Brendanb4321<br />
<br />
Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so <br />
that you have a rectangle. The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABE</math> and <math>\triangle DCE</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>E</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABE</math>.<br />
<cmath>\frac{7}{3}=\frac{y}{x}</cmath><br />
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath><br />
<cmath>7x=24-3x</cmath><br />
<cmath>10x=24</cmath><br />
<cmath>x=\frac{12}{5}</cmath><br />
<br />
This means that the area is <math>A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}</math>. This gets us <math>54+5=\boxed{59}.</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_II_2012_Problems&diff=104667Mock AIME II 2012 Problems2019-03-18T19:51:09Z<p>Mathwizard07: /* Problem 6 */</p>
<hr />
<div>==Problem 1==<br />
Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 1 |Solution]]<br />
<br />
==Problem 2==<br />
Let <math>\{a_n\}</math> be a recursion defined such that <math>a_1=1, a_2=20</math>, and <math>a_n=\sqrt{\left| a_{n-1}^2-a_{n-2}^2 \right|}</math> where <math>n\ge 3</math>, and <math>n</math> is an integer. If <math>a_m=k</math> for <math>k</math> being a positive integer greater than <math>1</math> and <math>m</math> being a positive integer greater than 2, find the smallest possible value of <math>m+k</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 2| Solution]]<br />
<br />
==Problem 3==<br />
The <math>\textit{digital root}</math> of a number is defined as the result obtained by repeatedly adding the digits of the number until a single digit remains. For example, the <math>\textit{digital root}</math> of <math>237</math> is <math>3</math> (<math>2+3+7=12, 1+2=3</math>). Find the <math>\textit{digital root}</math> of <math>2012^{2012^{2012}}</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 3| Solution]]<br />
<br />
==Problem 4==<br />
Let <math>\triangle ABC</math> be a triangle, and let <math>I_A</math>, <math>I_B</math>, and <math>I_C</math> be the points where the angle bisectors of <math>A</math>, <math>B</math>, and <math>C</math>, respectfully, intersect the sides opposite them. Given that <math>AI_B=5</math>, <math>CI_B=4</math>, and <math>CI_A=3</math>, then the ratio <math>AI_C:BI_C</math> can be written in the form <math>m/n</math> where <math>m</math> and <math>n</math> are positive relatively prime integers. Find <math>m+n</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 4| Solution]]<br />
<br />
==Problem 5==<br />
A fair die with <math>12</math> sides numbered <math>1</math> through <math>12</math> inclusive is rolled <math>n</math> times. The probability that the sum of the rolls is <math>2012</math> is nonzero and is equivalent to the probability that a sum of <math>k</math> is rolled. Find the minimum value of k.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 5| Solution]]<br />
<br />
<br />
==Problem 6==<br />
A circle with radius <math>5</math> and center in the first quadrant is placed so that it is tangent to the <math>y</math>-axis. If the line passing through the origin that is tangent to the circle has slope <math>\dfrac{1}{2}</math>, then the <math>y</math>-coordinate of the center of the circle can be written in the form <math>\dfrac{m+\sqrt{n}}{p}</math> where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, and <math>m</math> and <math>p</math> are relatively prime. Find <math>m+n+p</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 6| Solution]]<br />
<br />
==Problem 7==<br />
Given <math> x, y </math> are positive real numbers that satisfy <math> 3x+4y+1=3\sqrt{x}+2\sqrt{y} </math>, then the value <math> xy </math> can be expressed as <math> \frac{m}{n} </math>, where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n </math>.<br />
[[Mock AIME II 2012 Problems/Problem 7| Solution]]<br />
<br />
==Problem 8==<br />
Let <math>A</math> be a point outside circle <math>\Omega</math> with center <math>O</math> and radius <math>9</math> such that the tangents from <math>A</math> to <math>\Omega</math>, <math>AB</math> and <math>AC</math>, form <math>\angle BAO=15^{\circ}</math>. Let <math>AO</math> first intersect the circle at <math>D</math>, and extend the parallel to <math>AB</math> from <math>D</math> to meet the circle at <math>E</math>. The length <math>EC^2=m+k\sqrt{n}</math>, where <math>m</math>,<math>n</math>, and <math>k</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 8| Solution]]<br />
<br />
==Problem 9==<br />
In <math>\triangle ABC</math>, <math>AB=12</math>, <math>AC=20</math>, and <math>\angle ABC=120^\circ</math>. <math>D, E,</math> and <math>F</math> lie on <math>\overline{AC}, \overline{AB}</math>, and <math>\overline{BC}</math>, respectively. If <math>AE=\frac{1}{4}AB, BF=\frac{1}{4}BC</math>, and <math>AD=\frac{1}{4}AC</math>, the area of <math>\triangle DEF</math> can be expressed in the form <math>\frac{a\sqrt{b}-c\sqrt{d}}{e}</math> where <math>a, b, c, d, e</math> are all positive integers, and <math>b</math> and <math>d</math> do not have any perfect squares greater than <math>1</math> as divisors. Find <math>a+b+c+d+e</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 9| Solution]]<br />
<br />
==Problem 10==<br />
Call a set of positive integers <math>\mathcal{S}</math> <math>\textit{lucky}</math> if it can be split into two nonempty disjoint subsets <math>\mathcal{A}</math> and <math>\mathcal{B}</math> with <math>A\cup B=S</math> such that the product of the elements in <math>\mathcal{A}</math> and the product of the elements in <math>\mathcal{B}</math> sum up to the cardinality of <math>\mathcal{S}</math>. Find the number of <math>\textit{lucky}</math> sets such that the largest element is less than <math>15</math>. (Disjoint subsets have no elements in common, and the cardinality of a set is the number of elements in the set.)<br />
<br />
[[Mock AIME II 2012 Problems/Problem 10| Solution]]<br />
<br />
==Problem 11==<br />
There exist real values of <math>a</math> and <math>b</math> such that <math>a+b=n</math>, <math>a^2+b^2=2n</math>, and <math>a^3+b^3=3n</math> for some value of <math>n</math>. Let <math>S</math> be the sum of all possible values of <math>a^4+b^4</math>. Find <math>S</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 11| Solution]]<br />
<br />
==Problem 12==<br />
Let <math>\log_{a}b=5\log_{b}ac^4=3\log_{c}a^2b</math>. Assume the value of <math>\log_ab</math> has three real solutions <math>x,y,z</math>. If <math>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m+n</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 12| Solution]]<br />
<br />
==Problem 13==<br />
Regular octahedron <math>ABCDEF</math> (such that points <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> are coplanar and form the vertices of a square) is divided along plane <math> \mathcal{P} </math>, parallel to line <math> BC </math>, into two polyhedra of equal volume. The cosine of the acute angle plane <math> \mathcal{P} </math> makes with plane <math> BCDE </math> is <math> \frac{1}{3} </math>. Given that <math> AB=30 </math>, find the area of the cross section made by plane <math> \mathcal{P} </math> with octahedron <math> ABCDEF </math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 13| Solution]]<br />
<br />
==Problem 14==<br />
Call a number a <math>\textit{near Carmichael number}</math> if for all prime divisors <math>p</math> of the <math>\textit{near Carmichael number}</math>, <math>n</math>, <math>(p-1)</math> divides <math>(n-1)</math> and <math>n</math> is not prime. Find the sum of all two digit <math>\textit{near Carmichael numbers}</math>.<br />
<br />
[[Mock AIME II 2012 Problems/Problem 14| Solution]]<br />
<br />
==Problem 15==<br />
Define <math>a_n=\sum_{i=0}^{n}f(i)</math> for <math>n\ge 0</math> and <math>a_n=0</math>. Given that <math>f(x)</math> is a polynomial, and <math>a_1, a_2+1, a_3+8, a_4+27, a_5+64, a_6+125, \cdots</math> is an arithmetic sequence, find the smallest positive integer value of <math>x</math> such that <math>f(x)<-2012</math>. <br />
<br />
[[Mock AIME II 2012 Problems/Problem 15| Solution]]</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2014_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_5&diff=1046112014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 52019-03-17T06:05:59Z<p>Mathwizard07: </p>
<hr />
<div>== Problem ==<br />
<br />
<math>5^n</math> is written on the blackboard. The sum of its digits is calculated. Then the sum of the<br />
digits of the result is calculated and so on until we have a single digit. If <math>n = 2014</math>, what<br />
is this digit?<br />
<br />
== Solution ==<br />
What you have to realize is that the sum of the digits of a number is the number <math>\pmod{9}</math>. We can prove this right now.<br />
\begin{equation}<br />
\end{equation}<br />
<br />
== See also ==<br />
{{UNM-PNM Math Contest box|year=2014|n=II|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2014_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_5&diff=1046102014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 52019-03-17T06:04:53Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<math>5^n</math> is written on the blackboard. The sum of its digits is calculated. Then the sum of the<br />
digits of the result is calculated and so on until we have a single digit. If <math>n = 2014</math>, what<br />
is this digit?<br />
<br />
== Solution ==<br />
What you have to realize is that the sum of the digits of a number is the number <math>\pmod{9}</math>. We can prove this right now.<br />
Any number can be represented in base-10 like this:<br />
<cmath>N=a_{1}\cdot10^n+a_{2}\cdot10^{n-1}+\cdots+a_{n-1}\cdot10^1+a_n,</cmath><br />
where <math>0<a_n<10</math>. Now realize <math>10\equiv1 \pmod{9}</math>, so you can use properties of mod to find <math>N \pmod{9}.</math><br />
<cmath>N \equiv a_1+a_2+\cdots+a_{n-1}+a_n \pmod{9}</cmath><br />
What is significant here? By repeatedly summing the digits, you are repeatedly looking for the remainder when that sum is divided by 9. This means that the answer will be the original number <math>\pmod{9}</math>. Either by using the Euler Theorem and the fact that <math>\phi(9)=6</math>, or just by finding a pattern, you see that <math>5^6 \equiv 1 \pmod{9}</math>. This means that <math>5^{2014}=(5^6)^{335}\cdot 5^4 \equiv 5^4 \pmod{9}</math>, which, if calculated properly (not that hard to do), gives you a digit of <math>\boxed{4}</math>.<br />
<br />
== See also ==<br />
{{UNM-PNM Math Contest box|year=2014|n=II|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2014_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_5&diff=1046092014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 52019-03-17T06:03:15Z<p>Mathwizard07: </p>
<hr />
<div>== Problem ==<br />
<br />
<math>5^n</math> is written on the blackboard. The sum of its digits is calculated. Then the sum of the<br />
digits of the result is calculated and so on until we have a single digit. If <math>n = 2014</math>, what<br />
is this digit?<br />
<br />
== Solution ==<br />
What you have to realize is that the sum of the digits of a number is the number <math>\pmod{9}</math>. We can prove this right now.<br />
Any number can be represented in base-10 like this:<br />
<cmath>N=a_{1}\cdot10^n+a_{2}\cdot10^{n-1}+\cdots+a_{n-1}\cdot10^1+a_n,</cmath><br />
where <math>0<a_n<10</math>. Now realize <math>10\equiv1 \pmod{9}</math>, so you can use properties of mod to find <math>N \pmod{9}.</math><br />
<cmath>N \equiv a_1+a_2+\cdots+a_{n-1}+a_n \pmod{9}</cmath><br />
What is significant here? By repeatedly summing the digits, you are repeatedly looking for the remainder when that sum is divided by 9. Either by using the Euler Theorem and the fact that <math>\phi(9)=6</math>, or just by finding a pattern, you see that <math>5^6 \equiv 1 \pmod{9}</math>. This means that <math>5^{2014}=(5^6)^{335}\cdot 5^4 \equiv 5^4 \pmod{9}</math>, which, if calculated properly (not that hard to do), gives you a digit of <math>\boxed{4}</math>.<br />
== See also ==<br />
{{UNM-PNM Math Contest box|year=2014|n=II|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1994_AJHSME_Problems&diff=1045751994 AJHSME Problems2019-03-16T16:11:39Z<p>Mathwizard07: /* Problem 25 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Which of the following is the largest?<br />
<br />
<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{3}{8} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{7}{24}</math><br />
<br />
[[1994 AJHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
<math>\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}+\dfrac{55}{10}=</math><br />
<br />
<math>\text{(A)}\ 4\dfrac{1}{2} \qquad \text{(B)}\ 6.4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math><br />
<br />
[[1994 AJHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
Each day Maria must work <math>8</math> hours. This does not include the <math>45</math> minutes she takes for lunch. If she begins working at <math>\text{7:25 A.M.}</math> and takes her lunch break at noon, then her working day will end at<br />
<br />
<math>\text{(A)}\ \text{3:40 P.M.} \qquad \text{(B)}\ \text{3:55 P.M.} \qquad \text{(C)}\ \text{4:10 P.M.} \qquad \text{(D)}\ \text{4:25 P.M.} \qquad \text{(E)}\ \text{4:40 P.M.}</math><br />
<br />
[[1994 AJHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
Which of the following represents the result when the figure shown below is rotated clockwise <math>120^\circ</math> around its center?<br />
<br />
<center><br />
<asy><br />
unitsize(6);<br />
draw(circle((0,0),5));<br />
draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle);<br />
draw(circle((-2.5,-1.5),1));<br />
draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle);<br />
</asy><br />
</center><br />
<br />
<asy><br />
unitsize(6);<br />
for (int i = 0; i < 5; ++i)<br />
{<br />
draw(circle((12*i,0),5));<br />
}<br />
draw((-1,2.5)--(1,2.5)--(0,2.5+sqrt(3))--cycle);<br />
draw(circle((-2.5,-1.5),1));<br />
draw((1.5,-1)--(3,0)--(4,-1.5)--(2.5,-2.5)--cycle);<br />
draw((14,-2)--(16,-2)--(15,-2+sqrt(3))--cycle);<br />
draw(circle((12,3),1));<br />
draw((10.5,-1)--(9,0)--(8,-1.5)--(9.5,-2.5)--cycle);<br />
draw((22,-2)--(20,-2)--(21,-2+sqrt(3))--cycle);<br />
draw(circle((27,-1),1));<br />
draw((24,1.5)--(22.75,2.75)--(24,4)--(25.25,2.75)--cycle);<br />
draw((35,2.5)--(37,2.5)--(36,2.5+sqrt(3))--cycle);<br />
draw(circle((39,-1),1));<br />
draw((34.5,-1)--(33,0)--(32,-1.5)--(33.5,-2.5)--cycle);<br />
draw((50,-2)--(52,-2)--(51,-2+sqrt(3))--cycle);<br />
draw(circle((45.5,-1.5),1));<br />
draw((48,1.5)--(46.75,2.75)--(48,4)--(49.25,2.75)--cycle);<br />
label("(A)",(0,5),N);<br />
label("(B)",(12,5),N);<br />
label("(C)",(24,5),N);<br />
label("(D)",(36,5),N);<br />
label("(E)",(48,5),N);<br />
</asy><br />
<br />
[[1994 AJHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Given that <math>\text{1 mile} = \text{8 furlongs}</math> and <math>\text{1 furlong} = \text{40 rods}</math>, the number of rods in one mile is<br />
<br />
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 320 \qquad \text{(C)}\ 660 \qquad \text{(D)}\ 1760 \qquad \text{(E)}\ 5280</math><br />
<br />
[[1994 AJHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math><br />
<br />
[[1994 AJHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
If <math>\angle A = 60^\circ </math>, <math>\angle E = 40^\circ </math> and <math>\angle C = 30^\circ </math>, then <math>\angle BDC = </math><br />
<br />
<center><br />
<asy><br />
pair A,B,C,D,EE;<br />
A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5);<br />
draw(A--C--D); draw(B--EE--A);<br />
dot(A); dot(B); dot(C); dot(D); dot(EE);<br />
label("$A$",A,SW);<br />
label("$B$",B,S);<br />
label("$C$",C,SE);<br />
label("$D$",D,NE);<br />
label("$E$",EE,N);<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ</math><br />
<br />
[[1994 AJHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
For how many three-digit whole numbers does the sum of the digits equal <math>25</math>?<br />
<br />
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math><br />
<br />
[[1994 AJHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
A shopper buys a <math>100</math> dollar coat on sale for <math>20\% </math> off. An additional <math>5</math> dollars are taken off the sale price by using a discount coupon. A sales tax of <math>8\% </math> is paid on the final selling price. The total amount the shopper pays for the coat is<br />
<br />
<math>\text{(A)}\ \text{81.00 dollars} \qquad \text{(B)}\ \text{81.40 dollars} \qquad \text{(C)}\ \text{82.00 dollars} \qquad \text{(D)}\ \text{82.08 dollars} \qquad \text{(E)}\ \text{82.40 dollars}</math><br />
<br />
[[1994 AJHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
For how many positive integer values of <math>N</math> is the expression <math>\dfrac{36}{N+2}</math> an integer?<br />
<br />
<math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12</math><br />
<br />
[[1994 AJHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
Last summer <math>100</math> students attended basketball camp. Of those attending, <math>52</math> were boys and <math>48</math> were girls. Also, <math>40</math> students were from Jonas Middle School and <math>60</math> were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?<br />
<br />
<math>\text{(A)}\ 20 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 52</math><br />
<br />
[[1994 AJHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
Each of the three large squares shown below is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?<br />
<br />
<center><br />
<asy><br />
unitsize(36);<br />
fill((0,0)--(1,0)--(1,1)--cycle,gray);<br />
fill((1,1)--(1,2)--(2,2)--cycle,gray);<br />
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);<br />
draw((1,0)--(1,2));<br />
draw((0,0)--(2,2));<br />
<br />
fill((3,1)--(4,1)--(4,2)--(3,2)--cycle,gray);<br />
draw((3,0)--(5,0)--(5,2)--(3,2)--cycle);<br />
draw((4,0)--(4,2));<br />
draw((3,1)--(5,1));<br />
<br />
fill((6,1)--(6.5,0.5)--(7,1)--(7.5,0.5)--(8,1)--(7.5,1.5)--(7,1)--(6.5,1.5)--cycle,gray);<br />
draw((6,0)--(8,0)--(8,2)--(6,2)--cycle);<br />
draw((6,0)--(8,2));<br />
draw((6,2)--(8,0));<br />
draw((7,0)--(6,1)--(7,2)--(8,1)--cycle);<br />
<br />
label("$I$",(1,2),N);<br />
label("$II$",(4,2),N);<br />
label("$III$",(7,2),N);<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ \text{The shaded areas in all three are equal.}</math><br />
<br />
<math>\text{(B)}\ \text{Only the shaded areas of }I\text{ and }II\text{ are equal.}</math><br />
<br />
<math>\text{(C)}\ \text{Only the shaded areas of }I\text{ and }III\text{ are equal.}</math><br />
<br />
<math>\text{(D)}\ \text{Only the shaded areas of }II\text{ and }III\text{ are equal.}</math><br />
<br />
<math>\text{(E)}\ \text{The shaded areas of }I, II\text{ and }III\text{ are all different.}</math><br />
<br />
[[1994 AJHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is<br />
<br />
<math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math><br />
<br />
[[1994 AJHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
Two children at a time can play pairball. For <math>90</math> minutes, with only two children playing at a time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is<br />
<br />
<math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36</math><br />
<br />
[[1994 AJHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
If this path is to continue in the same pattern:<br />
<br />
<center><br />
<asy><br />
unitsize(24);<br />
draw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,0)--(5,0)--(5,1)--(6,1));<br />
draw((2/3,1/5)--(1,0)--(2/3,-1/5)); draw((4/5,2/3)--(1,1)--(6/5,2/3)); draw((5/3,6/5)--(2,1)--(5/3,4/5)); draw((9/5,1/3)--(2,0)--(11/5,1/3));<br />
draw((8/3,1/5)--(3,0)--(8/3,-1/5)); draw((14/5,2/3)--(3,1)--(16/5,2/3)); draw((11/3,6/5)--(4,1)--(11/3,4/5)); draw((19/5,1/3)--(4,0)--(21/5,1/3));<br />
draw((14/3,1/5)--(5,0)--(14/3,-1/5)); draw((24/5,2/3)--(5,1)--(26/5,2/3)); draw((17/3,6/5)--(6,1)--(17/3,4/5));<br />
dot((0,0)); dot((1,0)); dot((1,1)); dot((2,1)); dot((2,0)); dot((3,0)); dot((3,1)); dot((4,1)); dot((4,0)); dot((5,0)); dot((5,1));<br />
label("$0$",(0,0),S);<br />
label("$1$",(1,0),S);<br />
label("$2$",(1,1),N);<br />
label("$3$",(2,1),N);<br />
label("$4$",(2,0),S);<br />
label("$5$",(3,0),S);<br />
label("$6$",(3,1),N);<br />
label("$7$",(4,1),N);<br />
label("$8$",(4,0),S);<br />
label("$9$",(5,0),S);<br />
label("$10$",(5,1),N);<br />
label("$\vdots$",(5.85,0.5),E);<br />
label("$\cdots$",(6.5,0.15),S);<br />
</asy><br />
</center><br />
<br />
then which sequence of arrows goes from point <math>425</math> to point <math>427</math>?<br />
<br />
<asy><br />
unitsize(24);<br />
dot((0,0)); dot((0,1)); dot((1,1));<br />
draw((0,0)--(0,1)--(1,1));<br />
draw((-1/5,2/3)--(0,1)--(1/5,2/3));<br />
draw((2/3,6/5)--(1,1)--(2/3,4/5));<br />
label("(A)",(-1/3,1/3),W);<br />
dot((4,0)); dot((5,0)); dot((5,1));<br />
draw((4,0)--(5,0)--(5,1));<br />
draw((14/3,1/5)--(5,0)--(14/3,-1/5));<br />
draw((24/5,2/3)--(5,1)--(26/5,2/3));<br />
label("(B)",(11/3,1/3),W);<br />
dot((8,1)); dot((8,0)); dot((9,0));<br />
draw((8,1)--(8,0)--(9,0));<br />
draw((39/5,1/3)--(8,0)--(41/5,1/3));<br />
draw((26/3,1/5)--(9,0)--(26/3,-1/5));<br />
label("(C)",(23/3,1/3),W);<br />
dot((12,1)); dot((13,1)); dot((13,0));<br />
draw((12,1)--(13,1)--(13,0));<br />
draw((38/3,6/5)--(13,1)--(38/3,4/5));<br />
draw((64/5,1/3)--(13,0)--(66/5,1/3));<br />
label("(D)",(35/3,1/3),W);<br />
dot((17,1)); dot((17,0)); dot((16,0));<br />
draw((17,1)--(17,0)--(16,0));<br />
draw((84/5,1/3)--(17,0)--(86/5,1/3));<br />
draw((49/3,1/5)--(16,0)--(49/3,-1/5));<br />
label("(E)",(47/3,1/3),W);<br />
</asy><br />
<br />
[[1994 AJHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
The perimeter of one square is <math>3</math> times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?<br />
<br />
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9</math><br />
<br />
[[1994 AJHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
Pauline Bunyan can shovel snow at the rate of <math>20</math> cubic yards for the first hour, <math>19</math> cubic yards for the second, <math>18</math> for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is <math>4</math> yards wide, <math>10</math> yards long, and covered with snow <math>3</math> yards deep, then the number of hours it will take her to shovel it clean is closest to<br />
<br />
<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 12</math><br />
<br />
[[1994 AJHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip?<br />
<br />
<center><br />
<asy><br />
import graph;<br />
unitsize(12);<br />
real a(real x) {return ((x-15)^2)/2;}<br />
real b(real x) {return ((x-25)^2)/2;}<br />
real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;}<br />
real d(real x) {return ((x-15)^2)*8/25-15;}<br />
real e(real x) {return ((x-25)^2)*8/25-15;}<br />
draw((0,9)--(0,0)--(11,0));<br />
draw((15,9)--(15,0)--(26,0));<br />
draw((30,9)--(30,0)--(41,0));<br />
draw((0,-6)--(0,-15)--(11,-15));<br />
draw((15,-6)--(15,-15)--(26,-15));<br />
draw((0,0)--(3,8)--(7,8)--(10,0));<br />
draw(graph(a,15,17));<br />
draw((17,2)--(18,8)--(22,8)--(23,2));<br />
draw(graph(b,23,25));<br />
draw(graph(c,30,40));<br />
draw((0,-15)--(5,-7)--(10,-15));<br />
draw(graph(d,15,20));<br />
draw(graph(e,20,25));<br />
for (int k=0; k<3; ++k)<br />
{<br />
label("d",(15*k-1,8),N); label("i",(15*k-1,7),N); label("s",(15*k-1,6),N); label("t",(15*k-1,5),N); label("a",(15*k-1,4),N); label("n",(15*k-1,3),N); label("c",(15*k-1,2),N); label("e",(15*k-1,1),N);<br />
label("time",(15*k+8,0),S);<br />
}<br />
for (int k=0; k<2; ++k)<br />
{<br />
label("d",(15*k-1,8-15),N); label("i",(15*k-1,7-15),N); label("s",(15*k-1,6-15),N); label("t",(15*k-1,5-15),N); label("a",(15*k-1,4-15),N); label("n",(15*k-1,3-15),N); label("c",(15*k-1,2-15),N); label("e",(15*k-1,1-15),N);<br />
label("time",(15*k+8,0-15),S);<br />
}<br />
label("(A)",(5,9),N); label("(B)",(20,9),N); label("(C)",(35,9),N); label("(D)",(5,-6),N); label("(E)",(20,-6),N);<br />
</asy><br />
</center><br />
<br />
[[1994 AJHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
Around the outside of a <math>4</math> by <math>4</math> square, construct four semicircles (as shown in the figure) with the four sides of the square as their diameters. Another square, <math>ABCD</math>, has its sides parallel to the corresponding sides of the original square, and each side of <math>ABCD</math> is tangent to one of the semicircles. The area of the square <math>ABCD</math> is<br />
<br />
<center><br />
<asy><br />
pair A,B,C,D;<br />
A = origin; B = (4,0); C = (4,4); D = (0,4);<br />
draw(A--B--C--D--cycle);<br />
draw(arc((2,1),(1,1),(3,1),CCW)--arc((3,2),(3,1),(3,3),CCW)--arc((2,3),(3,3),(1,3),CCW)--arc((1,2),(1,3),(1,1),CCW));<br />
draw((1,1)--(3,1)--(3,3)--(1,3)--cycle);<br />
dot(A); dot(B); dot(C); dot(D); dot((1,1)); dot((3,1)); dot((1,3)); dot((3,3));<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,NE);<br />
label("$D$",D,NW);<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ 16 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 64</math><br />
<br />
[[1994 AJHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
Let <math>W,X,Y</math> and <math>Z</math> be four different digits selected from the set<br />
<br />
<center><math>\{ 1,2,3,4,5,6,7,8,9\}.</math></center><br />
<br />
If the sum <math>\dfrac{W}{X} + \dfrac{Y}{Z}</math> is to be as small as possible, then <math>\dfrac{W}{X} + \dfrac{Y}{Z}</math> must equal<br />
<br />
<math>\text{(A)}\ \dfrac{2}{17} \qquad \text{(B)}\ \dfrac{3}{17} \qquad \text{(C)}\ \dfrac{17}{72} \qquad \text{(D)}\ \dfrac{25}{72} \qquad \text{(E)}\ \dfrac{13}{36}</math><br />
<br />
[[1994 AJHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
A gumball machine contains <math>9</math> red, <math>7</math> white, and <math>8</math> blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is<br />
<br />
<math>\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math><br />
<br />
[[1994 AJHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
The two wheels shown below are spun and the two resulting numbers are added. The probability that the sum is even is<br />
<br />
<center><br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((7,0),3));<br />
draw((0,0)--(3,0));<br />
draw((0,-3)--(0,3));<br />
draw((7,3)--(7,0)--(7+3*sqrt(3)/2,-3/2));<br />
draw((7,0)--(7-3*sqrt(3)/2,-3/2));<br />
draw((0,5)--(0,3.5)--(-0.5,4));<br />
draw((0,3.5)--(0.5,4));<br />
draw((7,5)--(7,3.5)--(6.5,4));<br />
draw((7,3.5)--(7.5,4));<br />
label("$3$",(-0.75,0),W);<br />
label("$1$",(0.75,0.75),NE);<br />
label("$2$",(0.75,-0.75),SE);<br />
label("$6$",(6,0.5),NNW);<br />
label("$5$",(7,-1),S);<br />
label("$4$",(8,0.5),NNE);<br />
</asy><br />
</center><br />
<br />
<math>\text{(A)}\ \dfrac{1}{6} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{1}{3} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{4}{9}</math><br />
<br />
[[1994 AJHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
If <math>X</math>, <math>Y</math> and <math>Z</math> are different digits, then the largest possible <math>3-</math>digit sum for<br />
<br />
<center><br />
<math>\begin{tabular}{ccc}<br />
X & X & X \\<br />
& Y & X \\<br />
+ & & X \\ \hline<br />
\end{tabular}</math><br />
</center><br />
<br />
has the form<br />
<br />
<math>\text{(A)}\ XXY \qquad \text{(B)}\ XYZ \qquad \text{(C)}\ YYX \qquad \text{(D)}\ YYZ \qquad \text{(E)}\ ZZY</math><br />
<br />
[[1994 AJHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
A <math>2</math> by <math>2</math> square is divided into four <math>1</math> by <math>1</math> squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.<br />
<br />
<math>\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16</math><br />
<br />
[[1994 AJHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
Find the sum of the digits in the answer to<br />
<br />
<center><br />
<math>\underbrace{9999\ldots 99}_{94\text{ nines}} \times \underbrace{4444\ldots 44}_{94\text{ fours}}</math><br />
</center><br />
<br />
where a string of <math>94</math> nines is multiplied by a string of <math>94</math> fours.<br />
<br />
<math>\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072</math><br />
<br />
[[1994 AJHSME Problems/Problem 25|Solution]]<br />
<br />
== See also ==<br />
{{AJHSME box|year=1994|before=[[1993 AJHSME Problems|1993 AJHSME]]|after=[[1995 AJHSME Problems|1995 AJHSME]]}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
<br />
{{MAA Notice}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1996_AJHSME_Problems/Problem_22&diff=1045741996 AJHSME Problems/Problem 222019-03-16T16:10:51Z<p>Mathwizard07: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle <math>ABC</math> is<br />
<br />
<asy><br />
for (int a = 0; a < 5; ++a)<br />
{<br />
for (int b = 0; b < 4; ++b)<br />
{<br />
dot((a,b));<br />
}<br />
}<br />
draw((0,0)--(3,2)--(4,3)--cycle);<br />
label("$A$",(0,0),SW);<br />
label("$B$",(3,2),SE);<br />
label("$C$",(4,3),NE);<br />
<br />
</asy><br />
<br />
<math>\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/2 \qquad \text{(C)}\ 3/4 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5/4</math><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
for (int a = 0; a < 5; ++a)<br />
{<br />
for (int b = 0; b < 4; ++b)<br />
{<br />
dot((a,b));<br />
}<br />
}<br />
draw((0,0)--(3,2)--(4,3)--cycle);<br />
draw((0,0)--(3,2)--(4,0)--cycle);<br />
draw((4,2)--(3,2)--(4,3)--cycle);<br />
draw((0,0)--(4,0)--(4,3)--cycle);<br />
draw((3,2)--(3,0)--cycle);<br />
label("$A$",(0,0),SW);<br />
label("$B$",(3,2),SE);<br />
label("$C$",(4,3),NE);<br />
label("$D$",(4,0),SE);<br />
label("$E$",(4,2),SE);<br />
label("$F$",(3,0),SE);<br />
</asy><br />
<br />
<math>\triangle ADC</math> takes up half of the 4x3 grid, so it has area of <math>6</math>.<br />
<br />
<math>\triangle ABD</math> has height of <math>BF = 2</math> and a base of <math>AD = 4</math>, for an area of <math>\frac{1}{2}\cdot 2 \cdot 4 = 4</math>.<br />
<br />
<math>\triangle CBD</math> has height of <math>BE = 1</math> and a base of <math>CD = 3</math>, for an area of <math>\frac{1}{2}\cdot 1 \cdot 3 = \frac{3}{2}</math><br />
<br />
Note that <math>\triangle ABC</math> can be found by taking <math>\triangle ADC</math>, and subtracting off <math>\triangle ABD</math> and <math>\triangle CBD</math>.<br />
<br />
Thus, the area of <math>\triangle ABC = 6 - 4 - \frac{3}{2} = \frac{1}{2}</math>, and the answer is <math>\boxed{B}</math>.<br />
<br />
There are other equivalent ways of dissecting the figure; right triangles <math>\triangle ABF, \triangle BCE</math> and rectangle <math>\square BEDF</math> can also be used. You can also use <math>\triangle{BEC}</math> and trapezoid <math>ADBE</math>.<br />
<br />
==Solution 2==<br />
<br />
Using the [[Shoelace Theorem]], and labelling the points <math>A(0,0), B(3,2), C(4,3)</math>, we find the area is:<br />
<br />
<math>(0,0)</math><br />
<br />
<math>(3,2)</math><br />
<br />
<math>(4,3)</math><br />
<br />
<math>(0,0)</math><br />
<br />
Area = <math>\frac{1}{2} = |(3\cdot 0 + 4\cdot 2 + 0\cdot 3) - (0\cdot 2 + 3\cdot 3 + 4\cdot 0)| = \frac{1}{2}</math>, which is option <math>\boxed{B}</math>.<br />
<br />
==See Also==<br />
{{AJHSME box|year=1996|num-b=21|num-a=23}}<br />
* [[AJHSME]]<br />
* [[AJHSME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_5&diff=1041032017 USAJMO Problems/Problem 52019-03-07T01:37:12Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>O</math> and <math>H</math> be the circumcenter and the orthocenter of an acute triangle <math>ABC</math>. Points <math>M</math> and <math>D</math> lie on side <math>BC</math> such that <math>BM = CM</math> and <math>\angle BAD = \angle CAD</math>. Ray <math>MO</math> intersects the circumcircle of triangle <math>BHC</math> in point <math>N</math>. Prove that <math>\angle ADO = \angle HAN</math>. <br />
<br />
==Solution==<br />
[asy] pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue);<br />
<br />
pair P = dir(-90); pair Q = dir(90); pair D = extension(A, P, B, C); pair O = origin; pair M = extension(B, C, O, P); pair N = 2*M-P;<br />
<br />
draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue);<br />
<br />
draw(A--D--N--O--cycle, red);<br />
<br />
dot("<math>A</math>", A, dir(A)); dot("<math>B</math>", B, dir(B)); dot("<math>C</math>", C, dir(C)); dot("<math>P</math>", P, dir(P)); dot("<math>Q</math>", Q, dir(Q)); dot("<math>D</math>", D, dir(225)); dot("<math>O</math>", O, dir(315)); dot("<math>M</math>", M, dir(315)); dot("<math>N</math>", N, dir(315));<br />
<br />
/* TSQ Source:<br />
<br />
A = dir 130 B = dir 220 C = dir 320 unitcircle 0.1 lightcyan / lightblue<br />
<br />
P = dir -90 Q = dir 90 D = extension A P B C R225 O = origin R315 M = extension B C O P R315 N = 2*M-P R315<br />
<br />
A--B--C--cycle lightblue A--P--Q lightblue A--N--D--O--A lightblue<br />
<br />
A--D--N--O--cycle 0.1 yellow / red<br />
[/asy]<br />
<br />
Suppose ray <math>OM</math> intersects the circumcircle of <math>BHC</math> at <math>N'</math>, and let the foot of the A-altitude of <math>ABC</math> be <math>E</math>. Note that <math>\angle BHE=90-\angle HBE=90-90+\angle C=\angle C</math>. Likewise, <math>\angle CHE=\angle B</math>. So, <math>\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C</math>.<br />
<math>BHCN'</math> is cyclic, so <math>\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A</math>. Also, <math>\angle BAC=\angle A</math>. These two angles are on different circles and have the same measure, but they point to the same line <math>BC</math>! Hence, the two circles must be congruent. (This is also a well-known result)<br />
<br />
We know, since <math>M</math> is the midpoint of <math>BC</math>, that <math>OM</math> is perpendicular to <math>BC</math>. <math>AH</math> is also perpendicular to <math>BC</math>, so the two lines are parallel. <math>AN</math> is a transversal, so <math>\angle HAN=\angle ANO</math>. We wish to prove that <math>\angle ANO=\angle ADO</math>, which is equivalent to <math>AOND</math> being cyclic.<br />
<br />
Now, assume that ray <math>OM</math> intersects the circumcircle of <math>ABC</math> at a point <math>P</math>. Point <math>P</math> must be the midpoint of <math>\stackrel{\frown}{BC}</math>. Also, since <math>AD</math> is an angle bisector, it must also hit the circle at the point <math>P</math>. The two circles are congruent, which implies <math>MN=MP\implies ND=DP\implies</math> NDP is isosceles. Angle ADN is an exterior angle, so <math>\angle ADN=\angle DNP+\angle DPO=2\angle DPO</math>. <br />
Assume WLOG that <math>\angle B>\angle C</math>. So, <math>\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}</math>.<br />
In addition, <math>\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A</math>. Combining these two equations, <math>\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180</math>. <br />
<br />
Opposite angles sum to <math>180</math>, so quadrilateral <math>AOND</math> is cyclic, and the condition is proved.<br />
<br />
-william122<br />
<br />
==See also==<br />
{{USAJMO newbox|year=2017|num-b=4|num-a=6}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2011_UNCO_Math_Contest_II_Problems/Problem_8&diff=1039732011 UNCO Math Contest II Problems/Problem 82019-03-02T01:51:15Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<br />
The integer <math>45</math> can be expressed as a sum of two squares as <math>45 = 3^2 + 6^2</math>.<br />
<br />
(a) Express <math>74</math> as the sum of two squares.<br />
<br />
(b) Express the product <math>45\cdot 74</math> as the sum of two squares.<br />
<br />
(c) Prove that the product of two sums of two squares, <math>(a^2+b^2)(c^2+d^2)</math> , can be represented<br />
as the sum of two squares.<br />
<br />
<br />
== Solution ==<br />
(a) By testing the perfect squares up to <math>74</math>, you can see that <math>\boxed{74=25+49=5^2+7^2}</math>.<br />
<br />
== See Also ==<br />
{{UNCO Math Contest box|n=II|year=2011|num-b=7|num-a=9}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2011_UNCO_Math_Contest_II_Problems/Problem_8&diff=1039722011 UNCO Math Contest II Problems/Problem 82019-03-02T01:50:50Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
<br />
The integer <math>45</math> can be expressed as a sum of two squares as <math>45 = 3^2 + 6^2</math>.<br />
<br />
(a) Express <math>74</math> as the sum of two squares.<br />
<br />
(b) Express the product <math>45\cdot 74</math> as the sum of two squares.<br />
<br />
(c) Prove that the product of two sums of two squares, <math>(a^2+b^2)(c^2+d^2)</math> , can be represented<br />
as the sum of two squares.<br />
<br />
<br />
== Solution ==<br />
(a) By testing the perfect squares up to <math>74</math>, you can see that <math>\boxed{74=25+49=5^2+7^2</math>}$.<br />
<br />
== See Also ==<br />
{{UNCO Math Contest box|n=II|year=2011|num-b=7|num-a=9}}</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_25&diff=1036431977 AHSME Problems/Problem 252019-02-20T23:41:15Z<p>Mathwizard07: /* Problem 25 */</p>
<hr />
<div>== Problem 25 ==<br />
Determine the largest positive integer <math>n</math> such that <math>1005!</math> is divisible by <math>10^n</math>.<br />
<math><br />
\textbf{(A) }102\qquad<br />
\textbf{(B) }112\qquad<br />
\textbf{(C) }249\qquad<br />
\textbf{(D) }502\qquad<br />
</math><br />
\\\\</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_25&diff=1036421977 AHSME Problems/Problem 252019-02-20T23:40:49Z<p>Mathwizard07: /* Problem 25 */</p>
<hr />
<div>== Problem 25 ==<br />
Determine the largest positive integer <math>n</math> such that <math>1005!</math> is divisible by <math>10^n</math>.<br />
<math><br />
\textbf{(A) }102\qquad<br />
\textbf{(B) }112\qquad<br />
\textbf{(C) }249\qquad<br />
\textbf{(D) }502\qquad<br />
\\\\<br />
</math></div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_25&diff=1036411977 AHSME Problems/Problem 252019-02-20T23:39:45Z<p>Mathwizard07: /* Problem 25 */</p>
<hr />
<div>== Problem 25 ==<br />
Determine the largest positive integer <math>n</math> such that <math>1005!</math> is divisible by <math>10^n</math>.<br />
\textbf{(A) }79\qquad<br />
\textbf{(B) }80\qquad<br />
\textbf{(C) }81\qquad<br />
\textbf{(D) }82\qquad<br />
\\\\</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=1977_AHSME_Problems/Problem_25&diff=1036401977 AHSME Problems/Problem 252019-02-20T23:37:46Z<p>Mathwizard07: Created page with "== Problem 25 =="</p>
<hr />
<div>== Problem 25 ==</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=1014492019 AMC 10A Problems/Problem 22019-02-09T21:29:08Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the hundreds digit of <math>(20!-15!)?</math><br />
<br />
<br />
<br />
== Solution ==<br />
<br />
The last three digits of <math>n!</math> for all <math>n\geq15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=1013782019 AMC 10A Problems/Problem 22019-02-09T20:14:35Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the hundreds digit of <math>(20!-15!)?</math><br />
<br />
<br />
<br />
== Solution ==<br />
<br />
The last three digits of <math>n!</math> for all <math>n>=15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=1013772019 AMC 10A Problems/Problem 22019-02-09T20:14:19Z<p>Mathwizard07: /* Solution */</p>
<hr />
<div>== Problem ==<br />
What is the hundreds digit of <math>(20!-15!)?</math><br />
<br />
<br />
<br />
== Solution ==<br />
<br />
The last three digit of <math>n!</math> for all <math>n>=15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=1013762019 AMC 10A Problems/Problem 22019-02-09T20:14:03Z<p>Mathwizard07: /* Problem */</p>
<hr />
<div>== Problem ==<br />
What is the hundreds digit of <math>(20!-15!)?</math><br />
<br />
<br />
<br />
== Solution ==<br />
<br />
The last three digit of <math>n!</math> for all <math>n>=15</math> is 000, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=1013702019 AMC 10A Problems/Problem 22019-02-09T20:11:31Z<p>Mathwizard07: Created page with "== Problem == What is the hundreds digit of <math>(20!-15!)?</math>"</p>
<hr />
<div>== Problem ==<br />
What is the hundreds digit of <math>(20!-15!)?</math></div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems&diff=1013682019 AMC 10A Problems2019-02-09T20:07:10Z<p>Mathwizard07: /* Problem 25 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
==Problem 2==<br />
What is the hundreds digit of <math>(20!-15!)\ ?</math><br />
<br />
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math><br />
<br />
==Problem 3==<br />
==Problem 4==<br />
==Problem 5==<br />
==Problem 6==<br />
==Problem 7==<br />
==Problem 8==<br />
==Problem 9==<br />
==Problem 10==<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
==Problem 14==<br />
==Problem 15==<br />
==Problem 16==<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
==Problem 20==<br />
==Problem 21==<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==<br />
For how many integers <math>n</math> between 1 and 50, inclusive, is<br />
<br />
<math>\frac{(n^2-1)!}{(n!)^{n}}</math><br />
<br />
an integer? (Recall that <math>0!=1</math>.)<br />
<br />
A. 31<br />
B. 32<br />
C. 33<br />
D. 34<br />
E. 35</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems&diff=1013652019 AMC 10A Problems2019-02-09T20:03:23Z<p>Mathwizard07: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
<br />
==Problem 2==<br />
What is the hundreds digit of <math>(20!-15!)</math>?<br />
<br />
A. 0 B. 1 C. 2 D. 3 E. 4<br />
<br />
==Problem 3==<br />
==Problem 4==<br />
==Problem 5==<br />
==Problem 6==<br />
==Problem 7==<br />
==Problem 8==<br />
==Problem 9==<br />
==Problem 10==<br />
==Problem 11==<br />
==Problem 12==<br />
==Problem 13==<br />
==Problem 14==<br />
==Problem 15==<br />
==Problem 16==<br />
==Problem 17==<br />
==Problem 18==<br />
==Problem 19==<br />
==Problem 20==<br />
==Problem 21==<br />
==Problem 22==<br />
==Problem 23==<br />
==Problem 24==<br />
==Problem 25==</div>Mathwizard07https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems&diff=1013642019 AMC 10A Problems2019-02-09T20:02:39Z<p>Mathwizard07: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9</cmath><br />
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math><br />
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==Problem 2==<br />
What is the hundreds digit of <math>(20!-15!)</math>?<br />
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==Problem 3==<br />
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==Problem 25==</div>Mathwizard07