https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Cheetah01&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T11:46:36ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_17&diff=841822017 AMC 10B Problems/Problem 172017-02-24T01:23:47Z<p>Cheetah01: /* Solution 3: Answer Choices */</p>
<hr />
<div>{{duplicate|[[2017 AMC 12B Problems|2017 AMC 12B #11]] and [[2017 AMC 10B Problems|2017 AMC 10B #17]]}}<br />
<br />
==Problem==<br />
Call a positive integer <math>monotonous</math> if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, <math>3</math>, <math>23578</math>, and <math>987620</math> are monotonous, but <math>88</math>, <math>7434</math>, and <math>23557</math> are not. How many monotonous positive integers are there?<br />
<br />
<math>\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048</math><br />
<br />
==Solution 1==<br />
<br />
Case 1: monotonous numbers with digits in ascending order<br />
<br />
There are <math>\Sigma_{n=1}^{9} \binom{9}{n}</math> ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. The sum is equivalent to <math>\Sigma_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.</math><br />
<br />
Case 2: monotonous numbers with digits in descending order<br />
<br />
There are <math>\Sigma_{n=1}^{10} \binom{10}{n}</math> ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. The sum is equivalent to <math>\Sigma_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.</math> We discard the number 0 since it is not positive. Thus there are <math>1022</math> here. <br />
<br />
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are <math>511+1022-9=\boxed{\textbf{B} 1524}</math> monotonous numbers.<br />
<br />
==Solution 2==<br />
Like Solution 1, divide the problem into an increasing and decreasing case:<br />
<br />
<math>\bullet</math> Case 1: Monotonous numbers with digits in ascending order.<br />
<br />
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.<br />
<br />
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are <math>2^9 = 512</math> ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get <math>512-1=511</math> monotonous numbers for this case.<br />
<br />
<math>\bullet</math> Case 2: Monotonous numbers with digits in descending order.<br />
<br />
This time, we arrange all 10 digits in decreasing order and repeat the process to find <math>2^{10} = 1024</math> ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get <math>1024-2=1022</math> monotonous numbers for this case.<br />
<br />
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.<br />
<br />
Thus our final answer is <math>511+1022-9 = \boxed{\textbf{(B) } 1524}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Cheetah01https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=805572015 AMC 8 Problems/Problem 252016-10-09T01:36:13Z<p>Cheetah01: /* Solution 1 */</p>
<hr />
<div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br />
<br />
<math>\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17</math><br />
<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
</asy><br />
<br />
===Solution 1===<br />
We can draw a diagram as shown.<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
path arc = arc((2.5,4),1.5,0,90);<br />
pair P = intersectionpoint(arc,(0,5)--(5,5));<br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br />
draw(P--Pp--Ppp--Pppp--cycle);<br />
</asy> <br />
Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the <math>4</math> big triangles by <math>AA.</math> Let the height of a big triangle be <math>x</math> then <math>\tfrac{x}{x-1}=\tfrac{5-x}{1}</math>.<br />
<cmath>x=-x^2+6x-5</cmath><br />
<cmath>x^2-5x+5=0</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath><br />
Thus <math>x=\dfrac{5-\sqrt{5}}{2}</math>, because by symmetry, <math>x < \dfrac52</math>.<br />
<br />
This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math><br />
This the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}</math><br />
<br />
===Solution 2=== <br />
<br />
We draw a square as shown:<br />
<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br />
path arc = arc((2.5,4),1.5,0,90); <br />
pair P = intersectionpoint(arc,(0,5)--(5,5)); <br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br />
draw(P--Pp--Ppp--Pppp--cycle); <br />
filldraw((1,4)--P--(4,4)--cycle,red);<br />
filldraw((4,4)--Pppp--(4,1)--cycle,red);<br />
filldraw((1,1)--Ppp--(4,1)--cycle,red);<br />
filldraw((1,1)--Pp--(1,4)--cycle,red);<br />
</asy><br />
<br />
<br />
We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. The area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=\boxed{{\textbf{(C)}}~15}</math>.<br />
<br />
===Solution 3===<br />
<br />
<br />
Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length <math>1</math>, and let's label the other legs <math>x</math> for one of the triangles and <math>y</math> for the other. Note that <math>x + y = 3</math>.<br />
The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are <math>4</math> of each. So now we need to find <math>4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)</math>.<br />
<br />
<math>(4)\frac{x}{2} + (4)\frac{y}{2}</math><br />
<math>\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)</math><br />
<math>\Rightarrow~~4\left(\frac{x+y}{2}\right)</math><br />
Remember that <math>x+y=3</math>, so substituting this in we find that the area of all of the triangles is <math>4\left(\frac{3}{2}\right) = 6</math>. <br />
The area of the <math>4</math> unit squares is <math>4</math>, so the area of the square we need is <math>25- (4+6) = \boxed{\textbf{(C)}~15}</math><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
Thank you for reading the solutions of the 2015 AMC 8 Problems made by people on AoPS.</div>Cheetah01https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=805562015 AMC 8 Problems/Problem 252016-10-09T01:35:59Z<p>Cheetah01: /* Solution 1 */</p>
<hr />
<div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br />
<br />
<math>\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17</math><br />
<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
</asy><br />
<br />
===Solution 1===<br />
We can draw a diagram as shown below.<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
path arc = arc((2.5,4),1.5,0,90);<br />
pair P = intersectionpoint(arc,(0,5)--(5,5));<br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br />
draw(P--Pp--Ppp--Pppp--cycle);<br />
</asy> <br />
Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the <math>4</math> big triangles by <math>AA.</math> Let the height of a big triangle be <math>x</math> then <math>\tfrac{x}{x-1}=\tfrac{5-x}{1}</math>.<br />
<cmath>x=-x^2+6x-5</cmath><br />
<cmath>x^2-5x+5=0</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath><br />
Thus <math>x=\dfrac{5-\sqrt{5}}{2}</math>, because by symmetry, <math>x < \dfrac52</math>.<br />
<br />
This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math><br />
This the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{\textbf{(C)}~15}</math><br />
<br />
===Solution 2=== <br />
<br />
We draw a square as shown:<br />
<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); <br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); <br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); <br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); <br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); <br />
path arc = arc((2.5,4),1.5,0,90); <br />
pair P = intersectionpoint(arc,(0,5)--(5,5)); <br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; <br />
draw(P--Pp--Ppp--Pppp--cycle); <br />
filldraw((1,4)--P--(4,4)--cycle,red);<br />
filldraw((4,4)--Pppp--(4,1)--cycle,red);<br />
filldraw((1,1)--Ppp--(4,1)--cycle,red);<br />
filldraw((1,1)--Pp--(1,4)--cycle,red);<br />
</asy><br />
<br />
<br />
We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. The area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=\boxed{{\textbf{(C)}}~15}</math>.<br />
<br />
===Solution 3===<br />
<br />
<br />
Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length <math>1</math>, and let's label the other legs <math>x</math> for one of the triangles and <math>y</math> for the other. Note that <math>x + y = 3</math>.<br />
The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are <math>4</math> of each. So now we need to find <math>4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)</math>.<br />
<br />
<math>(4)\frac{x}{2} + (4)\frac{y}{2}</math><br />
<math>\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)</math><br />
<math>\Rightarrow~~4\left(\frac{x+y}{2}\right)</math><br />
Remember that <math>x+y=3</math>, so substituting this in we find that the area of all of the triangles is <math>4\left(\frac{3}{2}\right) = 6</math>. <br />
The area of the <math>4</math> unit squares is <math>4</math>, so the area of the square we need is <math>25- (4+6) = \boxed{\textbf{(C)}~15}</math><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}<br />
Thank you for reading the solutions of the 2015 AMC 8 Problems made by people on AoPS.</div>Cheetah01