https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Skittlesftw&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T14:25:48ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_1&diff=1008582004 AIME I Problems/Problem 12019-01-26T14:30:19Z<p>Skittlesftw: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The digits of a positive integer <math> n </math> are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when <math> n </math> is divided by <math>37</math>?<br />
<br />
== Solution ==<br />
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form <math>{\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} </math><math>= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.<br />
<br />
Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>.<br />
<br />
Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>, our answer.<br />
<br />
<br />
==Solution 2 ==<br />
<br />
For any n we are told that it is four consecutive integers in decreasing order when read from left to right. Thus for any number(<math>d</math>) <math>0-6</math>(<math>7-9</math> do not work because a digit cannot be greater than 9), <math>n</math> is equal to <math>(d)+10(d+1)+100(d+2)+1000(d+3)</math> or <math>1111d +3210</math>. Now we try this number for <math>d=0</math>. When <math>d=0</math>, <math>n=3210</math> and <math>3210</math> when divided by <math>37</math> has a remainder of 28. We now notice that every time you increase <math>d</math> by <math>1</math> you increase <math>n</math> by <math>1111</math> and <math>1111</math> has remainder <math>1</math> when divided by <math>37</math>. Thus, the remainder increases by <math>1</math> every time you increase <math>d</math> by <math>1</math>. Thus,<br />
<br />
When <math>d=0</math>, the remainder equals 28<br />
<br />
When <math>d=1</math>, the remainder equals 29<br />
<br />
When <math>d=2</math>, the remainder equals 30<br />
<br />
When <math>d=3</math>, the remainder equals 31<br />
<br />
When <math>d=4</math>, the remainder equals 32<br />
<br />
When <math>d=5</math>, the remainder equals 33<br />
<br />
When <math>d=6</math>, the remainder equals 34<br />
<br />
<br />
Thus the sum of the remainders is equal to <math>28+29+30+31+32+33+34</math> which is equal to <math>217</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=I|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Skittlesftwhttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_1&diff=1008572004 AIME I Problems/Problem 12019-01-26T14:29:51Z<p>Skittlesftw: </p>
<hr />
<div>== Problem ==<br />
The digits of a positive integer <math> n </math> are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when <math> n </math> is divided by <math>37</math>?<br />
<br />
== Solution ==<br />
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form <math>{\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} </math><math>= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.<br />
<br />
Now, note that <math>3\cdot 37 = 111</math> so <math>30 \cdot 37 = 1110</math>, and <math>90 \cdot 37 = 3330</math> so <math>87 \cdot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <math>n</math>, so in fact the remainders must equal <math>n + 28</math>.<br />
<br />
Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>, our answer.<br />
<br />
<br />
==Solution 2 ==<br />
<br />
For any n we are told that it is four consecutive integers in decreasing order when read from left to right. Thus for any number(<math>d</math>) <math>0-6</math>(<math>7-9</math> do not work because a digit cannot be greater than 9), <math>n</math> is equal to <math>(d)+10(d+1)+100(d+2)+1000(d+3)</math> or <math>1111d +3210</math>. Now we try this number for <math>d=0</math>. When <math>d=0</math>, <math>n=3210</math> and <math>3210</math> when divided by <math>37</math> has a remainder of 28. We now notice that every time you increase <math>d</math> by <math>1</math> you increase <math>n</math> by <math>1111</math> and <math>1111</math> has remainder <math>1</math> when divided by <math>37</math>. Thus, the remainder increases by <math>1</math> every time you increase <math>d</math> by <math>1</math>. Thus,<br />
<br />
When <math>d=0</math>, the remainder equals 28<br />
<br />
When <math>d=1</math>, the remainder equals 29<br />
<br />
When <math>d=2</math>, the remainder equals 30<br />
<br />
When <math>d=3</math>, the remainder equals 31<br />
<br />
When <math>d=4</math>, the remainder equals 32<br />
<br />
When <math>d=5</math>, the remainder equals 33<br />
<br />
When <math>d=6</math>, the remainder equals 34<br />
<br />
<br />
Thus the sum of the remainders is equal to <math>28+29+30+31+32+33+34</math> which is equal to <math>217</math>.<br />
<br />
-David Camacho<br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=I|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Skittlesftwhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=921422016 AMC 8 Problems/Problem 232018-02-20T19:01:59Z<p>Skittlesftw: /* Solution 1 */</p>
<hr />
<div>Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?<br />
<br />
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math><br />
<br />
Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB-60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>.<br />
<br />
==Solution 2==<br />
As in Solution 1, observe that <math>\triangle{EAB}</math> is equilateral. Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</math>. <br />
<br />
Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.<br />
<br />
<br />
<br />
{{AMC8 box|year=2016|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Skittlesftwhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=921412016 AMC 8 Problems/Problem 232018-02-20T19:01:32Z<p>Skittlesftw: /* Solution 1 */</p>
<hr />
<div>Two congruent circles centered at points <math>A</math> and <math>B</math> each pass through the other circle's center. The line containing both <math>A</math> and <math>B</math> is extended to intersect the circles at points <math>C</math> and <math>D</math>. The circles intersect at two points, one of which is <math>E</math>. What is the degree measure of <math>\angle CED</math>?<br />
<br />
<math>\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150</math><br />
<br />
==Solution 1==<br />
[asy]<br />
pair A=(0,0);<br />
pair B=(1,0);<br />
draw(circle((0,0),1));<br />
draw(circle((1,0),1));<br />
dot("A", A, dir(-45));<br />
dot("B", B, dir(225));<br />
pair C=(-1,0);<br />
pair D=(2,0);<br />
dot("C", C, dir(-45));<br />
dot("D", D);<br />
pair E=(.5,.86602540378);<br />
dot("E", E, dir(90));<br />
draw(E--A);<br />
draw(E--B);<br />
draw(A--B);<br />
draw(C--A);<br />
draw(B--D);<br />
draw(C--E);<br />
draw(E--D);<br />
[/asy]<br />
Drawing the diagram[SOMEONE DRAW IT PLEASE], we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB-60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>.<br />
<br />
==Solution 2==<br />
As in Solution 1, observe that <math>\triangle{EAB}</math> is equilateral. Therefore, <math>m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}</math>. Since <math>CD</math> is a straight line, we conclude that <math>m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}</math>. Since <math>BE=BD</math> (both are radii of the same circle), <math>\triangle{BED}</math> is isosceles, meaning that <math>m\angle{BED}=m\angle{BDE}=30^{\circ}</math>. Similarly, <math>m\angle{AEC}=m\angle{ACE}=30^{\circ}</math>. <br />
<br />
Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>.<br />
<br />
<br />
<br />
{{AMC8 box|year=2016|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Skittlesftwhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=886162000 AIME I Problems/Problem 62017-11-23T20:20:46Z<p>Skittlesftw: /* Problem */</p>
<hr />
<div>== Problem ==<br />
For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>?<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<cmath>\begin{eqnarray*}<br />
\frac{x+y}{2} &=& \sqrt{xy} + 2\\<br />
x+y-4 &=& 2\sqrt{xy}\\<br />
y - 2\sqrt{xy} + x &=& 4\\<br />
\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath><br />
<br />
Because <math>y > x</math>, we only consider <math>+2</math>.<br />
<br />
For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.<br />
<br />
The maximum that <math>\sqrt{y}</math> can be is <math>10^3 - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>.<br />
<!-- solution lost in edit conflict - azjps -<br />
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>.<br />
--><br />
<br />
=== Solution 2 ===<br />
<br />
<br />
Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math><br />
<br />
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath><br />
<cmath>a^2 + b^2 = 2ab + 4</cmath><br />
<cmath>(a-b)^2 = 4</cmath><br />
<cmath>(a-b) = \pm 2</cmath><br />
<br />
This makes counting a lot easier since now we just have to find all pairs <math>(a,b)</math> that differ by 2.<br />
<br />
<br />
Because <math>\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>.<br />
<br />
<br />
[[Without loss of generality]], let's say <math>a < b</math>.<br />
<br />
<br />
We can count even and odd pairs separately to make things easier*:<br />
<br />
<br />
Odd: <cmath>(1,3) , (3,5) , (5,7) . . . (997,999)</cmath><br />
<br />
<br />
Even: <cmath>(2,4) , (4,6) , (6,8) . . . (996,998)</cmath><br />
<br />
<br />
This makes 499 odd pairs and 498 even pairs, for a total of <math>\boxed{997}</math> pairs.<br />
<br />
<br />
<br />
<math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find.<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Skittlesftwhttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_6&diff=886152000 AIME I Problems/Problem 62017-11-23T20:20:10Z<p>Skittlesftw: /* Problem */</p>
<hr />
<div>== Problem ==<br />
For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math><br />
and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>?<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<cmath>\begin{eqnarray*}<br />
\frac{x+y}{2} &=& \sqrt{xy} + 2\\<br />
x+y-4 &=& 2\sqrt{xy}\\<br />
y - 2\sqrt{xy} + x &=& 4\\<br />
\sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}</cmath><br />
<br />
Because <math>y > x</math>, we only consider <math>+2</math>.<br />
<br />
For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation.<br />
<br />
The maximum that <math>\sqrt{y}</math> can be is <math>10^3 - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>.<br />
<!-- solution lost in edit conflict - azjps -<br />
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>.<br />
--><br />
<br />
=== Solution 2 ===<br />
<br />
<br />
Let <math>a^2</math> = <math>x</math> and <math>b^2</math> = <math>y</math><br />
<br />
Then <cmath>\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2</cmath><br />
<cmath>a^2 + b^2 = 2ab + 4</cmath><br />
<cmath>(a-b)^2 = 4</cmath><br />
<cmath>(a-b) = \pm 2</cmath><br />
<br />
This makes counting a lot easier since now we just have to find all pairs <math>(a,b)</math> that differ by 2.<br />
<br />
<br />
Because <math>\sqrt{10^6} = 10^3</math>, then we can use all positive integers less than 1000 for <math>a</math> and <math>b</math>.<br />
<br />
<br />
[[Without loss of generality]], let's say <math>a < b</math>.<br />
<br />
<br />
We can count even and odd pairs separately to make things easier*:<br />
<br />
<br />
Odd: <cmath>(1,3) , (3,5) , (5,7) . . . (997,999)</cmath><br />
<br />
<br />
Even: <cmath>(2,4) , (4,6) , (6,8) . . . (996,998)</cmath><br />
<br />
<br />
This makes 499 odd pairs and 498 even pairs, for a total of <math>\boxed{997}</math> pairs.<br />
<br />
<br />
<br />
<math>*</math>Note: We are counting the pairs for the values of <math>a</math> and <math>b</math>, which, when squared, translate to the pairs of <math>(x,y)</math> we are trying to find.<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Skittlesftw