https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Azax1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T09:20:56ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506232005 AMC 12B Problems/Problem 152013-01-31T20:53:46Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 1==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\boxed{d = 4}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506222005 AMC 12B Problems/Problem 152013-01-31T20:50:57Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 2==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\boxed{d = 4}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506212005 AMC 12B Problems/Problem 152013-01-31T20:50:46Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 2==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\boxed{\mathrm{(d = 4)}\}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506202005 AMC 12B Problems/Problem 152013-01-31T20:50:17Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 2==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>d = 4</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506192005 AMC 12B Problems/Problem 152013-01-31T20:49:49Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 2==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits mod 9, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\fbox{</math>d = 4<math>}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_15&diff=506182005 AMC 12B Problems/Problem 152013-01-31T20:49:36Z<p>Azax1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math> and no two of them are the same. Which of the following is '''not''' included among the eight digits?<br />
<br />
<math><br />
\mathrm{(A)}\ 1 \qquad<br />
\mathrm{(B)}\ 2 \qquad<br />
\mathrm{(C)}\ 3 \qquad<br />
\mathrm{(D)}\ 4 \qquad<br />
\mathrm{(E)}\ 5<br />
</math><br />
<br />
== Solution 2==<br />
<math>221</math> can be written as the sum of eight two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</math>, <math>\overline{cg}</math>, and <math>\overline{dh}</math>. Then <math>221= 10(a+b+c+d)+(e+f+g+h)</math>. The last digit of <math>221</math> is <math>1</math>, and <math>10(a+b+c+d)</math> won't affect the units digits, so <math>(e+f+g+h)</math> must end with <math>1</math>. The smallest value <math>(e+f+g+h)</math> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</math> or <math>21</math>. <br />
<br />
Case 1: <math>(e+f+g+h)=11</math><br />
<br />
The only distinct positive integers that can add up to <math>11</math> is <math>(1+2+3+5)</math>. So, <math>a</math>,<math>b</math>,<math>c</math>, and <math>d</math> must include four of the five numbers <math>(4,6,7,8,9)</math>. We have <math>10(a+b+c+d)=221-11=210</math>, or <math>a+b+c+d=21</math>. We can add all of <math>4+6+7+8+9=34</math>, and try subtracting one number to get to <math>21</math>, but to no avail. Therefore, <math>(e+f+g+h)</math> cannot add up to <math>11</math>.<br />
<br />
Case 2: <math>(e+f+g+h)=21</math><br />
<br />
Checking all the values for <math>e</math>,<math>f</math>,<math>g</math>,and <math>h</math> each individually may be time-consuming, instead of only having <math>1</math> solution like Case 1. We can try a different approach by looking at <math>(a+b+c+d)</math> first. If <math>(e+f+g+h)=21</math>, <math>10(a+b+c+d)=221-21=200</math>, or <math>(a+b+c+d)=20</math>. That means <math>(a+b+c+d)+(e+f+g+h)=21+20=41</math>. We know <math>(1+2+3+4+5+6+7+8+9)=45</math>, so the missing digit is <math>45-41=\boxed{\mathrm{(D)}\ 4}</math><br />
<br />
== Solution 2 ==<br />
Alternatively, we know that a number is congruent to the sum of its digits <math>\pmod 9</math>, so <math>221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d</math>, where <math>d</math> is some digit. Clearly, <math>\fbox{</math>d = 4<math>}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=B|num-b=14|num-a=16}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=488452012 AIME I Problems/Problem 62012-10-14T02:43:49Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>==Problem 6==<br />
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math><br />
<br />
===Solution 2===<br />
Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so each is of the form <math>\sin(\frac{2 \pi k}{142})</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive. This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=488442012 AIME I Problems/Problem 62012-10-14T02:43:27Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>==Problem 6==<br />
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math><br />
<br />
===Solution 2===<br />
Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so each is of the form <math>\sin(2 \pi k/142)</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive. This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=488432012 AIME I Problems/Problem 62012-10-14T02:43:12Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>==Problem 6==<br />
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math><br />
<br />
===Solution 2===<br />
Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so each is of the form sin<math>(2 \pi k/142)</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive. This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&diff=488422012 AIME I Problems/Problem 62012-10-14T02:42:56Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>==Problem 6==<br />
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math><br />
<br />
==Solutions==<br />
<br />
===Solution 1===<br />
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math><br />
<br />
===Solution 2===<br />
Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so each is of the form <math>sin(2 \pi k/142)</math> where <math>k</math> is a positive integer between <math>0</math> and <math>141</math> inclusive. This simplifies to <math>sin(pi*k/71)</math>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=481981983 AIME Problems/Problem 132012-09-04T00:22:50Z<p>Azax1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets, an alternating sum is defined as follows. Arrange the number in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for <math>\{1, 2, 4, 6,9\}</math> is <math>9-6+4-2+1=6</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<br />
<br />
== Solution 1==<br />
Let <math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>. <br />
<br />
Then the alternating sum of <math>S</math> plus the alternating sum of <math>S</math> with 7 included is 7. In mathematical terms, <math>S+ (S\cup 7)=7</math>. This is true because when we take an alternating sum, each term of <math>S</math> has the opposite sign of each corresponding term of <math>S\cup 7</math>.<br />
<br />
Because there are <math>63</math> of these pairs, the sum of all possible subsets of our given set is <math>63*7</math>. However, we forgot to include the subset that only contains <math>7</math>, so our answer is <math>64\cdot 7=\boxed{448}</math>.<br />
<br />
== Solution 2==<br />
<br />
Consider a given subset <math>T</math> of <math>S</math> that contains 7; then there is a subset <math>T'</math> which contains all the elements of <math>T</math> except for 7, and only those. Since each element of <math>T'</math> has one element fewer preceding it than it does in <math>T</math>, their signs are opposite; so the sum of the alternating sums of <math>T</math> and <math>T'</math> is equal to 7. There are <math>2^6</math> subsets containing 7, so our answer is <math>7 * 2^6 = \boxed{448}</math>.<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=12|num-a=14}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453752010 AIME I Problems/Problem 62012-03-11T21:54:39Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that equality holds when <math>y = 1</math> and therefore when <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(1) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,<br />
<br />
<math>\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1</math>,<br />
<br />
and thus by the [[sandwich theorem]] <math>\lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453742010 AIME I Problems/Problem 62012-03-11T21:54:02Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that equality holds when <math>y = 1</math> and therefore when <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(1) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,<br />
<br />
<math>\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1</math>,<br />
<br />
and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453732010 AIME I Problems/Problem 62012-03-11T21:53:23Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,<br />
<br />
<math>\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1</math>,<br />
<br />
and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453722010 AIME I Problems/Problem 62012-03-11T21:52:59Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,<br />
<br />
<math>\lim_{x \to 1} x - 2 = -1</math> and<br />
<math>\lim_{x \to 1} 2x - 3 = -1</math>,<br />
<br />
and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453712010 AIME I Problems/Problem 62012-03-11T21:51:32Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,<br />
<br />
<math>\lim_{x \to 1} x - 2 =</math> <math>lim_{x \to 1}</math> <math>2x - 3 = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453702010 AIME I Problems/Problem 62012-03-11T21:50:34Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x \to 1} x - 2</math> <math>=</math> <math>lim_{x \to 1} 2x - 3</math> <math>= -1</math>, and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453692010 AIME I Problems/Problem 62012-03-11T21:49:32Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x \to 1} x - 2 = lim_{x \to 1} 2x - 3 = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453682010 AIME I Problems/Problem 62012-03-11T21:48:26Z<p>Azax1: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x - 2 \to 1} = lim_{2x - 3 \to 1} = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{Q'(x) \to +\1} = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453672010 AIME I Problems/Problem 62012-03-11T21:47:40Z<p>Azax1: Another solution</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x - 2 \to +\1} = lim_{2x - 3 \to +\1} = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{Q'(x) \to +\1} = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=453662010 AIME I Problems/Problem 62012-03-11T21:47:22Z<p>Azax1: Another solution</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
<center><asy><br />
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br />
<br />
real P(real x) { return 8*(x-1)^2/5+1; }<br />
real Q(real x) { return (x-1)^2+1; }<br />
real R(real x) { return 2*(x-1)^2+1; }<br />
draw(graph(P,min,max),dark);<br />
draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));<br />
draw(graph(R,min,max),linetype("6 2")+linewidth(0.7));<br />
dot((1,1));<br />
label("$P(x)$",(max,P(max)),E,fontsize(10));<br />
label("$Q(x)$",(max,Q(max)),E,fontsize(10));<br />
label("$R(x)$",(max,R(max)),E,fontsize(10));<br />
<br />
/* axes */<br />
Label f; f.p=fontsize(8);<br />
xaxis(-2, 3, Ticks(f, 5, 1));<br />
yaxis(-1, 5, Ticks(f, 6, 1)); <br />
</asy></center><br />
<br />
Let <math>Q(x) = x^2 - 2x + 2</math>, <math>R(x) = 2x^2 - 4x + 3</math>. [[Completing the square]], we have <math>Q(x) = (x-1)^2 + 1</math>, and <math>R(x) = 2(x-1)^2 + 1</math>, so it follows that <math>P(x) \ge Q(x) \ge 1</math> for all <math>x</math> (by the [[Trivial Inequality]]). <br />
<br />
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.<br />
<br />
=== Solution 2 ===<br />
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:<br />
<br />
<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center><br />
<br />
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
<br />
=== Solution 3 ===<br />
<br />
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that <math>y = 1</math> and therefore <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(x) = 1</math>.<br />
<br />
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:<br />
<br />
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math><br />
<br />
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus<br />
<br />
<math>x - 2 \le Q'(x) \le 2x - 3</math><br />
<br />
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However, <math>\lim_{x - 2 \to +\1} = lim_{2x - 3 \to +\1} = -1</math>, and thus by the [[sandwich theorem]] <math>lim_{Q'(x) \to +\1} = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&diff=446692012 AMC 10A Problems/Problem 222012-02-11T17:06:17Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem 22 ==<br />
<br />
The sum of the first <math>m</math> positive odd integers is 212 more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math><br />
<br />
== Solution 1==<br />
<br />
The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>.<br />
<br />
Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>.<br />
<br />
Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), <math>4m^2 - 847</math> must be odd.<br />
<br />
Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7*11^2</math>.<br />
<br />
Our goal is to find possible values for <math>a</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.</math> We have three pairs of factors, <math>847*1, 7*121, and 11*77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2a</math>. Thus the possibilities for <math>a</math> are <math>423</math>, <math>57</math>, and <math>33</math>. <br />
<br />
Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\qquad\textbf{(B)}</math>.<br />
<br />
==Solution 2==<br />
<br />
As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have:<br />
<br />
<math>(n+a)^2 = n(n+1) + 212</math>.<br />
Expanding, grouping like terms and factoring, we get:<br />
<math>n = (212 - a^2)/(2a - 1)</math>.<br />
<br />
We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225</math>). Plugging in, the only values of <math>a</math> that give integral solutions are <math>1, 4,</math> and <math>6</math>. These gives <math>n</math> values of <math>211, 28,</math> and <math>16</math>, respectively. <math>211 + 28 + 16 = 255</math>. Hence, the answer is <math>\qquad\textbf{(B)}</math>.</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&diff=446682012 AMC 10A Problems/Problem 222012-02-11T17:05:56Z<p>Azax1: /* Solution 2 */</p>
<hr />
<div>== Problem 22 ==<br />
<br />
The sum of the first <math>m</math> positive odd integers is 212 more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math><br />
<br />
== Solution 1==<br />
<br />
The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>.<br />
<br />
Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>.<br />
<br />
Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), <math>4m^2 - 847</math> must be odd.<br />
<br />
Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7*11^2</math>.<br />
<br />
Our goal is to find possible values for <math>a</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.</math> We have three pairs of factors, <math>847*1, 7*121, and 11*77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2a</math>. Thus the possibilities for <math>a</math> are <math>423</math>, <math>57</math>, and <math>33</math>. <br />
<br />
Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\qquad\textbf{(B)}</math>.<br />
<br />
==Solution 2==<br />
<br />
As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have:<br />
<br />
<math>(n+a)^2 = n(n+1) + 212</math>.<br />
Expanding, grouping like terms and factoring, we get:<br />
<math>n = (212 - a^2)/(2a - 1)</math>.<br />
<br />
We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225). Plugging in, the only values of </math>a<math> that give integral solutions are </math>1, 4,<math> and </math>6<math>. These gives </math>n<math> values of </math>211, 28,<math> and </math>16<math>, respectively. </math>211 + 28 + 16 = 255<math>. Hence, the answer is </math>\qquad\textbf{(B)}$.</div>Azax1https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_22&diff=446672012 AMC 10A Problems/Problem 222012-02-11T17:05:02Z<p>Azax1: /* Solution */</p>
<hr />
<div>== Problem 22 ==<br />
<br />
The sum of the first <math>m</math> positive odd integers is 212 more than the sum of the first <math>n</math> positive even integers. What is the sum of all possible values of <math>n</math>?<br />
<br />
<math> \textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259 </math><br />
<br />
== Solution 1==<br />
<br />
The sum of the first <math>m</math> odd integers is given by <math>m^2</math>. The sum of the first <math>n</math> even integers is given by <math>n(n+1)</math>.<br />
<br />
Thus, <math>m^2 = n^2 + n + 212</math>. Since we want to solve for n, rearrange as a quadratic equation: <math>n^2 + n + (212 - m^2) = 0</math>.<br />
<br />
Use the quadratic formula: <math>n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}</math>. <math>n</math> is clearly an integer, so <math>1 - 4(212 - m^2) = 4m^2 - 847</math> must be not only a perfect square, but also an odd perfect square. This is because the entire expression must be an integer, and for the numerator to be even (divisible by 2), <math>4m^2 - 847</math> must be odd.<br />
<br />
Let <math>x</math> = <math>\sqrt{4m^2 - 847}</math>. (Note that this means that <math>n = \frac{-1 + x}{2}</math>.) This can be rewritten as <math>x^2 = 4m^2 - 847</math>, which can then be rewritten to <math>4m^2 - x^2 = 847</math>. Factor the left side by using the difference of squares. <math>(2m + x)(2m - x) = 847 = 7*11^2</math>.<br />
<br />
Our goal is to find possible values for <math>a</math>, then use the equation above to find <math>n</math>. The difference between the factors is <math>(2m + a) - (2m - a) = 2m + a - 2m + a = 2a.</math> We have three pairs of factors, <math>847*1, 7*121, and 11*77</math>. The differences between these factors are <math>846</math>, <math>114</math>, and <math>66</math> - those are all possible values for <math>2a</math>. Thus the possibilities for <math>a</math> are <math>423</math>, <math>57</math>, and <math>33</math>. <br />
<br />
Now plug in these values into the equation <math>n = \frac{-1 + x}{2}</math>. <math>n</math> can equal <math>211</math>, <math>28</math>, or <math>16</math>. Add <math>211 + 28 + 16 = 255</math>. The answer is <math>\qquad\textbf{(B)}</math>.<br />
<br />
==Solution 2==<br />
<br />
As above, start off by noting that the sum of the first <math>m</math> odd integers <math>= m^2</math> and the sum of the first <math>n</math> even integers <math>= n(n+1)</math>. Clearly <math>m > n</math>, so let <math>m = n + a</math>, where <math>a</math> is some positive integer. We have:<br />
<br />
<math>(n+a)^2 = n(n+1) + 212</math><br />
<math>n^2 + 2an + a^2 = n^2 + n + 212</math><br />
<math>2an + a^2 = n + 212</math><br />
<math>2an - n = 212 - a^2</math><br />
<math>n(2a - 1) = 212 - a^2</math><br />
<math>n = (212 - a^2)/(2a - 1)</math>.<br />
<br />
We know that <math>n</math> and <math>a</math> are both positive integers, so we need only check values of <math>a</math> from <math>1</math> to <math>14</math> (<math>14^2 = 196 < 212 < 15^2 = 225). Plugging in, the only values of </math>a<math> that give integral solutions are </math>1, 4,<math> and </math>6<math>. These gives </math>n<math> values of </math>211, 28,<math> and </math>16<math>, respectively. </math>211 + 28 + 16 = 255<math>. Hence, the answer is </math>\qquad\textbf{(B)}$.</div>Azax1