https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ryjs&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T06:14:02ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_24&diff=1210861986 AJHSME Problems/Problem 242020-04-17T18:05:58Z<p>Ryjs: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
The <math>600</math> students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately<br />
<br />
<math>\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}</math><br />
<br />
==Solution==<br />
<br />
Imagine that we run the computer many times. In roughly <math>1/3</math> of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly <math>1/3</math> cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is <math>\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}</math>, or <math>\boxed{\text{B}}</math>.<br />
<br />
(The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.)<br />
==Solution 2==<br />
There are <math>\binom{3}{1}</math> ways to choose which group the three kids are in and the chance that all three are in the same group is <math>\frac{1}{27}</math>. Hence <math>\frac{1}{9}</math> or <math>\boxed {B}</math>.<br />
<br />
==Solution 3==<br />
One of the statements, that there are <math>600</math> students in the school is redundant. Taking that there are <math>3</math> students and there are <math>3</math> groups, we can easily deduce there are <math>81</math> ways to group the <math>3</math> students, and there are <math>3</math> ways to group them in the same <math>1</math> group, so we might think <math>\frac{3}{54}=\frac{1}{27}</math> is the answer but as there are 3 groups we do <math>\frac{1}{27} (3)=\frac{1}{9}</math> which is <math>\boxed{\text{(B)}}</math>.<br />
<br />
==See Also==<br />
<br />
{{AJHSME box|year=1986|num-b=23|num-a=25}}<br />
[[Category:Introductory Probability Problems]]<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208502000 AMC 8 Problems/Problem 142020-04-12T04:23:09Z<p>Ryjs: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
-ryjs<br />
<br />
==Solution 3==<br />
Experimentation gives<br />
<cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath><br />
<br />
<cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath><br />
<br />
Using this we have<br />
<cmath>19^{19} + 99^{99}</cmath><br />
<cmath>9^{19} + 9^{99}</cmath><br />
<br />
Both <math>19</math> and <math>99</math> are odd, so we are left with<br />
<cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math><br />
-ryjs<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208492000 AMC 8 Problems/Problem 142020-04-12T04:22:47Z<p>Ryjs: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
-ryjs<br />
<br />
==Solution 3==<br />
Experimentation gives<br />
<cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath><br />
<br />
<cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath><br />
<br />
Using this we have<br />
\begin{align*}<br />
19^{19} + 99^{99} \\<br />
9^{19} + 9^{99} \\<br />
\end{align*}<br />
<br />
Both <math>19</math> and <math>99</math> are odd, so we are left with<br />
<cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math><br />
-ryjs<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208482000 AMC 8 Problems/Problem 142020-04-12T04:19:13Z<p>Ryjs: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
-ryjs<br />
<br />
==Solution 3==<br />
We find a pattern: both of our numbers have units digit <math>9.</math><br />
<br />
Experimentation gives that powers of <math>9</math> alternate between units digits <math>9</math> (for odd powers) and <math>1</math> (for even powers).<br />
<br />
Since both of our powers (<math>19</math> and <math>99</math>) are odd, we are left with <math>9+9=18,</math> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math><br />
<br />
-ryjs<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208472000 AMC 8 Problems/Problem 142020-04-12T04:18:27Z<p>Ryjs: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
-ryjs<br />
<br />
==Solution 3==<br />
We find a pattern: both of our numbers have units digit <math>9.</math><br />
<br />
Experimentation gives that powers of <math>9</math> alternate between units digits <math>9</math> (for odd powers) and <math>1</math> (for even powers).<br />
<br />
Since both of our powers (<math>19</math> and <math>99</math>) are odd, we are left with <math>9+9=18,</math> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math><br />
<br />
-ryjs<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208462000 AMC 8 Problems/Problem 142020-04-12T04:18:02Z<p>Ryjs: added solution 3</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
-ryjs<br />
<br />
==Solution 3==<br />
We find a pattern: both of our numbers have units digit <math>9.</math><br />
<br />
Experimentation gives that powers of <math>9</math> alternate between units digits <math>9</math> (for odd powers) and <math>1</math> (for even powers).<br />
<br />
Since both <math>19</math> and <math>99</math> are odd, we are left with <math>9+9=18,</math> which has units digit \boxed{(\textbf{D}) \ 8}.<br />
<br />
-ryjs<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208452000 AMC 8 Problems/Problem 142020-04-12T04:16:05Z<p>Ryjs: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_14&diff=1208442000 AMC 8 Problems/Problem 142020-04-12T04:15:47Z<p>Ryjs: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the units digit of <math>19^{19} + 99^{99}</math>?<br />
<br />
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math><br />
<br />
==Solution==<br />
<br />
Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>. <br />
<br />
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.<br />
<br />
==Solution 2==<br />
Using modular arithmetic:<br />
<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
Similarly,<br />
<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath><br />
<br />
We have <br />
<cmath>-1^{19} + -1^{99} = -1 + -1 \equiv \boxed{(\text{D })8} \pmod{10}</cmath><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=120714User:Piphi2020-04-08T17:01:44Z<p>Ryjs: /* User Count */</p>
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|}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_13&diff=1203682016 AMC 8 Problems/Problem 132020-04-01T03:09:37Z<p>Ryjs: /* Solution 3(Complementary Counting) */</p>
<hr />
<div>Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?<br />
<br />
<math>\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}</math><br />
<br />
==Solution 1==<br />
The product can only be <math>0</math> if one of the numbers is 0. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}</math> so the answer is <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.<br />
<br />
==Solution 2==<br />
There are a total of <math>30</math> possibilities, because the numbers are different. We want <math>0</math> to be the product so one of the numbers is <math>0</math>. There are <math>5</math> possibilities where <math>0</math> is chosen for the first number and there are <math>5</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.<br />
==Solution 3 (Complementary Counting)==<br />
Because the only way the product of the two numbers is <math>0</math> is if one of the numbers we choose is <math>0,</math> we calculate the probability of NOT choosing a <math>0.</math> We get <math>\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.</math> Therefore our answer is <math>1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.</math><br />
<br />
<br />
{{AMC8 box|year=2016|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_24&diff=1200211997 AHSME Problems/Problem 242020-03-23T22:52:44Z<p>Ryjs: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A rising number, such as <math>34689</math>, is a positive integer each digit of which is larger than each of the digits to its left. There are <math>\binom{9}{5} = 126</math> five-digit rising numbers. When these numbers are arranged from smallest to largest, the <math>97^{\text{th}}</math> number in the list does not contain the digit<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
<br />
The list starts with <math>12345</math>. There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.<br />
<br />
Thus, the <math>71^{\text{st}}</math> number is <math>23456</math>. There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.<br />
<br />
Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>. Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{\text{th}}</math> number, which does not contain the digit <math>\boxed{\textbf{(B)} \ 5}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_24&diff=1200201997 AHSME Problems/Problem 242020-03-23T22:52:04Z<p>Ryjs: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A rising number, such as <math>34689</math>, is a positive integer each digit of which is larger than each of the digits to its left. There are <math>\binom{9}{5} = 126</math> five-digit rising numbers. When these numbers are arranged from smallest to largest, the <math>97^{\text{th}}</math> number in the list does not contain the digit<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
<br />
The list starts with <math>12345</math>. There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.<br />
<br />
Thus, the <math>71^{\text{st}}</math> number is <math>23456</math>. There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.<br />
<br />
Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>. Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{th}</math> number, which does not contain the digit <math>\boxed{\textbf{(B)} \ 5}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_24&diff=1200191997 AHSME Problems/Problem 242020-03-23T22:51:41Z<p>Ryjs: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A rising number, such as <math>34689</math>, is a positive integer each digit of which is larger than each of the digits to its left. There are <math>\binom{9}{5} = 126</math> five-digit rising numbers. When these numbers are arranged from smallest to largest, the <math>97^{\text{th}}</math> number in the list does not contain the digit<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
<br />
The list starts with <math>12345</math>. There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.<br />
<br />
Thus, the <math>71^{\text{st}}</math> number is <math>23456</math>. There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.<br />
<br />
Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>. Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{th}</math> number, which does not contain the digit <math>5</math>. The answer is <math>\boxed{\qquad\textbf{(B)} \ 5}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_24&diff=1200181997 AHSME Problems/Problem 242020-03-23T22:51:13Z<p>Ryjs: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
A rising number, such as <math>34689</math>, is a positive integer each digit of which is larger than each of the digits to its left. There are <math>\binom{9}{5} = 126</math> five-digit rising numbers. When these numbers are arranged from smallest to largest, the <math>97^{\text{th}}</math> number in the list does not contain the digit<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
<br />
The list starts with <math>12345</math>. There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.<br />
<br />
Thus, the <math>71^{\text{st}}</math> number is <math>23456</math>. There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.<br />
<br />
Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>. Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{th}</math> number, which does not contain the digit <math>5</math>. The answer is <math>\boxed{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_24&diff=1200171997 AHSME Problems/Problem 242020-03-23T22:50:59Z<p>Ryjs: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A rising number, such as <math>34689</math>, is a positive integer each digit of which is larger than each of the digits to its left. There are <math>\binom{9}{5} = 126</math> five-digit rising numbers. When these numbers are arranged from smallest to largest, the <math>97^{th}</math> number in the list does not contain the digit<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math><br />
<br />
==Solution==<br />
<br />
The list starts with <math>12345</math>. There are <math>\binom{8}{4} = 70</math> four-digit rising numbers that do not begin with <math>1</math>, and thus also <math>70</math> five digit rising numbers that do begin with <math>1</math> that are formed by simply putting a <math>1</math> before the four digit number.<br />
<br />
Thus, the <math>71^{\text{st}}</math> number is <math>23456</math>. There are <math>\binom{6}{3} = 20</math> three-digit rising numbers that do not begin with a <math>1,2</math> or <math>3</math>, and thus <math>20</math> five digit rising numbers that begin with a <math>23</math>.<br />
<br />
Thus, the <math>91^{\text{st}}</math> number is <math>24567</math>. Counting up, <math>24568, 24569, 24578, 24579, 24589, 24678</math> is the <math>97^{th}</math> number, which does not contain the digit <math>5</math>. The answer is <math>\boxed{B}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1997|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_22&diff=1167491978 AHSME Problems/Problem 222020-02-03T17:38:21Z<p>Ryjs: /* Solution */</p>
<hr />
<div>The following four statements, and only these are found on a card:<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,1);<br />
B=(0,5);<br />
C=(11,5);<br />
D=(11,1);<br />
E=(0,4);<br />
F=(0,3);<br />
G=(0,2);<br />
draw(A--B--C--D--cycle);<br />
label("On this card exactly one statement is false.", B, SE);<br />
label("On this card exactly two statements are false.", E, SE);<br />
label("On this card exactly three statements are false.", F, SE);<br />
label("On this card exactly four statements are false.", G, SE);<br />
</asy><br />
<br />
(Assume each statement is either true or false.) Among them the number of false statements is exactly<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 4 </math><br />
<br />
== Solution ==<br />
<br />
There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math>. If there are <math>0</math> true on the card, statement <math>4</math> ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is <math>\boxed{\textbf{(D) } 3}</math>, since <math>3</math> are false and only the third statement ("On this card exactly three statements are false") is correct.</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_22&diff=1137311978 AHSME Problems/Problem 222019-12-30T01:10:37Z<p>Ryjs: </p>
<hr />
<div>The following four statements, and only these are found on a card:<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,1);<br />
B=(0,5);<br />
C=(11,5);<br />
D=(11,1);<br />
E=(0,4);<br />
F=(0,3);<br />
G=(0,2);<br />
draw(A--B--C--D--cycle);<br />
label("On this card exactly one statement is false.", B, SE);<br />
label("On this card exactly two statements are false.", E, SE);<br />
label("On this card exactly three statements are false.", F, SE);<br />
label("On this card exactly four statements are false.", G, SE);<br />
</asy><br />
<br />
(Assume each statement is either true or false.) Among them the number of false statements is exactly<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 4 </math><br />
<br />
== Solution ==<br />
<br />
There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math>. If there are <math>0</math> true on the card, statement <math>4</math> ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is <math>\textbf{(D)}\ 3</math>, since <math>3</math> are false and only the third statement ("On this card exactly three statements are false") is correct.</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems&diff=1137301978 AHSME Problems2019-12-30T01:09:52Z<p>Ryjs: /* Problem 22 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
If <math>1-\frac{4}{x}+\frac{4}{x^2}=0</math>, then <math>\frac{2}{x}</math> equals<br />
<br />
<math>\textbf{(A) }-1\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }-1\text{ or }2\qquad <br />
\textbf{(E) }-1\text{ or }-2 </math><br />
<br />
[[1978 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is <br />
<br />
<math>\textbf{(A) }\frac{1}{\pi^2}\qquad<br />
\textbf{(B) }\frac{1}{\pi}\qquad<br />
\textbf{(C) }1\qquad<br />
\textbf{(D) }\pi\qquad<br />
\textbf{(E) }\pi^2 </math> <br />
<br />
[[1978 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
For all non-zero numbers <math>x</math> and <math>y</math> such that <math>x = 1/y</math>, <math>\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)</math> equals<br />
<br />
<math>\textbf{(A) }2x^2\qquad<br />
\textbf{(B) }2y^2\qquad<br />
\textbf{(C) }x^2+y^2\qquad<br />
\textbf{(D) }x^2-y^2\qquad <br />
\textbf{(E) }y^2-x^2 </math><br />
<br />
[[1978 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</math> is equal to<br />
<br />
<math>\textbf{(A) }1111\qquad<br />
\textbf{(B) }2222\qquad<br />
\textbf{(C) }3333\qquad<br />
\textbf{(D) }1212\qquad <br />
\textbf{(E) }4242 </math> <br />
<br />
[[1978 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Four boys bought a boat for <math>\textdollar 60</math>. The first boy paid one half of the sum of the amounts paid by the other boys; <br />
the second boy paid one third of the sum of the amounts paid by the other boys; <br />
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay? <br />
<br />
<math>\textbf{(A) }\textdollar 10\qquad<br />
\textbf{(B) }\textdollar 12\qquad<br />
\textbf{(C) }\textdollar 13\qquad<br />
\textbf{(D) }\textdollar 14\qquad<br />
\textbf{(E) }\textdollar 15</math> <br />
<br />
[[1978 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The number of distinct pairs <math>(x,y)</math> of real numbers satisfying both of the following equations: <br />
<br />
<cmath>x=x^2+y^2 \\ y=2xy</cmath> <br />
is<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }3\qquad <br />
\textbf{(E) }4 </math><br />
<br />
[[1978 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
Opposite sides of a regular hexagon are <math>12</math> inches apart. The length of each side, in inches, is<br />
<br />
<math>\textbf{(A) }7.5\qquad<br />
\textbf{(B) }6\sqrt{2}\qquad<br />
\textbf{(C) }5\sqrt{2}\qquad<br />
\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad <br />
\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad<br />
\textbf{(E) }4\sqrt{3} </math><br />
<br />
[[1978 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
If <math>x\neq y</math> and the sequences <math>x,a_1,a_2,y</math> and <math>x,b_1,b_2,b_3,y</math> each are in arithmetic progression, then <math>(a_2-a_1)/(b_2-b_1)</math> equals<br />
<br />
<math>\textbf{(A) }\frac{2}{3}\qquad<br />
\textbf{(B) }\frac{3}{4}\qquad<br />
\textbf{(C) }1\qquad<br />
\textbf{(D) }\frac{4}{3}\qquad<br />
\textbf{(E) }\frac{3}{2} </math> <br />
<br />
[[1978 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
If <math>x<0</math>, then <math>\left|x-\sqrt{(x-1)^2}\right|</math> equals<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }1-2x\qquad<br />
\textbf{(C) }-2x-1\qquad<br />
\textbf{(D) }1+2x\qquad <br />
\textbf{(E) }2x-1 </math><br />
<br />
[[1978 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
If <math>\mathit{B}</math> is a point on circle <math>\mathit{C}</math> with center <math>\mathit{P}</math>, then the set of all points <math>\mathit{A}</math> in the plane of circle <br />
<math>\mathit{C}</math> such that the distance between <math>\mathit{A}</math> and <math>\mathit{B}</math> is less than or equal to the distance between <math>\mathit{A}</math> <br />
and any other point on circle <math>\mathit{C}</math> is<br />
<br />
<math>\textbf{(A) }\text{the line segment from }P \text{ to }B\qquad\\<br />
\textbf{(B) }\text{the ray beginning at }P \text{ and passing through }B\qquad\\<br />
\textbf{(C) }\text{a ray beginning at }B\qquad\\<br />
\textbf{(D) }\text{a circle whose center is }P\qquad\\<br />
\textbf{(E) }\text{a circle whose center is }B </math> <br />
<br />
[[1978 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
If <math>r</math> is positive and the line whose equation is <math>x + y = r</math> is tangent to the circle whose equation is <math>x^2 + y ^2 = r</math>, then <math>r</math> equals<br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }\sqrt{2}\qquad <br />
\textbf{(E) }2\sqrt{2} </math><br />
<br />
[[1978 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
In <math>\triangle ADE</math>, <math>\measuredangle ADE=140^\circ</math>, points <math>B</math> and <math>C</math> lie on sides <math>AD</math> and <math>AE</math>, <br />
respectively, and points <math>A,~B,~C,~D,~E</math> are distinct.* If lengths <math>AB,~BC,~CD</math>, and <math>DE</math> are all equal, <br />
then the measure of <math>\measuredangle EAD</math> is<br />
* The specification that points <math>A,B,C,D,E</math> be distinct was not included in the original statement of the problem. <br />
If <math>B=D</math>, then <math>C=E</math> and <math>\measuredangle EAD=20^\circ</math>. <br />
<br />
<math>\textbf{(A) }5^\circ\qquad<br />
\textbf{(B) }6^\circ\qquad<br />
\textbf{(C) }7.5^\circ\qquad<br />
\textbf{(D) }8^\circ\qquad<br />
\textbf{(E) }10^\circ</math><br />
<br />
<br />
[[1978 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
If <math>a,b,c</math>, and <math>d</math> are non-zero numbers such that <math>c</math> and <math>d</math> are the solutions of <math>x^2+ax+b=0</math> and <math>a</math> and <math>b</math> are <br />
the solutions of <math>x^2+cx+d=0</math>, then <math>a+b+c+d</math> equals<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }-2\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }4\qquad <br />
\textbf{(E) }(-1+\sqrt{5})/2 </math> <br />
<br />
[[1978 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
If an integer <math>n > 8</math> is a solution of the equation <math>x^2 - ax+b=0</math> and the representation of <math>a</math> in the base-<math>n</math> number system is <math>18</math>, <br />
then the base-n representation of <math>b</math> is<br />
<br />
<math>\textbf{(A)}\ 18 \qquad<br />
\textbf{(B)}\ 20 \qquad<br />
\textbf{(C)}\ 80 \qquad<br />
\textbf{(D)}\ 81 \qquad<br />
\textbf{(E)}\ 280 </math><br />
<br />
[[1978 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is<br />
<br />
<math>\textbf{(A) }-\frac{4}{3}\qquad<br />
\textbf{(B) }-\frac{3}{4}\qquad<br />
\textbf{(C) }\frac{3}{4}\qquad<br />
\textbf{(D) }\frac{4}{3}\qquad\\<br />
\textbf{(E) }\text{not completely determined by the given information} </math><br />
<br />
[[1978 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
In a room containing <math>N</math> people, <math>N > 3</math>, at least one person has not shaken hands with everyone else in the room. <br />
What is the maximum number of people in the room that could have shaken hands with everyone else?<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }N-1\qquad<br />
\textbf{(D) }N\qquad <br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
[[1978 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
If <math>k</math> is a positive number and <math>f</math> is a function such that, for every positive number <math>x</math>, <math>\left[f(x^2+1)\right]^{\sqrt{x}}=k</math>; <br />
then, for every positive number <math>y</math>, <math>\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}</math> is equal to<br />
<br />
<math>\textbf{(A) }\sqrt{k}\qquad<br />
\textbf{(B) }2k\qquad<br />
\textbf{(C) }k\sqrt{k}\qquad<br />
\textbf{(D) }k^2\qquad <br />
\textbf{(E) }y\sqrt{k} </math> <br />
<br />
[[1978 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
What is the smallest positive integer <math>n</math> such that <math>\sqrt{n}-\sqrt{n-1}<.01</math>?<br />
<br />
<math>\textbf{(A) }2499\qquad<br />
\textbf{(B) }2500\qquad<br />
\textbf{(C) }2501\qquad<br />
\textbf{(D) }10,000\qquad <br />
\textbf{(E) }\text{There is no such integer}</math> <br />
<br />
[[1978 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
A positive integer <math>n</math> not exceeding <math>100</math> is chosen in such a way that if <math>n\le 50</math>, then the probability of choosing <br />
<math>n</math> is <math>p</math>, and if <math>n > 50</math>, then the probability of choosing <math>n</math> is <math>3p</math>. The probability that a perfect square is chosen is<br />
<br />
<math>\textbf{(A) }.05\qquad<br />
\textbf{(B) }.065\qquad<br />
\textbf{(C) }.08\qquad<br />
\textbf{(D) }.09\qquad <br />
\textbf{(E) }.1 </math><br />
<br />
[[1978 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
If <math>a,b,c</math> are non-zero real numbers such that <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}</math>, <br />
and <math>x=\frac{(a+b)(b+c)(c+a)}{abc}</math>, and <math>x<0</math>, then <math>x</math> equals<br />
<br />
<math>\textbf{(A) }-1\qquad<br />
\textbf{(B) }-2\qquad<br />
\textbf{(C) }-4\qquad<br />
\textbf{(D) }-6\qquad <br />
\textbf{(E) }-8 </math> <br />
<br />
[[1978 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
For all positive numbers <math>x</math> distinct from <math>1</math>,<br />
<br />
<cmath>\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}</cmath><br />
<br />
equals<br />
<br />
<math>\text{(A) }\frac{1}{\log_{60}(x)}\qquad\\<br />
\text{(B) }\frac{1}{\log_{x}(60)}\qquad\\<br />
\text{(C) }\frac{1}{(\log_{3}(x))(\log_{4}(x))(\log_{5}(x))}\qquad\\<br />
\text{(D) }\frac{12}{\log_{3}(x)+\log_{4}(x)+\log_{5}(x)}\qquad\\<br />
\text{(E) }\frac{\log_{2}(x)}{\log_{3}(x)\log_{5}(x)}+\frac{\log_{3}(x)}{\log_{2}(x)\log_{5}(x)}+\frac{\log_{5}(x)}{\log_{2}(x)\log_{3}(x)}</math><br />
<br />
[[1978 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
The following four statements, and only these are found on a card:<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,1);<br />
B=(0,5);<br />
C=(11,5);<br />
D=(11,1);<br />
E=(0,4);<br />
F=(0,3);<br />
G=(0,2);<br />
draw(A--B--C--D--cycle);<br />
label("On this card exactly one statement is false.", B, SE);<br />
label("On this card exactly two statements are false.", E, SE);<br />
label("On this card exactly three statements are false.", F, SE);<br />
label("On this card exactly four statements are false.", G, SE);<br />
</asy><br />
<br />
(Assume each statement is either true or false.) Among them the number of false statements is exactly<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 4 </math><br />
<br />
[[1978 AHSME Problems/Problem 22|Solution]]<br />
<br />
== Problem 23 ==<br />
<br />
<asy><br />
size(100);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
draw((0,1)--(1,0));<br />
draw((0,0)--(.5,sqrt(3)/2)--(1,0));<br />
label("$A$",(0,0),SW);<br />
label("$B$",(1,0),SE);<br />
label("$C$",(1,1),NE);<br />
label("$D$",(0,1),NW);<br />
label("$E$",(.5,sqrt(3)/2),E);<br />
label("$F$",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W);<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
Vertex <math>E</math> of equilateral <math>\triangle ABE</math> is in the interior of square <math>ABCD</math>, and <math>F</math> is the point of intersection <br />
of diagonal <math>BD</math> and line segment <math>AE</math>. If length <math>AB</math> is <math>\sqrt{1+\sqrt{3}}</math> then the area of <math>\triangle ABF</math> is<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }\frac{\sqrt{2}}{2}\qquad<br />
\textbf{(C) }\frac{\sqrt{3}}{2}\qquad<br />
\textbf{(D) }4-2\sqrt{3}\qquad <br />
\textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4} </math> <br />
<br />
[[1978 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
If the distinct non-zero numbers <math>x ( y - z),~ y(z - x),~ z(x - y )</math> form a geometric progression with common ratio <math>r</math>, <br />
then <math>r</math> satisfies the equation<br />
<br />
<math>\textbf{(A) }r^2+r+1=0\qquad<br />
\textbf{(B) }r^2-r+1=0\qquad<br />
\textbf{(C) }r^4+r^2-1=0\qquad\\<br />
\textbf{(D) }(r+1)^4+r=0\qquad <br />
\textbf{(E) }(r-1)^4+r=0 </math><br />
<br />
[[1978 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
Let <math>u</math> be a positive number. Consider the set <math>S</math> of all points whose rectangular coordinates <math>(x, y )</math> satisfy all of the following conditions:<br />
<br />
<math> \text{(i) }\frac{a}{2}\le x\le 2a\qquad<br />
\text{(ii) }\frac{a}{2}\le y\le 2a\qquad<br />
\text{(iii) }x+y\ge a\\ \\ \qquad<br />
\text{(iv) }x+a\ge y\qquad<br />
\text{(v) }y+a\ge x </math><br />
<br />
The boundary of set S is a polygon with<br />
<br />
<math>\textbf{(A) }3\text{ sides}\qquad<br />
\textbf{(B) }4\text{ sides}\qquad<br />
\textbf{(C) }5\text{ sides}\qquad<br />
\textbf{(D) }6\text{ sides}\qquad<br />
\textbf{(E) }7\text{ sides} </math><br />
<br />
[[1978 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
<asy><br />
size(100);<br />
real a=4, b=3;<br />
//<br />
import cse5;<br />
pathpen=black;<br />
pair A=(a,0), B=(0,b), C=(0,0);<br />
D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle);<br />
pair X=IP(B--A,(0,0)--(b,a));<br />
D(CP((X+C)/2,C));<br />
D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0))));<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
<br />
In <math>\triangle ABC, AB = 10~ AC = 8</math> and <math>BC = 6</math>. Circle <math>P</math> is the circle with smallest radius which passes through <math>C</math> <br />
and is tangent to <math>AB</math>. Let <math>Q</math> and <math>R</math> be the points of intersection, distinct from <math>C</math> , of circle <math>P</math> with sides <math>AC</math> <br />
and <math>BC</math>, respectively. The length of segment <math>QR</math> is<br />
<br />
<math>\textbf{(A) }4.75\qquad<br />
\textbf{(B) }4.8\qquad<br />
\textbf{(C) }5\qquad<br />
\textbf{(D) }4\sqrt{2}\qquad <br />
\textbf{(E) }3\sqrt{3} </math><br />
<br />
[[1978 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
There is more than one integer greater than <math>1</math> which, when divided by any integer <math>k</math> such that <math>2 \le k \le 11</math>, has a remainder of <math>1</math>. <br />
What is the difference between the two smallest such integers?<br />
<br />
<math>\textbf{(A) }2310\qquad<br />
\textbf{(B) }2311\qquad<br />
\textbf{(C) }27,720\qquad<br />
\textbf{(D) }27,721\qquad <br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
[[1978 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
<br />
<asy><br />
size(100);<br />
import cse5;<br />
pathpen=black;<br />
pair A1=(0,0), A2=(1,0), A3=(0.5,sqrt(3)/2);<br />
D(MP("A_1",A1)--MP("A_2",A2)--MP("A_3",A3,N)--cycle);<br />
pair A4=(A1+A2)/2, A5 = (A3+A2)/2, A6 = (A4+A3)/2;<br />
D(MP("A_4",A4,S)--MP("A_6",A6,W)--A3);<br />
D(A6--MP("A_5",A5,NE)--A4);<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
<br />
If <math>\triangle A_1A_2A_3</math> is equilateral and <math>A_{n+3}</math> is the midpoint of line segment <math>A_nA_{n+1}</math> for all positive integers <math>n</math>, <br />
then the measure of <math>\measuredangle A_{44}A_{45}A_{43}</math> equals<br />
<br />
<math>\textbf{(A) }30^\circ\qquad<br />
\textbf{(B) }45^\circ\qquad<br />
\textbf{(C) }60^\circ\qquad<br />
\textbf{(D) }90^\circ\qquad <br />
\textbf{(E) }120^\circ </math><br />
<br />
[[1978 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. <br />
Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is<br />
<br />
<math>\textbf{(A) }20\qquad<br />
\textbf{(B) }40\qquad<br />
\textbf{(C) }45\qquad<br />
\textbf{(D) }50\qquad <br />
\textbf{(E) }60 </math><br />
<br />
[[1978 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
In a tennis tournament, <math>n</math> women and <math>2n</math> men play, and each player plays exactly one match with every other player. <br />
If there are no ties and the ratio of the number of matches won by women to the number of matches won by men is <math>7/5</math>, then <math>n</math> equals<br />
<br />
<math>\textbf{(A) }2\qquad<br />
\textbf{(B) }4\qquad<br />
\textbf{(C) }6\qquad<br />
\textbf{(D) }7\qquad <br />
\textbf{(E) }\text{none of these} </math><br />
<br />
[[1978 AHSME Problems/Problem 30|Solution]]<br />
<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1978|before=[[1977 AHSME]]|after=[[1979 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems&diff=1137281978 AHSME Problems2019-12-30T01:09:39Z<p>Ryjs: /* Problem 22 */</p>
<hr />
<div>== Problem 1 ==<br />
<br />
If <math>1-\frac{4}{x}+\frac{4}{x^2}=0</math>, then <math>\frac{2}{x}</math> equals<br />
<br />
<math>\textbf{(A) }-1\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }-1\text{ or }2\qquad <br />
\textbf{(E) }-1\text{ or }-2 </math><br />
<br />
[[1978 AHSME Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
<br />
If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is <br />
<br />
<math>\textbf{(A) }\frac{1}{\pi^2}\qquad<br />
\textbf{(B) }\frac{1}{\pi}\qquad<br />
\textbf{(C) }1\qquad<br />
\textbf{(D) }\pi\qquad<br />
\textbf{(E) }\pi^2 </math> <br />
<br />
[[1978 AHSME Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
<br />
For all non-zero numbers <math>x</math> and <math>y</math> such that <math>x = 1/y</math>, <math>\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)</math> equals<br />
<br />
<math>\textbf{(A) }2x^2\qquad<br />
\textbf{(B) }2y^2\qquad<br />
\textbf{(C) }x^2+y^2\qquad<br />
\textbf{(D) }x^2-y^2\qquad <br />
\textbf{(E) }y^2-x^2 </math><br />
<br />
[[1978 AHSME Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</math> is equal to<br />
<br />
<math>\textbf{(A) }1111\qquad<br />
\textbf{(B) }2222\qquad<br />
\textbf{(C) }3333\qquad<br />
\textbf{(D) }1212\qquad <br />
\textbf{(E) }4242 </math> <br />
<br />
[[1978 AHSME Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Four boys bought a boat for <math>\textdollar 60</math>. The first boy paid one half of the sum of the amounts paid by the other boys; <br />
the second boy paid one third of the sum of the amounts paid by the other boys; <br />
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay? <br />
<br />
<math>\textbf{(A) }\textdollar 10\qquad<br />
\textbf{(B) }\textdollar 12\qquad<br />
\textbf{(C) }\textdollar 13\qquad<br />
\textbf{(D) }\textdollar 14\qquad<br />
\textbf{(E) }\textdollar 15</math> <br />
<br />
[[1978 AHSME Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
The number of distinct pairs <math>(x,y)</math> of real numbers satisfying both of the following equations: <br />
<br />
<cmath>x=x^2+y^2 \\ y=2xy</cmath> <br />
is<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }3\qquad <br />
\textbf{(E) }4 </math><br />
<br />
[[1978 AHSME Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
Opposite sides of a regular hexagon are <math>12</math> inches apart. The length of each side, in inches, is<br />
<br />
<math>\textbf{(A) }7.5\qquad<br />
\textbf{(B) }6\sqrt{2}\qquad<br />
\textbf{(C) }5\sqrt{2}\qquad<br />
\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad <br />
\textbf{(D) }\frac{9}{2}\sqrt{3}\qquad<br />
\textbf{(E) }4\sqrt{3} </math><br />
<br />
[[1978 AHSME Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
If <math>x\neq y</math> and the sequences <math>x,a_1,a_2,y</math> and <math>x,b_1,b_2,b_3,y</math> each are in arithmetic progression, then <math>(a_2-a_1)/(b_2-b_1)</math> equals<br />
<br />
<math>\textbf{(A) }\frac{2}{3}\qquad<br />
\textbf{(B) }\frac{3}{4}\qquad<br />
\textbf{(C) }1\qquad<br />
\textbf{(D) }\frac{4}{3}\qquad<br />
\textbf{(E) }\frac{3}{2} </math> <br />
<br />
[[1978 AHSME Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
If <math>x<0</math>, then <math>\left|x-\sqrt{(x-1)^2}\right|</math> equals<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }1-2x\qquad<br />
\textbf{(C) }-2x-1\qquad<br />
\textbf{(D) }1+2x\qquad <br />
\textbf{(E) }2x-1 </math><br />
<br />
[[1978 AHSME Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
If <math>\mathit{B}</math> is a point on circle <math>\mathit{C}</math> with center <math>\mathit{P}</math>, then the set of all points <math>\mathit{A}</math> in the plane of circle <br />
<math>\mathit{C}</math> such that the distance between <math>\mathit{A}</math> and <math>\mathit{B}</math> is less than or equal to the distance between <math>\mathit{A}</math> <br />
and any other point on circle <math>\mathit{C}</math> is<br />
<br />
<math>\textbf{(A) }\text{the line segment from }P \text{ to }B\qquad\\<br />
\textbf{(B) }\text{the ray beginning at }P \text{ and passing through }B\qquad\\<br />
\textbf{(C) }\text{a ray beginning at }B\qquad\\<br />
\textbf{(D) }\text{a circle whose center is }P\qquad\\<br />
\textbf{(E) }\text{a circle whose center is }B </math> <br />
<br />
[[1978 AHSME Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
If <math>r</math> is positive and the line whose equation is <math>x + y = r</math> is tangent to the circle whose equation is <math>x^2 + y ^2 = r</math>, then <math>r</math> equals<br />
<br />
<math>\textbf{(A) }\frac{1}{2}\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }\sqrt{2}\qquad <br />
\textbf{(E) }2\sqrt{2} </math><br />
<br />
[[1978 AHSME Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
In <math>\triangle ADE</math>, <math>\measuredangle ADE=140^\circ</math>, points <math>B</math> and <math>C</math> lie on sides <math>AD</math> and <math>AE</math>, <br />
respectively, and points <math>A,~B,~C,~D,~E</math> are distinct.* If lengths <math>AB,~BC,~CD</math>, and <math>DE</math> are all equal, <br />
then the measure of <math>\measuredangle EAD</math> is<br />
* The specification that points <math>A,B,C,D,E</math> be distinct was not included in the original statement of the problem. <br />
If <math>B=D</math>, then <math>C=E</math> and <math>\measuredangle EAD=20^\circ</math>. <br />
<br />
<math>\textbf{(A) }5^\circ\qquad<br />
\textbf{(B) }6^\circ\qquad<br />
\textbf{(C) }7.5^\circ\qquad<br />
\textbf{(D) }8^\circ\qquad<br />
\textbf{(E) }10^\circ</math><br />
<br />
<br />
[[1978 AHSME Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
<br />
If <math>a,b,c</math>, and <math>d</math> are non-zero numbers such that <math>c</math> and <math>d</math> are the solutions of <math>x^2+ax+b=0</math> and <math>a</math> and <math>b</math> are <br />
the solutions of <math>x^2+cx+d=0</math>, then <math>a+b+c+d</math> equals<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }-2\qquad<br />
\textbf{(C) }2\qquad<br />
\textbf{(D) }4\qquad <br />
\textbf{(E) }(-1+\sqrt{5})/2 </math> <br />
<br />
[[1978 AHSME Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
If an integer <math>n > 8</math> is a solution of the equation <math>x^2 - ax+b=0</math> and the representation of <math>a</math> in the base-<math>n</math> number system is <math>18</math>, <br />
then the base-n representation of <math>b</math> is<br />
<br />
<math>\textbf{(A)}\ 18 \qquad<br />
\textbf{(B)}\ 20 \qquad<br />
\textbf{(C)}\ 80 \qquad<br />
\textbf{(D)}\ 81 \qquad<br />
\textbf{(E)}\ 280 </math><br />
<br />
[[1978 AHSME Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is<br />
<br />
<math>\textbf{(A) }-\frac{4}{3}\qquad<br />
\textbf{(B) }-\frac{3}{4}\qquad<br />
\textbf{(C) }\frac{3}{4}\qquad<br />
\textbf{(D) }\frac{4}{3}\qquad\\<br />
\textbf{(E) }\text{not completely determined by the given information} </math><br />
<br />
[[1978 AHSME Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
In a room containing <math>N</math> people, <math>N > 3</math>, at least one person has not shaken hands with everyone else in the room. <br />
What is the maximum number of people in the room that could have shaken hands with everyone else?<br />
<br />
<math>\textbf{(A) }0\qquad<br />
\textbf{(B) }1\qquad<br />
\textbf{(C) }N-1\qquad<br />
\textbf{(D) }N\qquad <br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
[[1978 AHSME Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
If <math>k</math> is a positive number and <math>f</math> is a function such that, for every positive number <math>x</math>, <math>\left[f(x^2+1)\right]^{\sqrt{x}}=k</math>; <br />
then, for every positive number <math>y</math>, <math>\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}</math> is equal to<br />
<br />
<math>\textbf{(A) }\sqrt{k}\qquad<br />
\textbf{(B) }2k\qquad<br />
\textbf{(C) }k\sqrt{k}\qquad<br />
\textbf{(D) }k^2\qquad <br />
\textbf{(E) }y\sqrt{k} </math> <br />
<br />
[[1978 AHSME Problems/Problem 17|Solution]]<br />
<br />
== Problem 18 ==<br />
<br />
What is the smallest positive integer <math>n</math> such that <math>\sqrt{n}-\sqrt{n-1}<.01</math>?<br />
<br />
<math>\textbf{(A) }2499\qquad<br />
\textbf{(B) }2500\qquad<br />
\textbf{(C) }2501\qquad<br />
\textbf{(D) }10,000\qquad <br />
\textbf{(E) }\text{There is no such integer}</math> <br />
<br />
[[1978 AHSME Problems/Problem 18|Solution]]<br />
<br />
== Problem 19 ==<br />
<br />
A positive integer <math>n</math> not exceeding <math>100</math> is chosen in such a way that if <math>n\le 50</math>, then the probability of choosing <br />
<math>n</math> is <math>p</math>, and if <math>n > 50</math>, then the probability of choosing <math>n</math> is <math>3p</math>. The probability that a perfect square is chosen is<br />
<br />
<math>\textbf{(A) }.05\qquad<br />
\textbf{(B) }.065\qquad<br />
\textbf{(C) }.08\qquad<br />
\textbf{(D) }.09\qquad <br />
\textbf{(E) }.1 </math><br />
<br />
[[1978 AHSME Problems/Problem 19|Solution]]<br />
<br />
== Problem 20 ==<br />
<br />
If <math>a,b,c</math> are non-zero real numbers such that <math>\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}</math>, <br />
and <math>x=\frac{(a+b)(b+c)(c+a)}{abc}</math>, and <math>x<0</math>, then <math>x</math> equals<br />
<br />
<math>\textbf{(A) }-1\qquad<br />
\textbf{(B) }-2\qquad<br />
\textbf{(C) }-4\qquad<br />
\textbf{(D) }-6\qquad <br />
\textbf{(E) }-8 </math> <br />
<br />
[[1978 AHSME Problems/Problem 20|Solution]]<br />
<br />
== Problem 21 ==<br />
<br />
For all positive numbers <math>x</math> distinct from <math>1</math>,<br />
<br />
<cmath>\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}</cmath><br />
<br />
equals<br />
<br />
<math>\text{(A) }\frac{1}{\log_{60}(x)}\qquad\\<br />
\text{(B) }\frac{1}{\log_{x}(60)}\qquad\\<br />
\text{(C) }\frac{1}{(\log_{3}(x))(\log_{4}(x))(\log_{5}(x))}\qquad\\<br />
\text{(D) }\frac{12}{\log_{3}(x)+\log_{4}(x)+\log_{5}(x)}\qquad\\<br />
\text{(E) }\frac{\log_{2}(x)}{\log_{3}(x)\log_{5}(x)}+\frac{\log_{3}(x)}{\log_{2}(x)\log_{5}(x)}+\frac{\log_{5}(x)}{\log_{2}(x)\log_{3}(x)}</math><br />
<br />
[[1978 AHSME Problems/Problem 21|Solution]]<br />
<br />
== Problem 22 ==<br />
<br />
The following four statements, and only these are found on a card:<br />
<asy><br />
pair A,B,C,D,E,F,G;<br />
A=(0,1);<br />
B=(0,5);<br />
C=(11,5);<br />
D=(11,1);<br />
E=(0,4);<br />
F=(0,3);<br />
G=(0,2);<br />
draw(A--B--C--D--cycle);<br />
label("On this card exactly one statement is false.", B, SE);<br />
label("On this card exactly two statements are false.", E, SE);<br />
label("On this card exactly three statements are false.", F, SE);<br />
label("On this card exactly four statements are false.", G, SE);<br />
</asy><br />
<br />
(Assume each statement is either true or false.) Among them the number of false statements is exactly<br />
<br />
<math>\textbf{(A)}\ 0 \qquad<br />
\textbf{(B)}\ 1 \qquad<br />
\textbf{(C)}\ 2 \qquad<br />
\textbf{(D)}\ 3 \qquad<br />
\textbf{(E)}\ 4 </math><br />
<br />
[[1978 AHSME Problems/Problem 22|Solution]]<br />
There can be at most one true statement on the card, eliminating and . If there are true on the card, statement ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is , since are false and only the third statement ("On this card exactly three statements are false") is correct.<br />
<br />
== Problem 23 ==<br />
<br />
<asy><br />
size(100);<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
draw((0,1)--(1,0));<br />
draw((0,0)--(.5,sqrt(3)/2)--(1,0));<br />
label("$A$",(0,0),SW);<br />
label("$B$",(1,0),SE);<br />
label("$C$",(1,1),NE);<br />
label("$D$",(0,1),NW);<br />
label("$E$",(.5,sqrt(3)/2),E);<br />
label("$F$",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W);<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
Vertex <math>E</math> of equilateral <math>\triangle ABE</math> is in the interior of square <math>ABCD</math>, and <math>F</math> is the point of intersection <br />
of diagonal <math>BD</math> and line segment <math>AE</math>. If length <math>AB</math> is <math>\sqrt{1+\sqrt{3}}</math> then the area of <math>\triangle ABF</math> is<br />
<br />
<math>\textbf{(A) }1\qquad<br />
\textbf{(B) }\frac{\sqrt{2}}{2}\qquad<br />
\textbf{(C) }\frac{\sqrt{3}}{2}\qquad<br />
\textbf{(D) }4-2\sqrt{3}\qquad <br />
\textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4} </math> <br />
<br />
[[1978 AHSME Problems/Problem 23|Solution]]<br />
<br />
== Problem 24 ==<br />
<br />
If the distinct non-zero numbers <math>x ( y - z),~ y(z - x),~ z(x - y )</math> form a geometric progression with common ratio <math>r</math>, <br />
then <math>r</math> satisfies the equation<br />
<br />
<math>\textbf{(A) }r^2+r+1=0\qquad<br />
\textbf{(B) }r^2-r+1=0\qquad<br />
\textbf{(C) }r^4+r^2-1=0\qquad\\<br />
\textbf{(D) }(r+1)^4+r=0\qquad <br />
\textbf{(E) }(r-1)^4+r=0 </math><br />
<br />
[[1978 AHSME Problems/Problem 24|Solution]]<br />
<br />
== Problem 25 ==<br />
<br />
Let <math>u</math> be a positive number. Consider the set <math>S</math> of all points whose rectangular coordinates <math>(x, y )</math> satisfy all of the following conditions:<br />
<br />
<math> \text{(i) }\frac{a}{2}\le x\le 2a\qquad<br />
\text{(ii) }\frac{a}{2}\le y\le 2a\qquad<br />
\text{(iii) }x+y\ge a\\ \\ \qquad<br />
\text{(iv) }x+a\ge y\qquad<br />
\text{(v) }y+a\ge x </math><br />
<br />
The boundary of set S is a polygon with<br />
<br />
<math>\textbf{(A) }3\text{ sides}\qquad<br />
\textbf{(B) }4\text{ sides}\qquad<br />
\textbf{(C) }5\text{ sides}\qquad<br />
\textbf{(D) }6\text{ sides}\qquad<br />
\textbf{(E) }7\text{ sides} </math><br />
<br />
[[1978 AHSME Problems/Problem 25|Solution]]<br />
<br />
== Problem 26 ==<br />
<br />
<asy><br />
size(100);<br />
real a=4, b=3;<br />
//<br />
import cse5;<br />
pathpen=black;<br />
pair A=(a,0), B=(0,b), C=(0,0);<br />
D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle);<br />
pair X=IP(B--A,(0,0)--(b,a));<br />
D(CP((X+C)/2,C));<br />
D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0))));<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
<br />
In <math>\triangle ABC, AB = 10~ AC = 8</math> and <math>BC = 6</math>. Circle <math>P</math> is the circle with smallest radius which passes through <math>C</math> <br />
and is tangent to <math>AB</math>. Let <math>Q</math> and <math>R</math> be the points of intersection, distinct from <math>C</math> , of circle <math>P</math> with sides <math>AC</math> <br />
and <math>BC</math>, respectively. The length of segment <math>QR</math> is<br />
<br />
<math>\textbf{(A) }4.75\qquad<br />
\textbf{(B) }4.8\qquad<br />
\textbf{(C) }5\qquad<br />
\textbf{(D) }4\sqrt{2}\qquad <br />
\textbf{(E) }3\sqrt{3} </math><br />
<br />
[[1978 AHSME Problems/Problem 26|Solution]]<br />
<br />
== Problem 27 ==<br />
<br />
There is more than one integer greater than <math>1</math> which, when divided by any integer <math>k</math> such that <math>2 \le k \le 11</math>, has a remainder of <math>1</math>. <br />
What is the difference between the two smallest such integers?<br />
<br />
<math>\textbf{(A) }2310\qquad<br />
\textbf{(B) }2311\qquad<br />
\textbf{(C) }27,720\qquad<br />
\textbf{(D) }27,721\qquad <br />
\textbf{(E) }\text{none of these} </math> <br />
<br />
[[1978 AHSME Problems/Problem 27|Solution]]<br />
<br />
== Problem 28 ==<br />
<br />
<asy><br />
size(100);<br />
import cse5;<br />
pathpen=black;<br />
pair A1=(0,0), A2=(1,0), A3=(0.5,sqrt(3)/2);<br />
D(MP("A_1",A1)--MP("A_2",A2)--MP("A_3",A3,N)--cycle);<br />
pair A4=(A1+A2)/2, A5 = (A3+A2)/2, A6 = (A4+A3)/2;<br />
D(MP("A_4",A4,S)--MP("A_6",A6,W)--A3);<br />
D(A6--MP("A_5",A5,NE)--A4);<br />
//Credit to chezbgone2 for the diagram<br />
</asy><br />
<br />
<br />
If <math>\triangle A_1A_2A_3</math> is equilateral and <math>A_{n+3}</math> is the midpoint of line segment <math>A_nA_{n+1}</math> for all positive integers <math>n</math>, <br />
then the measure of <math>\measuredangle A_{44}A_{45}A_{43}</math> equals<br />
<br />
<math>\textbf{(A) }30^\circ\qquad<br />
\textbf{(B) }45^\circ\qquad<br />
\textbf{(C) }60^\circ\qquad<br />
\textbf{(D) }90^\circ\qquad <br />
\textbf{(E) }120^\circ </math><br />
<br />
[[1978 AHSME Problems/Problem 28|Solution]]<br />
<br />
== Problem 29 ==<br />
<br />
Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. <br />
Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is<br />
<br />
<math>\textbf{(A) }20\qquad<br />
\textbf{(B) }40\qquad<br />
\textbf{(C) }45\qquad<br />
\textbf{(D) }50\qquad <br />
\textbf{(E) }60 </math><br />
<br />
[[1978 AHSME Problems/Problem 29|Solution]]<br />
<br />
== Problem 30 ==<br />
<br />
In a tennis tournament, <math>n</math> women and <math>2n</math> men play, and each player plays exactly one match with every other player. <br />
If there are no ties and the ratio of the number of matches won by women to the number of matches won by men is <math>7/5</math>, then <math>n</math> equals<br />
<br />
<math>\textbf{(A) }2\qquad<br />
\textbf{(B) }4\qquad<br />
\textbf{(C) }6\qquad<br />
\textbf{(D) }7\qquad <br />
\textbf{(E) }\text{none of these} </math><br />
<br />
[[1978 AHSME Problems/Problem 30|Solution]]<br />
<br />
<br />
== See also ==<br />
<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
<br />
{{AHSME box|year=1978|before=[[1977 AHSME]]|after=[[1979 AHSME]]}} <br />
<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_22&diff=1137271978 AHSME Problems/Problem 222019-12-30T01:07:27Z<p>Ryjs: solution</p>
<hr />
<div>There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and <math>\textbf{(C)}</math>. If there are <math>0</math> true on the card, statement <math>4</math> ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is <math>\textbf{(D)}\ 3</math>, since <math>3</math> are false and only the third statement ("On this card exactly three statements are false") is correct.</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_22&diff=1137261978 AHSME Problems/Problem 222019-12-30T01:06:36Z<p>Ryjs: solution</p>
<hr />
<div>There can be at most one true statement on the card, eliminating <math>\textbf{(A)}, \textbf{(B)},</math> and <math></math>\textbf{(C)}<math>. If there are </math>0<math> true on the card, statement </math>4<math> ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is </math>\qquad \textbf{(D)}\ 3<math>, since </math>3$ are false and only the third statement ("On this card exactly three statements are false") is correct.</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_20&diff=1133482012 AMC 8 Problems/Problem 202019-12-24T07:18:39Z<p>Ryjs: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
What is the correct ordering of the three numbers <math> \frac{5}{19} </math>, <math> \frac{7}{21} </math>, and <math> \frac{9}{23} </math>, in increasing order?<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21} </math> <br />
<br />
<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math><br />
<br />
==Solution 1==<br />
The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator.<br />
<br />
<math> \frac{5}{19} \implies \frac{345}{1311} </math><br />
<br />
<math> \frac{1}{3} \implies \frac{437}{1311} </math><br />
<br />
<math> \frac{9}{23} \implies \frac{513}{1311} </math><br />
<br />
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
==Solution 2==<br />
Instead of finding the LCD, we can subtract each fraction from <math>1</math> to get a common numerator. Thus,<br />
<br />
<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math><br />
<br />
<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math><br />
<br />
<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math><br />
<br />
All three fractions have common numerator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}</math>. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
<br />
==Solution 3==<br />
Change <math>7/21</math> into <math>1/3</math>;<br />
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath><br />
<cmath>\frac{5}{15}>\frac{5}{19}</cmath><br />
<cmath>\frac{7}{21}>\frac{5}{19}</cmath><br />
And<br />
<cmath>\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}</cmath><br />
<cmath>\frac{9}{27}<\frac{9}{23}</cmath><br />
<cmath>\frac{7}{21}<\frac{9}{23}</cmath><br />
Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
==Solution 4==<br />
When <math>\frac{x}{y}<1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}>\frac{x}{y}</math>. Hence, the answer is <math>{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}</math>.<br />
~ ryjs<br />
<br />
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2012|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_20&diff=1133472012 AMC 8 Problems/Problem 202019-12-24T07:18:14Z<p>Ryjs: </p>
<hr />
<div>==Problem==<br />
What is the correct ordering of the three numbers <math> \frac{5}{19} </math>, <math> \frac{7}{21} </math>, and <math> \frac{9}{23} </math>, in increasing order?<br />
<br />
<math> \textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21} </math> <br />
<br />
<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math><br />
<br />
==Solution 1==<br />
The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator.<br />
<br />
<math> \frac{5}{19} \implies \frac{345}{1311} </math><br />
<br />
<math> \frac{1}{3} \implies \frac{437}{1311} </math><br />
<br />
<math> \frac{9}{23} \implies \frac{513}{1311} </math><br />
<br />
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
==Solution 2==<br />
Instead of finding the LCD, we can subtract each fraction from <math>1</math> to get a common numerator. Thus,<br />
<br />
<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math><br />
<br />
<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math><br />
<br />
<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math><br />
<br />
All three fractions have common numerator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}</math>. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
<br />
==Solution 3==<br />
Change <math>7/21</math> into <math>1/3</math>;<br />
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath><br />
<cmath>\frac{5}{15}>\frac{5}{19}</cmath><br />
<cmath>\frac{7}{21}>\frac{5}{19}</cmath><br />
And<br />
<cmath>\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}</cmath><br />
<cmath>\frac{9}{27}<\frac{9}{23}</cmath><br />
<cmath>\frac{7}{21}<\frac{9}{23}</cmath><br />
Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
<br />
==Solution 4==<br />
When <math>\frac{x}{y}<1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}>\frac{x}{y}</math>. Hence, the answer is {\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} $.<br />
~ ryjs<br />
<br />
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2012|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=1133462019 AMC 8 Problems/Problem 32019-12-24T07:15:00Z<p>Ryjs: /* Solution 3 */</p>
<hr />
<div>==Problem 3==<br />
Which of the following is the correct order of the fractions <math>\frac{15}{11},\frac{19}{15},</math> and <math>\frac{17}{13},</math> from least to greatest? <br />
<br />
<math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math><br />
<br />
==Solution 1==<br />
Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math><br />
<br />
-will3145<br />
<br />
==Solution 2==<br />
We take a common denominator:<br />
<cmath>\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.</cmath><br />
<br />
Since <math>2717<2805<2925</math> it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
<br />
-xMidnightFirex<br />
<br />
~ dolphin7 - I took your idea and made it an explanation.<br />
<br />
==Solution 3==<br />
When <math>\frac{x}{y}>1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}<\frac{x}{y}</math>. Hence, the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
~ ryjs<br />
<br />
This is also similar to Problem 20 on the AMC 2012.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=2|num-a=4}}<br />
<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=1133442019 AMC 8 Problems/Problem 32019-12-24T04:53:53Z<p>Ryjs: /* Solution 3 */</p>
<hr />
<div>==Problem 3==<br />
Which of the following is the correct order of the fractions <math>\frac{15}{11},\frac{19}{15},</math> and <math>\frac{17}{13},</math> from least to greatest? <br />
<br />
<math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math><br />
<br />
==Solution 1==<br />
Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math><br />
<br />
-will3145<br />
<br />
==Solution 2==<br />
We take a common denominator:<br />
<cmath>\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.</cmath><br />
<br />
Since <math>2717<2805<2925</math> it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
<br />
-xMidnightFirex<br />
<br />
~ dolphin7 - I took your idea and made it an explanation.<br />
<br />
==Solution 3==<br />
When <math>\frac{x}{y}>1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}<\frac{x}{y}</math>. Hence, the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
~ ryjs<br />
<br />
==See Also==<br />
{{AMC8 box|year=2019|num-b=2|num-a=4}}<br />
<br />
{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_3&diff=1133432019 AMC 8 Problems/Problem 32019-12-24T04:29:22Z<p>Ryjs: </p>
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<div>==Problem 3==<br />
Which of the following is the correct order of the fractions <math>\frac{15}{11},\frac{19}{15},</math> and <math>\frac{17}{13},</math> from least to greatest? <br />
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<math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math><br />
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==Solution 1==<br />
Consider subtracting 1 from each of the fractions. Our new fractions would then be <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math>. Since <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>, it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math><br />
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-will3145<br />
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==Solution 2==<br />
We take a common denominator:<br />
<cmath>\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.</cmath><br />
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Since <math>2717<2805<2925</math> it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
<br />
-xMidnightFirex<br />
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~ dolphin7 - I took your idea and made it an explanation.<br />
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==Solution 3==<br />
When <math>\frac{x}{y}>1</math> and <math>z</math> is a positive integer, <math>\frac{x+z}{y+z}<\frac{x}{y}</math>. Hence, the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>.<br />
~ ryjs<br />
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==See Also==<br />
{{AMC8 box|year=2019|num-b=2|num-a=4}}<br />
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{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_20&diff=1118392019 AMC 8 Problems/Problem 202019-11-21T00:41:51Z<p>Ryjs: /* Solution 1 */</p>
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<div>==Problem 20==<br />
How many different real numbers <math>x</math> satisfy the equation <cmath>(x^{2}-5)^{2}=16?</cmath><br />
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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }8</math><br />
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==Solution 1==<br />
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We know that to get <math>16</math>, you can square <math>4</math> or <math>-4</math>. Thus, <math>x^2 - 5</math> can have <math>2</math> possibilities. <math>x^2</math> is either <math>9</math> or <math>1</math>, leaving <math>x</math> with possiblities <math>3,-3, 1,</math> and <math>-1,</math> so <math>\boxed{(D)}</math>.<br />
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==See Also==<br />
{{AMC8 box|year=2019|num-b=19|num-a=21}}<br />
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{{MAA Notice}}</div>Ryjshttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_8_Problems/Problem_20&diff=1115662012 AMC 8 Problems/Problem 202019-11-20T02:00:45Z<p>Ryjs: /* Solution 3 */</p>
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<div>==Problem==<br />
What is the correct ordering of the three numbers <math> \frac{5}{19} </math>, <math> \frac{7}{21} </math>, and <math> \frac{9}{23} </math>, in increasing order?<br />
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<math> \textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21} </math> <br />
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<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math><br />
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==Solution 1==<br />
The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator.<br />
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<math> \frac{5}{19} \implies \frac{345}{1311} </math><br />
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<math> \frac{1}{3} \implies \frac{437}{1311} </math><br />
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<math> \frac{9}{23} \implies \frac{513}{1311} </math><br />
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Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
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==Solution 2==<br />
Instead of finding the LCD, we can subtract each fraction from <math>1</math> to get a common numerator. Thus,<br />
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<math>1-\dfrac{5}{19}=\dfrac{14}{19}</math><br />
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<math>1-\dfrac{7}{21}=\dfrac{14}{21}</math><br />
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<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math><br />
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All three fractions have common numerator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}</math>. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
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==Solution 3==<br />
Change <math>7/21</math> into <math>1/3</math>;<br />
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath><br />
<cmath>\frac{5}{15}>\frac{5}{19}</cmath><br />
<cmath>\frac{7}{21}>\frac{5}{19}</cmath><br />
And<br />
<cmath>\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}</cmath><br />
<cmath>\frac{9}{27}<\frac{9}{23}</cmath><br />
<cmath>\frac{7}{21}<\frac{9}{23}</cmath><br />
Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>.<br />
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==See Also==<br />
{{AMC8 box|year=2012|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Ryjs