https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Zachc16&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T01:53:30ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_5&diff=1022722019 AMC 12B Problems/Problem 52019-02-14T17:37:51Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Exact change to buy 12 red candy, 14 green candy, 15 blue candy, or n purple candy. Purple candy cost 20 cents, what is the minimum number of n?<br />
<br />
==Solution==<br />
We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of <math>\boxed{n = 21}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022712019 AMC 12B Problems/Problem 82019-02-14T17:37:35Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{A}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022702019 AMC 12B Problems/Problem 82019-02-14T17:36:40Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{A}</math>. -zachc16<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022692019 AMC 12B Problems/Problem 82019-02-14T17:36:11Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{A}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022682019 AMC 12B Problems/Problem 82019-02-14T17:35:52Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{(A) 0}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022672019 AMC 12B Problems/Problem 82019-02-14T17:35:34Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is <math>\boxed{(A) 0}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022662019 AMC 12B Problems/Problem 82019-02-14T17:35:18Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is $\boxed{A}<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022652019 AMC 12B Problems/Problem 82019-02-14T17:35:03Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is $\boxed{(A) 0}<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022642019 AMC 12B Problems/Problem 82019-02-14T17:34:17Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
Note that <math>f(x) = f(1-x)</math>. We can see from this that the terms cancel and the answer is<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_8&diff=1022622019 AMC 12B Problems/Problem 82019-02-14T17:33:16Z<p>Zachc16: /* Problem */</p>
<hr />
<div>==Problem==<br />
Let <math>f(x) = x^{2}(1-x)^{2}</math>. What is the value of the sum<br />
<math>f(\frac{1}{2019})-f(\frac{2}{2019})+f(\frac{3}{2019})-f(\frac{4}{2019})+\cdots </math><br />
<br />
<math>+ f(\frac{2017}{2019}) - f(\frac{2018}{2019})</math>?<br />
<br />
(A) <math>0</math>, (B) <math>\frac{1}{2019^{4}}</math>, (C) <math>\frac{2018^{2}}{2019^{4}}</math>, (D) <math>\frac{2020^{2}}{2019^{4}}</math>, (E) <math>1</math>.<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_16&diff=1022462019 AMC 12B Problems/Problem 162019-02-14T17:26:50Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_16&diff=1022452019 AMC 12B Problems/Problem 162019-02-14T17:26:37Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
==Solution==<br />
<math>\boxed{15/256}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=15|num-a=17}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_5&diff=1022442019 AMC 12B Problems/Problem 52019-02-14T17:25:42Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Exact change to buy 12 red candy, 14 green candy, 15 blue candy, or n purple candy. Purple candy cost 20 cents, what is the minimum number of n?<br />
<br />
==Solution==<br />
We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy, so the minimum value of <math>\boxed{n = 21}</math>. -zachc16<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_5&diff=1022422019 AMC 12B Problems/Problem 52019-02-14T17:24:52Z<p>Zachc16: /* Solution */</p>
<hr />
<div>==Problem==<br />
Exact change to buy 12 red candy, 14 green candy, 15 blue candy, or n purple candy. Purple candy cost 20 cents, what is the minimum number of n?<br />
<br />
==Solution==<br />
We simply need to find a value of 20*n that divides 12, 14, and 15. 20*18 divides 12 and 15, but not 14. 20*21 successfully divides 12, 14 and 15, meaning that we have exact change (in this case, 420 cents) to buy each type of candy so the minimum value of <math>\boxed{n = 21}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zachc16https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_5&diff=1022342019 AMC 12B Problems/Problem 52019-02-14T17:21:35Z<p>Zachc16: /* Problem */</p>
<hr />
<div>==Problem==<br />
Exact change to buy 12 red candy, 14 green candy, 15 blue candy, or n purple candy. Purple candy cost 20 cents, what is the minimum number of n?<br />
<br />
==Solution==<br />
n = 18 (SuperWill)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Zachc16