https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Epicmonster&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-22T10:46:17Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_23&diff=98615 2012 AMC 12B Problems/Problem 23 2018-11-12T03:32:40Z <p>Epicmonster: /* Solution 2 */</p> <hr /> <div>== Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution (dubious) ==<br /> <br /> Since &lt;math&gt;z_0&lt;/math&gt; is a root of &lt;math&gt;P&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; has integer coefficients, &lt;math&gt;z_0&lt;/math&gt; must be algebraic. Since &lt;math&gt;z_0&lt;/math&gt; is algebraic and lies on the unit circle, &lt;math&gt;z_0&lt;/math&gt; must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since &lt;math&gt;P&lt;/math&gt; has degree 4, it seems reasonable (and we will assume this only temporarily) that &lt;math&gt;z_0&lt;/math&gt; must be a 2nd, 3rd, or 4th root of unity. These are among the set &lt;math&gt;\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}&lt;/math&gt;. Since complex roots of polynomials come in conjugate pairs, we have that &lt;math&gt;P&lt;/math&gt; has one (or more) of the following factors: &lt;math&gt;z+1&lt;/math&gt;, &lt;math&gt;z-1&lt;/math&gt;, &lt;math&gt;z^2+1&lt;/math&gt;, or &lt;math&gt;z^2+z+1&lt;/math&gt;. If &lt;math&gt;z=1&lt;/math&gt; then &lt;math&gt;a+b+c+d+4=0&lt;/math&gt;; a contradiction since &lt;math&gt;a,b,c,d&lt;/math&gt; are non-negative. On the other hand, suppose &lt;math&gt;z=-1&lt;/math&gt;. Then &lt;math&gt;(a+c)-(b+d)=4&lt;/math&gt;. This implies &lt;math&gt;a+b=8,7,6,5,4&lt;/math&gt; while &lt;math&gt;b+d=4,3,2,1,0&lt;/math&gt; correspondingly. After listing cases, the only such valid &lt;math&gt;a,b,c,d&lt;/math&gt; are &lt;math&gt;4,4,4,0&lt;/math&gt;, &lt;math&gt;4,3,3,0&lt;/math&gt;, &lt;math&gt;4,2,2,0&lt;/math&gt;, &lt;math&gt;4,1,1,0&lt;/math&gt;, and &lt;math&gt;4,0,0,0&lt;/math&gt;. <br /> <br /> Now suppose &lt;math&gt;z=i&lt;/math&gt;. Then &lt;math&gt;4=(a-c)i+(b-d)&lt;/math&gt; whereupon &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b-d=4&lt;/math&gt;. But then &lt;math&gt;a=b=c&lt;/math&gt; and &lt;math&gt;d=a-4&lt;/math&gt;. This gives only the cases &lt;math&gt;a,b,c,d&lt;/math&gt; equals &lt;math&gt;4,4,4,0&lt;/math&gt;, which we have already counted in a previous case.<br /> <br /> Suppose &lt;math&gt;z=-i&lt;/math&gt;. Then &lt;math&gt;4=i(c-a)+(b-d)&lt;/math&gt; so that &lt;math&gt;a=c&lt;/math&gt; and &lt;math&gt;b=4+d&lt;/math&gt;. This only gives rise to &lt;math&gt;a,b,c,d&lt;/math&gt; equal &lt;math&gt;4,4,4,0&lt;/math&gt; which we have previously counted. <br /> <br /> Finally suppose &lt;math&gt;z^2+z+1&lt;/math&gt; divides &lt;math&gt;P&lt;/math&gt;. Using polynomial division ((or that &lt;math&gt;z^3=1&lt;/math&gt; to make the same deductions) we ultimately obtain that &lt;math&gt;b=4+c&lt;/math&gt;. This can only happen if &lt;math&gt;a,b,c,d&lt;/math&gt; is &lt;math&gt;4,4,0,0&lt;/math&gt;. <br /> <br /> Hence we've the polynomials<br /> &lt;cmath&gt;4x^4+4x^3+4x^2+4x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+3x^2+3x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+2x^2+2x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+x^2+x&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3&lt;/cmath&gt;<br /> &lt;cmath&gt;4x^4+4x^3+4x^2&lt;/cmath&gt;<br /> However, by inspection &lt;math&gt;4x^4+4x^3+4x^2+4x+4&lt;/math&gt; has roots on the unit circle, because &lt;math&gt;x^4+x^3+x^2+x+1=(x^5-1)/(x-1)&lt;/math&gt; which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that &lt;math&gt;z_0&lt;/math&gt; is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that &lt;math&gt;z_0&lt;/math&gt; in an &lt;math&gt;n&lt;/math&gt;th root of unity where &lt;math&gt;n&gt;5&lt;/math&gt;, and &lt;math&gt;z_0&lt;/math&gt; is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If &lt;math&gt;n&lt;/math&gt; is prime, then \textit{every} &lt;math&gt;n&lt;/math&gt;th root of unity except 1 must satisfy our polynomial, but since &lt;math&gt;n&gt;5&lt;/math&gt; and the degree of our polynomial is 4, this is impossible. Suppose &lt;math&gt;n&lt;/math&gt; is composite. If it has a prime factor &lt;math&gt;p&lt;/math&gt; greater than 5 then again every &lt;math&gt;p&lt;/math&gt;th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose &lt;math&gt;n&lt;/math&gt; is divisible only by 2,3,or 5. Since by hypothesis &lt;math&gt;z_0&lt;/math&gt; is not a 2nd or 3rd root of unity, &lt;math&gt;z_0&lt;/math&gt; must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy &lt;math&gt;P(z_0)=0&lt;/math&gt;. But &lt;math&gt;(x^5-1)/(x-1)&lt;/math&gt; has exactly all 5th roots of unity excluding 1, and &lt;math&gt;(x^5-1)/(x-1)=x^4+x^3+x^2+x+1&lt;/math&gt;. Thus this must divide &lt;math&gt;P&lt;/math&gt; which implies &lt;math&gt;P(x)=4(x^4+x^3+x^2+x+1)&lt;/math&gt;. This completes the proof.<br /> <br /> == Solution 2 ==<br /> First, assume that &lt;math&gt;z_0\in \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. &lt;math&gt;1&lt;/math&gt; does not work because &lt;math&gt;P(1)\geq 4&lt;/math&gt;. Assume that &lt;math&gt;z_0=-1&lt;/math&gt;. Then &lt;math&gt;0=P(-1)=4-a+b-c+d&lt;/math&gt;, we have &lt;math&gt;4+b+d=a+c\leq 4+b&lt;/math&gt;, so &lt;math&gt;d=0&lt;/math&gt;. Also, &lt;math&gt;a=4&lt;/math&gt; has to be true since &lt;math&gt;4+b=a+c \leq a+b&lt;/math&gt;. Now &lt;math&gt;4+b=4+c&lt;/math&gt; gives &lt;math&gt;b=c&lt;/math&gt;, therefore the only possible choices for &lt;math&gt;(a,b,c,d)&lt;/math&gt; are &lt;math&gt;(4,t,t,0)&lt;/math&gt;. In these cases, &lt;math&gt;P(-1)=4-4+t-t+0=0&lt;/math&gt;. The sum of &lt;math&gt;P(1)&lt;/math&gt; over these cases is &lt;math&gt;\sum_{t=0}^{4} (4+4+t+t) = 40+20=60&lt;/math&gt;.<br /> <br /> Second, assume that &lt;math&gt;z_0\in \mathbb{C} \backslash \mathbb{R}&lt;/math&gt;, so &lt;math&gt;z_0=x_0+iy_0&lt;/math&gt; for some real &lt;math&gt;x_0, y_0&lt;/math&gt;, &lt;math&gt;|x_0|&lt;1&lt;/math&gt;. By conjugate roots theorem we have that &lt;math&gt;P(z_0)=P(z_0^{*})=0&lt;/math&gt;, therefore &lt;math&gt;(z-z_0)(z-z_0^{*}) = (z^2 - 2x_0*z + 1)&lt;/math&gt; is a factor of &lt;math&gt;P(z)&lt;/math&gt;, and we may assume that <br /> <br /> &lt;cmath&gt;P(z) = (z^2-2x_0 z + 1)(4z^2 + pz + d)&lt;/cmath&gt;<br /> <br /> for some real &lt;math&gt;p&lt;/math&gt;. Expanding this polynomial and comparing the coefficients, we have the following equations:<br /> <br /> &lt;cmath&gt;p-8x_0 = a&lt;/cmath&gt;<br /> &lt;cmath&gt;d+4-2px_0 = b&lt;/cmath&gt;<br /> &lt;cmath&gt;p-2dx_0 = c&lt;/cmath&gt;<br /> <br /> From the first and the third we may deduce that &lt;math&gt;2x_0 = \frac{a-c}{d-4}&lt;/math&gt; and that &lt;math&gt;p=\frac{da-4c}{d-4}&lt;/math&gt;, if &lt;math&gt;d\neq 4&lt;/math&gt; (we will consider &lt;math&gt;d=4&lt;/math&gt; by the end). Let &lt;math&gt;k=2px_0=\frac{(a-c)(da-4c)}{(4-d)^2}&lt;/math&gt;. From the second equation, we know that &lt;math&gt;k=d+4-b&lt;/math&gt; is non-negative.<br /> <br /> Consider the following cases:<br /> <br /> Case 1: &lt;math&gt;a=c&lt;/math&gt;. Then &lt;math&gt;k=0&lt;/math&gt;, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=c=4&lt;/math&gt;, &lt;math&gt;d=0&lt;/math&gt;. However, this has already been found (i.e. the form of &lt;math&gt;(4,t,t,0)&lt;/math&gt;).<br /> <br /> Case 2: &lt;math&gt;a&gt;c\geq 0&lt;/math&gt;. Then since &lt;math&gt;k\geq 0&lt;/math&gt;, we have &lt;math&gt;da-4c\geq 0&lt;/math&gt;. However, &lt;math&gt;da \leq 4c&lt;/math&gt;, therefore &lt;math&gt;da-4c=0&lt;/math&gt;. This is true only when &lt;math&gt;d=c&lt;/math&gt;. Also, we get &lt;math&gt;k=0&lt;/math&gt; again. In this case, &lt;math&gt;b=d+4&lt;/math&gt;, so &lt;math&gt;a=b=4&lt;/math&gt;, &lt;math&gt;c=d=0&lt;/math&gt;, &lt;math&gt;x_0=-1/2&lt;/math&gt;. &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/3}&lt;/math&gt;. &lt;math&gt;P(1)=12&lt;/math&gt;.<br /> <br /> Last case: &lt;math&gt;d=4&lt;/math&gt;. We have &lt;math&gt;a=b=c=d=4&lt;/math&gt; and that &lt;math&gt;P(z)&lt;/math&gt; has a root &lt;math&gt;z_0=e^{i2\pi/5}&lt;/math&gt;. &lt;math&gt;P(1)=20&lt;/math&gt;.<br /> <br /> Therefore the desired sum is &lt;math&gt;60+12+20=92 ...\framebox{B}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_11&diff=98505 2001 AMC 12 Problems/Problem 11 2018-11-04T23:33:24Z <p>Epicmonster: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #11]] and [[2001 AMC 10 Problems|2001 AMC 10 #23]]}}<br /> <br /> == Problem ==<br /> <br /> A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?<br /> <br /> &lt;math&gt;<br /> \text{(A) }\frac {3}{10}<br /> \qquad<br /> \text{(B) }\frac {2}{5}<br /> \qquad<br /> \text{(C) }\frac {1}{2}<br /> \qquad<br /> \text{(D) }\frac {3}{5}<br /> \qquad<br /> \text{(E) }\frac {7}{10}<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is &lt;math&gt;\boxed{(\text{D}) \frac {3}{5}}&lt;/math&gt;.<br /> <br /> ==Solution 2 ==<br /> <br /> Let's assume we don't stop picking until all of the balls are picked. To satisfy this condition, we have to arrange the letters: ''W, W, R, R, R'' such that both ''R's'' appear in the first 4. We find the number of ways to arrange the red balls in the first 4 and divide that by the total ways to chose all the balls. The probability of this occurring is &lt;math&gt;\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2001|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_11&diff=98481 2001 AMC 12 Problems/Problem 11 2018-11-03T20:19:43Z <p>Epicmonster: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #11]] and [[2001 AMC 10 Problems|2001 AMC 10 #23]]}}<br /> <br /> == Problem ==<br /> <br /> A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?<br /> <br /> &lt;math&gt;<br /> \text{(A) }\frac {3}{10}<br /> \qquad<br /> \text{(B) }\frac {2}{5}<br /> \qquad<br /> \text{(C) }\frac {1}{2}<br /> \qquad<br /> \text{(D) }\frac {3}{5}<br /> \qquad<br /> \text{(E) }\frac {7}{10}<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is &lt;math&gt;\boxed{(\text{D}) \frac {3}{5}}&lt;/math&gt;.<br /> <br /> ==Solution 2 ==<br /> <br /> We wish to arrange the letters: ''W,W, R, R, R'' such that ''R'' appears last. The probability of this occurring is simply &lt;math&gt;\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2001|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=98253 2015 AMC 8 Problems/Problem 17 2018-10-24T02:30:22Z <p>Epicmonster: /* Solution 4 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> [INCOMPLETE]<br /> <br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have 2x/3 = 18 miles per hour. Solving for x gives us 27 miles per hour. Because 20 minutes is a third of an hour, the distance would then be 9 miles. (D)9<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=98252 2015 AMC 8 Problems/Problem 17 2018-10-24T02:29:50Z <p>Epicmonster: /* Solution 4 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> [INCOMPLETE]<br /> <br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have \dfrac{2x}{3} = 18 miles per hour. Solving for x gives us 27 miles per hour. Because 20 minutes is a third of an hour, the distance would then be 9 miles. (D)9<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=98251 2015 AMC 8 Problems/Problem 17 2018-10-24T02:27:25Z <p>Epicmonster: /* Solution 4 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> [INCOMPLETE]<br /> <br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have \dfrac{2x}{3} = 18 miles per hour. Solving for x gives us 27 miles per hour. Because 20 minutes is a third of an hour, the distance would then be \boxed{\textbf{(D)}~9}$<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_17&diff=98250 2015 AMC 8 Problems/Problem 17 2018-10-24T02:25:25Z <p>Epicmonster: /* Solution 3 */</p> <hr /> <div>Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 6 \qquad<br /> \textbf{(C) } 8 \qquad<br /> \textbf{(D) } 9 \qquad<br /> \textbf{(E) } 12<br /> &lt;/math&gt;<br /> <br /> ===Solution 1===<br /> [INCOMPLETE]<br /> <br /> So &lt;math&gt;\frac{d}{v}=\frac{1}{3}&lt;/math&gt; and &lt;math&gt;\frac{d}{v+18}=\frac{1}{5}&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;d=\frac{1}{5}v+3.6=\frac{1}{3}v&lt;/math&gt;, which gives &lt;math&gt;v=27&lt;/math&gt;, which then gives &lt;math&gt;d=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> &lt;math&gt;d = rt&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; &lt;math&gt;\frac{1}{3} \times r = \frac{1}{5} \times (r + 18)&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{r}{3} = \frac{r}{5} + \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2r}{15} = \frac{18}{5}&lt;/math&gt;<br /> <br /> &lt;math&gt;10r = 270&lt;/math&gt; so &lt;math&gt;r = 27&lt;/math&gt;, plug into the first one and it's &lt;math&gt;\boxed{\textbf{(D)}~9}&lt;/math&gt; miles to school<br /> <br /> ===Solution 3===<br /> We set up an equation in terms of &lt;math&gt;d&lt;/math&gt; the distance and &lt;math&gt;x&lt;/math&gt; the speed In miles per hour. We have &lt;math&gt;d=\left (\dfrac{1}{3}\right )(x)=\left (\dfrac{1}{5}\right) (x+18)&lt;/math&gt;<br /> &lt;cmath&gt;d=(5)(x)=(3)(x+18)&lt;/cmath&gt;<br /> &lt;cmath&gt;5x=3x+54&lt;/cmath&gt;<br /> &lt;cmath&gt;2x=54&lt;/cmath&gt;<br /> &lt;cmath&gt;x=27&lt;/cmath&gt;<br /> <br /> So &lt;math&gt;d=\dfrac{27}{3}=\boxed{\textbf{(D)}~9}&lt;/math&gt;<br /> <br /> ===Solution 4===<br /> Since it takes 3/5 of the original time for him to get to school when there is no traffic, the speed must be 5/3 of the speed in traffic or 2/3 more. Letting x be the distance he can drive with traffic in 1 hour, we have {2}/3x) = 18 miles per hour. Solving for x gives us 27 miles per hour. Because 20 minutes is a third of an hour, the distance would then be \boxed{\textbf{(D)}~9}$<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=91763 2018 AMC 10B Problems/Problem 9 2018-02-17T03:01:12Z <p>Epicmonster: /* Solution 3 (Simple Logic) */</p> <hr /> <div>The faces of each of &lt;math&gt;7&lt;/math&gt; standard dice are labeled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be the probabilities that when all &lt;math&gt;7&lt;/math&gt; dice are rolled, the sum of the numbers on the top faces is &lt;math&gt;10&lt;/math&gt;. What other sum occurs with the same probability as &lt;math&gt;p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br /> <br /> So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br /> <br /> (7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br /> <br /> However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is &lt;math&gt;\textbf{(D)} \text{ 39}&lt;/math&gt;, and we are done.<br /> <br /> Written By: Archimedes15<br /> <br /> ==Solution 2==<br /> <br /> Let's call the unknown value &lt;math&gt;x&lt;/math&gt;. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and &lt;math&gt;x&lt;/math&gt;. So, <br /> <br /> &lt;math&gt;10 - 7 = 42- x &lt;/math&gt;<br /> <br /> &lt;math&gt; x = 39 &lt;/math&gt; and our answer is &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> <br /> By: Soccer_JAMS<br /> <br /> ==Solution 3 (Simple Logic)==<br /> <br /> For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 7 = 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 - 10 = 39 so the answer is &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> <br /> By: epicmonster<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Epicmonster https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_9&diff=91762 2018 AMC 10B Problems/Problem 9 2018-02-17T03:00:08Z <p>Epicmonster: /* Solution 2 */</p> <hr /> <div>The faces of each of &lt;math&gt;7&lt;/math&gt; standard dice are labeled with the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be the probabilities that when all &lt;math&gt;7&lt;/math&gt; dice are rolled, the sum of the numbers on the top faces is &lt;math&gt;10&lt;/math&gt;. What other sum occurs with the same probability as &lt;math&gt;p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7 6s.<br /> <br /> So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows:<br /> <br /> (7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...<br /> <br /> However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is &lt;math&gt;\textbf{(D)} \text{ 39}&lt;/math&gt;, and we are done.<br /> <br /> Written By: Archimedes15<br /> <br /> ==Solution 2==<br /> <br /> Let's call the unknown value &lt;math&gt;x&lt;/math&gt;. By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and &lt;math&gt;x&lt;/math&gt;. So, <br /> <br /> &lt;math&gt;10 - 7 = 42- x &lt;/math&gt;<br /> <br /> &lt;math&gt; x = 39 &lt;/math&gt; and our answer is &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> <br /> By: Soccer_JAMS<br /> <br /> ==Solution 3 (Simple Logic)==<br /> <br /> For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 7 = 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 - 10 = 39 so the answer is &lt;math&gt;\textbf{(D)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Epicmonster