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1,301
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_14
| 1
|
The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$
|
First, we notice that $3^0$ is congruent to $1 \ \text{(mod 10)}$ $3^1$ is $3 \ \text{(mod 10)}$ $3^2$ is $9 \ \text{(mod 10)}$ $3^3$ is $7 \ \text{(mod 10)}$ $3^4$ is $1 \ \text{(mod 10)}$ , and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \ \text{(mod 10)}$
The number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \ \text{(mod 10)}$ . Finally, $13^{1003}$ is congruent to $3^{1003} \equiv 7 \ \text{(mod 10)}$ . Thus the required units digit is $3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}$ , so the answer is $\boxed{9}$
| 9
|
1,302
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_16
| 1
|
Let $x = .123456789101112....998999$ , where the digits are obtained by writing the integers $1$ through $999$ in order.
The $1983$ rd digit to the right of the decimal point is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$
|
We consider the first $1983$ digits, letting the $1983$ rd digit be $z$ . We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$ -digit numbers), let $B$ denote $1011...9899$ (the $2$ -digit numbers), and let $C$ denote $100101...z$ (the $3$ -digit numbers). Clearly there are $9$ digits in $A$ ; in $B$ , there are $99-10+1 = 90$ numbers, so $90 \cdot 2 = 180$ digits. This leaves $1983 - 9 - 180 = 1794$ digits in $C$ . Notice that $1794 = 3 \cdot 598$ with no remainder, so $C$ consists of precisely the first $598$ $3$ -digit numbers. Since the first $3$ -digit number is $100$ , the $598$ th is $100 + 598 - 1 = 697$ , so as $z$ is the last digit, the answer is $\boxed{7}$
| 7
|
1,303
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_19
| 1
|
Point $D$ is on side $CB$ of triangle $ABC$ . If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$ ,
then the length of $AD$ is
$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$
|
Let $AD = y$ . Since $AD$ bisects $\angle{BAC}$ , the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$ , so let $CD = x$ and $DB = 2x$ . Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$ , and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$ . Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$ , so $y$ is $0$ or $2$ . But since $y \neq 0$ $y$ is the length of a side of a triangle), $y$ must be $2$ , so the answer is $\boxed{2}$
| 2
|
1,304
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_20
| 1
|
If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$ , and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$ , then $rs$ is necessarily
$\textbf{(A)} \ pq \qquad \textbf{(B)} \ \frac{1}{pq} \qquad \textbf{(C)} \ \frac{p}{q^2} \qquad \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)}\ \frac{p}{q}$
|
By Vieta's Formulae, we have $\tan(\alpha)\tan(\beta)=q$ and $\cot(\alpha)\cot(\beta)=s$ . Recalling that $\cot\theta=\frac{1}{\tan\theta}$ , we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$
Also by Vieta's Formulae, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$ , and again using $\cot\theta=\frac{1}{\tan\theta}$ , we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$ . Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$ , we therefore deduce that $r=\frac{p}{q}$ , which yields $rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$
Thus, the answer is $\boxed{2}$
| 2
|
1,305
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_23
| 1
|
In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$ .
If the radius of the largest circle is $18$ and that of the smallest one is $8$ , then the radius of the middle circle is
[asy] size(250);defaultpen(linewidth(0.7)); real alpha=5.797939254, x=71.191836; int i; for(i=0; i<5; i=i+1) { real r=8*(sqrt(6)/2)^i; draw(Circle((x+r)*dir(alpha), r)); x=x+2r; } real x=71.191836+40+20*sqrt(6), r=18; pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2); pair A1=300*dir(origin--A), B1=300*dir(origin--B); draw(B1--origin--A1); pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X, Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y, Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z; clip(X--Y--Y1--X1--cycle); label("$L_2$", Z, S); label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]
$\textbf{(A)} \ 12 \qquad \textbf{(B)} \ 12.5 \qquad \textbf{(C)} \ 13 \qquad \textbf{(D)} \ 13.5 \qquad \textbf{(E)} \ 14$
|
Pdfresizer.com-pdf-convert-q23.png
Consider three consecutive circles, as shown in the diagram above; observe that their centres $P$ $Q$ , and $R$ are collinear by symmetry. Let $A$ $B$ , and $C$ be the points of tangency, and let $PS$ and $QT$ be segments parallel to the upper tangent (i.e. $L_1$ ), as also shown. Since $PQ$ is parallel to $QR$ (the three points are collinear), $PS$ is parallel to $QT$ (as both are parallel to $L_1$ ), and $SQ$ is parallel to $TR$ (as both are perpendicular to $L_1$ , due to the tangent being perpendicular to the radius), we have $\triangle PQS \sim \triangle QRT$
Now, if we let $x, y$ , and $z$ be the radii of the three circles (from smallest to largest), then $QS = y - x$ and $RT = z - y$ . Thus, from the similarity that we just proved, $\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}$ (where e.g. $PQ = x + y$ because of collinearity). This equation reduces to $y^2 = zx$ , i.e. $\frac{y}{x} = \frac{z}{y}$ , so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is $8$ and, four radii later, the radius is $18$ , this constant ratio is $\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}$ . Therefore the middle radius is $8 \cdot {\left(\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}\right)}^{2} = 8 \sqrt{\frac{18}{8}} = \sqrt{18 \cdot 8} = \sqrt{144} = 12$ , which is choice $\boxed{12}$
| 12
|
1,306
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_25
| 1
|
If $60^a=3$ and $60^b=5$ , then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$
|
We have that $12=\frac{60}{5}$ . We can substitute our value for 5, to get \[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\] Hence \[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\] Since $4=\frac{60}{3\cdot 5}$ , we have \[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\] Therefore, we have \[60^{(1-a-b)/2}=4^{1/2}=\boxed{2}\]
| 2
|
1,307
|
https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_30
| 1
|
Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$ .
The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$ . If $\stackrel{\frown}{MA} = 40^{\circ}$ , then $\stackrel{\frown}{BN}$ equals
[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B)); draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B); label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]
$\textbf{(A)}\ 10^{\circ}\qquad \textbf{(B)}\ 15^{\circ}\qquad \textbf{(C)}\ 20^{\circ}\qquad \textbf{(D)}\ 25^{\circ}\qquad \textbf{(E)}\ 30^{\circ}$
|
Since $\angle CAP = \angle CBP = 10^\circ$ , quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem "angles inscribed in the same arc are equal".
[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B)); draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B); draw(A--B); draw(circumcircle(A,B,C),dashed); label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]
Since $\angle ACM = 40^\circ$ $\angle ACP = 140^\circ$ , so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\circ}$ , we have $\angle ABP = 40^\circ$ . Hence $\angle ABC = \angle ABP - \angle CBP = 40^ \circ - 10^\circ = 30^\circ$
Since $CA = CB$ , triangle $ABC$ is isosceles, with $\angle BAC = \angle ABC = 30^\circ$ . Now, $\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ$ . Finally, again using the fact that angles inscribed in the same arc are equal, we have $\angle BCP = \angle BAP = \boxed{20}$
| 20
|
1,308
|
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_3
| 1
|
Evaluate $(x^x)^{(x^x)}$ at $x = 2$
$\text{(A)} \ 16 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 256 \qquad \text{(D)} \ 1024 \qquad \text{(E)} \ 65,536$
|
Plugging in $2$ as $x$ gives $4^4$ , which is merely $\boxed{256}$
| 256
|
1,309
|
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_6
| 1
|
The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$ . The remaining angle is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)}\ 130^\circ\qquad \text{(E)}\ 144^\circ$
|
Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\circ$ , and each interior angle belongs to $[0, 180^\circ)$ . Therefore, we must have $n - 2 = \lfloor \frac{2570^\circ}{180^\circ} \rfloor = 15$ . Then the missing angle must be $180*15^\circ - 2570^\circ = 130^\circ$ , so our answer is $\boxed{130}$ and we are done.
| 130
|
1,310
|
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_17
| 1
|
How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^{x}+3=0$
$\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3 \qquad \text {(E)} 4$
|
Let $a = 3^x$ . Then the preceding equation can be expressed as the quadratic, \[9a^2-28a+3 = 0\] Solving the quadratic yields the roots $3$ and $1/9$ . Setting these equal to $3^x$ , we can immediately see that there are $\boxed{2}$ real values of $x$ that satisfy the equation.
| 2
|
1,311
|
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_19
| 1
|
Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$ . The sum of the largest and smallest values of $f(x)$ is
$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$
|
Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$
Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(2,0)$",A[0],(0,-1.5),UnFill); label("$(3,2)$",A[1],(0,1.5),UnFill); label("$(4,0)$",A[2],(0,-1.5),UnFill); label("$(8,0)$",A[3],(0,-1.5),UnFill); [/asy] The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{2}.$
| 2
|
1,312
|
https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_30
| 1
|
Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{none of these}$
|
Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
We have \begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*} Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\] We add the two equations and take the sum modulo $10:$ \begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*} It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{9}.$
| 9
|
1,313
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_1
| 1
|
If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$
|
If we square both sides of the $\sqrt{x+2} = 2$ , we will get $x+2 = 4$ , if we square that again, we get $(x+2)^2 = \boxed{16}$
| 16
|
1,314
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_1
| 2
|
If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$
|
We can immediately get that $x = 2$ , after we square $(2+2)$ , we get $\boxed{16}$
| 16
|
1,315
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_2
| 1
|
Point $E$ is on side $AB$ of square $ABCD$ . If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$
|
Note that $\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\sqrt{3}$ . Since this is the side length of the square, the area of $ABCD$ is $\boxed{3}$
| 3
|
1,316
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_14
| 1
|
In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$ , and the sum of the first $6$ terms is $91$ . The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$
|
Denote the sum of the first $2$ terms as $x$ . Since we know that the sum of the first $6$ terms is $91$ which is $7 \cdot 13$ , we have $x$ $xy$ $xy^2$ $13x$ because it is a geometric series. We can quickly see that $y$ $3$ , and therefore, the sum of the first $4$ terms is $4x = 4 \cdot 7 = \boxed{28}$
| 28
|
1,317
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_20
| 1
|
A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$ ) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$ ). If $\measuredangle CDA=8^\circ$ , what is the largest value $n$ can have?
[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$
|
Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$ , but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$ , the reflection would give us a triangle with angles $16, 90$ , and $74$ . Then, when we reflect again, we will have $180 - 74 - 74$ $32$ . Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle.
Notice that we keep decreasing by increments of $8$ . This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$ , we must decrease by $8$ every time. This is the most optimal path of the light beam.
The pattern of light will be $82-74-66-58-50-42-34-26-18-10$ . When we get to the angle of $2$ degrees, we have reached angle $A$ . Therefore, we don't count the $2$ , so our total number of reflections between $CD$ and $AD$ is $\boxed{10}$
| 10
|
1,318
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_21
| 1
|
In a triangle with sides of lengths $a$ $b$ , and $c$ $(a+b+c)(a+b-c) = 3ab$ . The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$
|
\[(a+b+c)(a+b-c)=3ab\] \[a^2+2ab+b^2-c^2=3ab\] \[a^2+b^2-c^2=ab\] \[c^2=a^2+b^2-ab\] This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cos{c}$ \[c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}\] \[ab=2ab\cos{c}\] \[\frac{1}{2}=\cos{c}\] $\cos{c}$ is $\frac{1}{2}$ , so the angle opposite side $c$ is $\boxed{60}$
| 60
|
1,319
|
https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_24
| 1
|
If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$ , then for each positive integer $n$ $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$
|
Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$ . Using the quadratic equation, we can solve for $x$ . After some simplifying:
\[x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}\] \[x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}\] \[x=\cos(\theta) + i\sin(\theta)\]
Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:
\[(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}\]
Using DeMoivre's Theorem:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)\]
Because $\cos$ is even and $\sin$ is odd:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\]
$=\boxed{2},$ (Error compiling LaTeX. Unknown error_msg)
| 2
|
1,320
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_1
| 1
|
The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$
|
We want to find the smallest integer $x$ so that $7x < 100$ . Dividing by 7 gets $x < 14\dfrac{2}{7}$ , so the answer is 14. $\boxed{14}$
| 14
|
1,321
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_2
| 1
|
The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$
|
It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\boxed{17}$
| 17
|
1,322
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_7
| 1
|
Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3, 4, 12, and 13, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy]
$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$
|
Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{36}$
| 36
|
1,323
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_11
| 1
|
If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$ ,
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$
|
Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.
Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$
Solving the system, we get $d=-\frac{11}{50}$ $a=\frac{1099}{100}$ . The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$
Therefore, $\boxed{110}$
| 110
|
1,324
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_12
| 1
|
The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$ , and that $L_1$ has 4 times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$
|
Solution by e_power_pi_times_i
$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \tan(\theta_1)$ . Similarly, $n = \tan(\theta_2)$ . Since $4n = m$ and $\theta_1 = 2\theta_2$ $\tan(2\theta_2) = 4\tan(\theta_2)$ . Using the angle addition formula for tangents, $\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)$ . Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$ . But line $L_1$ is not horizontal, so therefore $(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})$ . Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{2}$
| 2
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1,325
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_16
| 1
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Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$
|
We assume the side length of the cube is $1$ . The side length of the tetrahedron is $\sqrt2$ , so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$ . The surface area of the cube is $6\times1\times1=6$ , so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{3}$
| 3
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1,326
|
https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_25
| 1
|
In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$ , and $d$ such that for all positive integers $n$ $a_n=b\lfloor \sqrt{n+c} \rfloor +d$ ,
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . The sum $b+c+d$ equals
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$
|
Solution by e_power_pi_times_i
Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$ . However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$ $a = 1$ . Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$ , and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{2}$
| 2
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1,327
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_1
| 1
|
[asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$
|
Solution by e_power_pi_times_i
Since the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\dfrac{1}{2}\cdot\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\dfrac{1}{4}\cdot72 = \boxed{18}$
| 18
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1,328
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_2
| 1
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For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$
|
Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \[(x+1)(y-1) = -1\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{1}$ as our final answer.
| 1
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1,329
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_2
| 2
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For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$
|
Notice that we can do $\frac{x-y}{xy} = \frac{xy}{xy}$ . We are left with $\frac{1}{y} - \frac{1}{x} = 1$ . Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{1}$
| 1
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1,330
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_3
| 1
|
[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ .
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
Solution by e_power_pi_times_i
Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{15}$
| 15
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1,331
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_3
| 2
|
[asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ .
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$
|
WLOG, let the side length of the square and the equilateral triangle be $1$ $\angle{DAE}=90^\circ+60^\circ=150^\circ$ . Apply the law of cosines then the law of sines, we find that $\angle{AED}=15^\circ$ . Select $\boxed{15}$
| 15
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1,332
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_5
| 1
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Find the sum of the digits of the largest even three digit number (in base ten representation)
which is not changed when its units and hundreds digits are interchanged.
$\textbf{(A) }22\qquad \textbf{(B) }23\qquad \textbf{(C) }24\qquad \textbf{(D) }25\qquad \textbf{(E) }26$
|
Solution by e_power_pi_times_i
Since the number doesn't change when the units and hundreds digits are switched, the number must be of the form $aba$ . We want to create the largest even $3$ -digit number, so $a = 8$ and $b = 9$ . The sum of the digits is $8+9+8 = \boxed{25}$
| 25
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1,333
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_11
| 1
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Find a positive integral solution to the equation $\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$
$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$
|
Solution by e_power_pi_times_i
Notice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$ , and the denominator is $n(n+1)$ . Then $\frac{n}{n+1} = \frac{115}{116}$ , so $n = \boxed{115}$
| 115
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1,334
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_12
| 1
|
[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$
|
Solution by e_power_pi_times_i
Because $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$ . Then $\measuredangle ABO = 180^\circ-2\theta$ , and $\measuredangle EBO = \measuredangle OEB = 2\theta$ , so $\measuredangle BOE = 180^\circ-4\theta$ . Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$ . Therefore $\theta+180-4\theta = 135^\circ$ , and $\theta = \boxed{15}$
| 15
|
1,335
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_12
| 2
|
[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$
|
Draw $BO$ . Let $y = \angle BAO$ . Since $AB = OD = BO$ , triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$ . Angle $\angle EBO$ is exterior to triangle $ABO$ , so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$
Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$ . Then $\angle EOD$ is external to triangle $AEO$ , so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$ . But $\angle EOD = 45^\circ$ , so $\angle BAO = y = 45^\circ/3 = \boxed{15}$
| 15
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1,336
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_16
| 1
|
A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2$ . If the radius of the larger circle is $3$ ,
and if $A_1 , A_2, A_1 + A_2$ is an arithmetic progression, then the radius of the smaller circle is
$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{2}{\sqrt{3}}\qquad \textbf{(D) }\frac{3}{2}\qquad \textbf{(E) }\sqrt{3}$
|
Solution by e_power_pi_times_i
The area of the larger circle is $A_1 + A_2 = 9\pi$ . Then $A_1 , 9\pi-A_1 , 9\pi$ are in an arithmetic progression. Thus $9\pi-(9\pi-A_1) = 9\pi-A_1-A_1$ . This simplifies to $3A_1 = 9\pi$ , or $A_1 = 3\pi$ . The radius of the smaller circle is $\boxed{3}$
| 3
|
1,337
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_19
| 1
|
Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$
|
Solution by e_power_pi_times_i
Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$ . However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$ , and dilating by $2$ gives $\pm2$ . The sum of the squares is $(2)^2+(-2)^2 = \boxed{8}$
| 8
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1,338
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_20
| 1
|
If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals
$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$
|
Solution by e_power_pi_times_i
Since $a=\frac{1}{2}$ $b=\frac{1}{3}$ . Now we evaluate $\arctan a$ and $\arctan b$ . Denote $x$ and $\theta$ such that $\arctan x = \theta$ . Then $\tan(\arctan(x)) = \tan(\theta)$ , and simplifying gives $x = \tan(\theta)$ . So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$ . The question asks for $\theta_a + \theta_b$ , so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$ . Using the angle addition formula for $\tan(\alpha+\beta)$ , we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$ . Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$ . Simplifying, $\tan(\theta_a + \theta_b) = 1$ , so $\theta_a + \theta_b$ in radians is $\boxed{4}$
| 4
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1,339
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_24
| 1
|
Sides $AB,~ BC$ , and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$ , and $20$ , respectively.
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$ , then side $AD$ has length
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$
|
We know that $\sin(C)=-\cos(B)=\frac{3}{5}$ . Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$ . It is known that $\sin(x)=\cos(90-x)$ , so $180-C=90-(180-C)=180-B$ . We simplify this as follows:
\[-90+C=180-B\]
\[B+C=270^{\circ}\]
Since $B+C=270^{\circ}$ , we know that $A+D=360-(B+C)=90^{\circ}$ . Now extend $AB$ and $CD$ as follows:
[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]
Let $AB$ and $CD$ intersect at $E$ . We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$
Since $\sin BCD = \frac{3}{5}$ , we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$ . Thus, $EB=3$ and $EC=4$ from simple sin application.
$AD$ is the hypotenuse of right $\triangle AED$ , with leg lengths $AB+BE=7$ and $EC+CD=24$ . Thus, $AD=\boxed{25}$
| 25
|
1,340
|
https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_29
| 1
|
For each positive number $x$ , let $f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$ .
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$
|
Let $a = \left( x + \frac{1}{x} \right)^3$ and $b = x^3 + \frac{1}{x^3}$ . Then \begin{align*} f(x) &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + 2 + \frac{1}{x^6})}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^3 + \frac{1}{x^3})^2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{a^2 - b^2}{a + b}. \end{align*}
By difference of squares, \begin{align*} f(x) &= \frac{(a - b)(a + b)}{a + b} \\ &= a - b \\ &= \left( x + \frac{1}{x} \right)^3 - \left( x^3 + \frac{1}{x^3} \right) \\ &= \left( x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \right) - \left( x^3 + \frac{1}{x^3} \right) \\ &= 3x + \frac{3}{x} \\ &= 3 \left( x + \frac{1}{x} \right). \end{align*}
By the AM-GM inequality, \[x + \frac{1}{x} \ge 2,\] so $f(x) \ge 6$ . Furthermore, when $x = 1$ $f(1) = 6$ , so the minimum value of $f(x)$ is $\boxed{6}$
| 6
|
1,341
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1
| 1
|
If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$
|
By guessing and checking, 2 works. $\frac{2}{x} = \boxed{1}$ ~awin
| 1
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1,342
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1
| 2
|
If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$
|
Multiplying each side by $x^2$ , we get $x^2-4x+4 = 0$ . Factoring, we get $(x-2)(x-2) = 0$ . Therefore, $x = 2$ $\frac{2}{x} = \boxed{1}$ ~awin
| 1
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1,343
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1
| 3
|
If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$
|
Directly factoring, we get $(1-\frac{2}{x})^2 = 0$ . Thus $\frac{2}{x}$ must equal $\boxed{1}$
| 1
|
1,344
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_2
| 1
|
If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$
|
Creating equations, we get $4\cdot\frac{1}{2\pi r} = 2r$ . Simplifying, we get $\frac{1}{\pi r} = r$ . Multiplying each side by $r$ , we get $\frac{1}{\pi} = r^2$ . Because the formula of the area of a circle is $\pi r^2$ , we multiply each side by $\pi$ to get $1 = \pi r^2$ .
Therefore, our answer is $\boxed{1}$
| 1
|
1,345
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_4
| 1
|
If $a = 1,~ b = 10, ~c = 100$ , and $d = 1000$ , then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad \textbf{(E) }4242$
|
Adding all four of the equations up, we can see that it equals \[3(a+b+c+d)\] This is equal to $3(1111) = \boxed{3333}$ ~awin
| 333
|
1,346
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_5
| 1
|
Four boys bought a boat for $\textdollar 60$ . The first boy paid one half of the sum of the amounts paid by the other boys;
the second boy paid one third of the sum of the amounts paid by the other boys;
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?
$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$
|
If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{3}$ of the total.
If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{4}$ of the total.
If the third boy paid one fourth of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{5}$ of the total.
Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$ .
Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{13}$ ~awin
| 13
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1,347
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_6
| 1
|
The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \ \ y=2xy\] is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }4$
|
If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.
Case 1: $y = 0$
If $y = 0$ , then $x = x^2$ and $0 = 0$
Therefore, $x = 0$ or $x = 1$
This yields 2 solutions
Case 2: $x = \frac{1}{2}$
If $x = \frac{1}{2}$ , this means that $y = y$ , and $\frac{1}{2} = \frac{1}{4} + y^2$
Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$
This yields another 2 solutions.
$2+2 = \boxed{4}$
| 4
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1,348
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_11
| 1
|
If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$ , then $r$ equals
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$
|
The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\sqrt{r}$ . Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$ , then the distance between $(0,0)$ and the line $x + y = r$ is $\sqrt{r}$
The distance between $(0,0)$ and the line $x + y = r$ is \[\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.\] Hence, \[\frac{r}{\sqrt{2}} = \sqrt{r}.\] Then $r = \sqrt{r} \cdot \sqrt{2}$ , so $\sqrt{r} = \sqrt{2}$ , which means $r = \boxed{2}$ or (B), $2$
| 2
|
1,349
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_13
| 1
|
If $a,b,c$ , and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are
the solutions of $x^2+cx+d=0$ , then $a+b+c+d$ equals
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }(-1+\sqrt{5})/2$
|
By Vieta's formulas, $c + d = -a$ $cd = b$ $a + b = -c$ , and $ab = d$ . From the equation $c + d = -a$ $d = -a - c$ , and from the equation $a + b = -c$ $b = -a - c$ , so $b = d$
Then from the equation $cd = b$ $cb = b$ . Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$ . Similarly, from the equation $ab = d$ $ab = b$ , so $a = 1$ . Then $b = d = -a - c = -2$ . Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{2}$ . The answer is (B).
| 2
|
1,350
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_14
| 1
|
If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base- $n$ number system is $18$ ,
then the base-n representation of $b$ is
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$
|
Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$
$n_n = 10_n$ , so $10_n + m_n = 18_n$
\[m_n = 8_n\]
Also by Vieta's formulas, $n_n \cdot m_n = b_n$ \[10_n \cdot 8_n = \boxed{80}\]
| 80
|
1,351
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_18
| 1
|
What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$
$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad \textbf{(E) }\text{There is no such integer}$
|
Adding $\sqrt{n - 1}$ to both sides, we get \[\sqrt{n} < \sqrt{n - 1} + 0.01.\] Squaring both sides, we get \[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\] which simplifies to \[0.9999 < 0.02 \sqrt{n - 1},\] or \[\sqrt{n - 1} > 49.995.\] Squaring both sides again, we get \[n - 1 > 2499.500025,\] so $n > 2500.500025$ . The smallest positive integer $n$ that satisfies this inequality is $\boxed{2501}$
| 501
|
1,352
|
https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_22
| 1
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The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
|
There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\boxed{3}$ , since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.
| 3
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1,353
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https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_22
| 2
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The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
|
If all of them are false, that would mean that the $4$ th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\boxed{3}$
| 3
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1,354
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_1
| 1
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If $y = 2x$ and $z = 2y$ , then $x + y + z$ equals
$\text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x$
|
Solution by e_power_pi_times_i
$x+y+z = x+(2x)+(4x) = \boxed{7}$
| 7
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1,355
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_9
| 1
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[asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]
In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$ , arc $BC$ , and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$
|
Solution by e_power_pi_times_i
If arcs $AB$ $BC$ , and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$ . Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$ , and $\measuredangle ADB = \measuredangle ACB = \theta$ . Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ $\theta = 55^\circ$ $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{15}$
| 15
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1,356
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_10
| 1
|
If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$ , then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$
|
Solution by e_power_pi_times_i
Notice that if $x=1$ , then $a_7x^7 + a_6x^6 + \cdots + a_0 = a_7 + a_6 + \cdots + a_0$ . Therefore the answer is $(3(1)-1)^7) = \boxed{128}$
| 128
|
1,357
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_12
| 1
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Al's age is $16$ more than the sum of Bob's age and Carl's age, and the square of Al's age is $1632$ more than the square of the sum of
Bob's age and Carl's age. What is the sum of the ages of Al, Bob, and Carl?
$\text{(A)}\ 64 \qquad \text{(B)}\ 94 \qquad \text{(C)}\ 96 \qquad \text{(D)}\ 102 \qquad \text{(E)}\ 140$
|
Solution by e_power_pi_times_i
Denote Al's age, Bob's age, and Carl's age by $a$ $b$ , and $c$ , respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$ . Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$ . Thus, $bc + 16b + 16c = 688$ , and $(b+16)(c+16) = 944$ . Since $944 = 2^4\cdot59$ $(b,c) = (0,43)$ or $(43,0)$ . Then $a + b + c = 2b + 2c + 16 = \boxed{102}$
| 102
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1,358
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_14
| 1
|
How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }\text{more than }4$
|
Solution by e_power_pi_times_i
If $m+n=mn$ $mn-m-n = (m-1)(n-1)-1 = 0$ . Then $(m-1)(n-1) = 1$ , and $(m,n) = (2,2) or (0,0)$ . The answer is $\boxed{2}$
| 2
|
1,359
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_18
| 1
|
If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then
$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$
|
Solution by e_power_pi_times_i
Note that $\log_{a}b = \dfrac{\log{b}}{\log{a}}$ . Then $y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{5}$
| 5
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1,360
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_25
| 1
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Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$
|
We first observe that since there will be more 2s than 5s in $1005!$ , we are looking for the largest $n$ such that $5^n$ divides $1005!$ . We will use the fact that:
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]
(This is an application of Legendre's formula).
From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$ . Thus, our calculation becomes
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]
\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]
\[n = 201 + 40 + 8 + 1 = \boxed{250}\]
| 250
|
1,361
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28
| 1
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Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$
|
Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$ . Then $r(x)$ is the unique polynomial such that \[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$ , and $\deg r(x) < 5$
Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$ . Also, \[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\] Each term is a multiple of $x^6 - 1$ . For example, \[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\] Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$ , which means that $g(x^{12}) - 6$ is a multiple of $g(x)$ . Therefore, the remainder is $\boxed{6}$ . The answer is (A).
| 6
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1,362
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28
| 2
|
Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$
|
We express the quotient and remainder as follows. \[g(x^{12}) = Q(x) g(x) + R(x)\] Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$ . Hence, we have $x^6 = 1$ , allowing us to set: \[g(x^{12}) = 6\] \[g(x) = 0\] We have $5$ values of $x$ that return $R(x) = 6$ . However, $g(x)$ is quintic, implying the remainder is of degree at most $4$ . Since there are $5$ solutions, the only possibility is that the remainder is a constant $\boxed{6}$
| 6
|
1,363
|
https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28
| 3
|
Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$
|
We can use the Chinese remainder theorem over $\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$ $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$ , we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$ $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$ , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$ , the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$ ). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well.
| 6
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1,364
|
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_24
| 1
|
In the adjoining figure, circle $K$ has diameter $AB$ ; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$ ; and circle $M$ tangent to circle $K$ , to circle $L$ and $AB$ . The ratio of the area of circle $K$ to the area of circle $M$ is [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(150); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); label("$A$", A, W); label("$K$", K, S); label("$B$", B, E); label("$L$", L); label("$M$", M); [/asy] $\textbf{(A) }12\qquad \textbf{(B) }14\qquad \textbf{(C) }16\qquad \textbf{(D) }18\qquad \textbf{(E) }\text{not an integer}$
|
Let $R$ and $r$ be the radius of $\odot K$ and the radius of $\odot M,$ respectively. It follows that the radius of $\odot L$ is $\frac{R}{2}.$
Suppose $P$ is the foot of the perpendicular from $M$ to $\overline{KL}.$ We construct the auxiliary lines, as shown below: [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(200); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25),I=(2*sqrt(2)/3,1/3),E=(sqrt(2)/3,1/3),P=(0,0.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); draw(L--K,red); draw(L--M,red); draw(K--I,red); draw(P--M,red); label("$A$", A, (-5/4,0)); label("$K$", K, (0,-5/4)); label("$B$", B, (5/4,0)); label("$L$", L, (0,5/4)); label("$M$", M, (0,5/4)); label("$P$", P, (-5/4,0)); dot(K,linewidth(4)); dot(L,linewidth(4)); dot(M,linewidth(4)); dot(I,linewidth(4)); dot(E,linewidth(4)); dot(P,linewidth(4)); [/asy] In right $\triangle KPM,$ we have $KP=r$ and $KM=R-r.$ By the Pythagorean Theorem, we get $PM^2=(R-r)^2-r^2.$
In right $\triangle LPM,$ we have $LP=\frac{R}{2}-r$ and $LM=\frac{R}{2}+r.$ By the Pythagorean Theorem, we get $PM^2=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2.$
We equate the expressions for $PM^2,$ then simplify: \begin{align*} (R-r)^2-r^2&=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2 \\ \left(R^2-2Rr+r^2\right)-r^2&=\left(\frac{R^2}{4}+Rr+r^2\right)-\left(\frac{R^2}{4}-Rr+r^2\right) \\ R^2-2Rr&=2Rr \\ R^2&=4Rr \\ R&=4r. \end{align*} Therefore, the ratio of the area of $\odot K$ to the area of $\odot M$ is $\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=\boxed{16}.$
| 16
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1,365
|
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_27
| 1
|
If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$
|
Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note that \begin{align*} x^2&=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2} \\ &=\frac{\left(\sqrt{5}+2\right)+2\cdot\left(\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ &=2. \end{align*} Since $x>0,$ we have $x=\sqrt{2}.$
On the other hand, note that \begin{align*} y^2&=3-2\sqrt{2} \\ &=2-2\sqrt{2}+1 \\ &=\left(\sqrt{2}-1\right)^2. \end{align*} Since $y>0,$ we have $y=\sqrt{2}-1.$
Finally, the answer is \[N=x-y=\boxed{1}.\]
| 1
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1,366
|
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_27
| 2
|
If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$
|
Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note that \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] We rewrite each term in the numerator separately:
Substituting these results into $(\bigstar),$ we have \[x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.\] On the other hand, we have \[y=\sqrt2-1\] by the argument of either Solution 1 or Solution 2.
Finally, the answer is \[N=x-y=\boxed{1}.\]
| 1
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1,367
|
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_28
| 1
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Lines $L_1,L_2,\dots,L_{100}$ are distinct. All lines $L_{4n}, n$ a positive integer, are parallel to each other.
All lines $L_{4n-3}, n$ a positive integer, pass through a given point $A.$ The maximum number of points of intersection of pairs of lines from the complete set $\{L_1,L_2,\dots,L_{100}\}$ is
$\textbf{(A) }4350\qquad \textbf{(B) }4351\qquad \textbf{(C) }4900\qquad \textbf{(D) }4901\qquad \textbf{(E) }9851$
|
We partition $\{L_1,L_2,\dots,L_{100}\}$ into three sets. Let \begin{align*} X &= \{L_n\mid n\equiv0\pmod{4}\}, \\ Y &= \{L_n\mid n\equiv1\pmod{4}\}, \\ Z &= \{L_n\mid n\equiv2,3\pmod{4}\}, \\ \end{align*} from which $|X|=|Y|=25$ and $|Z|=50.$
Any two distinct lines can intersect at most once. To maximize the number of points of intersection, note that each point must be passed by exactly two lines. If three or more lines pass through the same point, then we can create more points of intersection by translating the lines.
We construct the sets one by one:
Together, the answer is $626+3725=\boxed{4351}.$
| 351
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1,368
|
https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_30
| 1
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How many distinct ordered triples $(x,y,z)$ satisfy the following equations? \begin{align*} x + 2y + 4z &= 12 \\ xy + 4yz + 2xz &= 22 \\ xyz &= 6 \end{align*} $\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$
|
The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac b2,\frac c4\right).$
We rewrite the given equations in terms of $a,b,$ and $c:$ \begin{align*} a + b + c &= 12, \\ \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ \frac{abc}{8} &= 6. \end{align*} We clear fractions in these equations: \begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*} By Vieta's Formulas, note that $a,b,$ and $c$ are the roots of the equation \[r^3 - 12r^2 + 44r - 48 = 0,\] which factors as \[(r - 2)(r - 4)(r - 6) = 0.\] It follows that $\{a,b,c\}=\{2,4,6\}.$ Since the substitution $(x,y,z)=\left(a,\frac b2,\frac c4\right)$ is not symmetric with respect to $x,y,$ and $z,$ we conclude that different ordered triples $(a,b,c)$ generate different ordered triples $(x,y,z),$ as shown below: \[\begin{array}{c|c|c||c|c|c} & & & & & \\ [-2.5ex] \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] \hline & & & & & \\ [-2ex] 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array}\] So, there are $\boxed{6}$ such ordered triples $(x,y,z).$
| 6
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1,369
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_2
| 1
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For which real values of m are the simultaneous equations
\begin{align*}y &= mx + 3 \\ y& = (2m - 1)x + 4\end{align*}
satisfied by at least one pair of real numbers $(x,y)$
$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\ \text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m$
|
Solution by e_power_pi_times_i
Solving the systems of equations, we find that $mx+3 = (2m-1)x+4$ , which simplifies to $(m-1)x+1 = 0$ . Therefore $x = \dfrac{1}{1-m}$ $x$ is only a real number if $\boxed{1}$
| 1
|
1,370
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_4
| 1
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If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$
|
Solution by e_power_pi_times_i
Denote the side of one square as $s$ . Then the diagonal of the second square is $s$ , so the side of the second square is $\dfrac{s\sqrt{2}}{2}$ . The area of the second square is $\dfrac{1}{2}s^2$ , so the ratio of the areas is $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{2}$
| 2
|
1,371
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_6
| 1
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The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$
|
Solution by e_power_pi_times_i
When the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{80}$
| 80
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1,372
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_10
| 1
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The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$ , where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2$
|
We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: \[110\text{ some number of zeros }011.\] This works because of the way they are multiplied. Therefore, the answer is $\boxed{4}$
| 4
|
1,373
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_15
| 1
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In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ -1$
|
First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$ th term is $-2$ and the sum of the all those previous terms is $0$ . Then, write the 97th to the 100th terms down: $1, 3, 2,-1$ and add them up to get the sum of $\boxed{5}$
| 5
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1,374
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27
| 1
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If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$
|
If $p$ is a root of $x^3 - x^2 + x - 2 = 0$ , then $p^3 - p^2 + p - 2 = 0$ , or \[p^3 = p^2 - p + 2.\] Similarly, $q^3 = q^2 - q + 2$ , and $r^3 = r^2 - r + 2$ , so \[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]
By Vieta's formulas $p + q + r = 1$ $pq + pr + qr = 1$ , and $pqr = 2$ . Squaring the equation $p + q + r = 1$ , we get \[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\] Subtracting $2pq + 2pr + 2qr = 2$ , we get \[p^2 + q^2 + r^2 = -1.\]
Therefore, $p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}$
| 4
|
1,375
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27
| 2
|
If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$
|
We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$ . By Vieta's formulas, $p+q+r=1$ $pqr=2$ , and $pq+qr+pr=1$ .
So if we can find $p^2+q^2+r^2$ , we are done. Notice that $(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr$ , so $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1$ , which means that $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{4}$
| 4
|
1,376
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27
| 3
|
If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$
|
Use Vieta's formulas to get $p+q+r=1$ $pq+qr+pr=1$ , and $pqr=2$
Square $p+q+r=1$ , and get $p^2+q^2+r^2+2pq+2pr+2qr=1$
Substitute $pq+qr+pr=1$ and simplify to get $-1=p^2+q^2+r^2$
After that, multiply both sides by $1=p+q+r$ , to get $-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q$
Then, factor out $pq$ $qr$ , and $pr$ $-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)$
Then, substitute the first equation into $p+q$ $q+r$ , and $p+r$ $-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)$
Then, multiply it out: $-1=p^3+q^3+r^3+pq+qr+pr-3pqr$
After that, substitute the equations $pq+qr+pr=1$ and $pqr=2$ $-1=p^3+q^3+r^3+1-6$
Solving that, you get $p^3+q^3+r^3=\boxed{4}$ ~EZ PZ Ms.Lemon SQUEEZY
| 4
|
1,377
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_29
| 1
|
What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$
|
$(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$ . Finally, multiply and add to find that the smallest integer higher is $\boxed{970}$
| 970
|
1,378
|
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_29
| 2
|
What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$
|
Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6$ . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\sqrt{3}-\sqrt{2})^6) < 1$ , the greatest integer greater than our original expression is $\boxed{970}$
| 970
|
1,379
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_2
| 1
|
Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$ $i=1, 2$ . Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \ } \frac{b}{3} \qquad \mathrm{(D) \ } 2b \qquad \mathrm{(E) \ }-\frac{b}{3}$
|
Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$ . By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{3}$
| 3
|
1,380
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_4
| 1
|
What is the remainder when $x^{51}+51$ is divided by $x+1$
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ }51$
|
From the Remainder Theorem , the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \boxed{50}$
| 50
|
1,381
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_5
| 1
|
Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$ , if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$ , find $\measuredangle EBC$
$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \ } 70^\circ \qquad \mathrm{(D) \ } 88^\circ \qquad \mathrm{(E) \ }92^\circ$
|
Since $ABCD$ is cyclic, opposite angles must sum to $180^\circ$ . Therefore, $\angle ADC+\angle ABC=180^\circ$ , and $\angle ABC=180^\circ-\angle ADC=180^\circ-68^\circ=112^\circ$ . Notice also that $\angle ABC$ and $\angle CBE$ form a linear pair, and so they sum to $180^\circ$ . Therefore, $\angle EBC=180^\circ-\angle ABC=180^\circ-112^\circ=68^\circ, \boxed{68}$ . Notice that the answer didn't even depend on $\angle BAD$
| 68
|
1,382
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_10
| 1
|
What is the smallest integral value of $k$ such that \[2x(kx-4)-x^2+6=0\] has no real roots?
$\mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }5$
|
Expanding, we have $2kx^2-8x-x^2+6=0$ , or $(2k-1)x^2-8x+6=0$ . For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6}$ . From here, we can easily see that the smallest integral value of $k$ is $2, \boxed{2}$
| 2
|
1,383
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_17
| 1
|
If $i^2=-1$ , then $(1+i)^{20}-(1-i)^{20}$ equals
$\mathrm{(A)\ } -1024 \qquad \mathrm{(B) \ }-1024i \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 1024 \qquad \mathrm{(E) \ }1024i$
|
Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,
\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \boxed{0}.\]
| 0
|
1,384
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_29
| 1
|
For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \ } 80400 \qquad \mathrm{(D) \ } 80600 \qquad \mathrm{(E) \ }80800$
|
The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$
We now want to evaluate $\sum_{p=1}^{10}(1600p-780)$ \[\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)\] \[=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{80200}.\]
| 200
|
1,385
|
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_30
| 1
|
A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If $R$ is the ratio of the lesser part to the greater part, then the value of
\[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}\]
is
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }2R \qquad \mathrm{(C) \ } R^{-1} \qquad \mathrm{(D) \ } 2+R^{-1} \qquad \mathrm{(E) \ }2+R$
|
Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\frac{w}{l}=\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0$ . Therefore, $R^2+R-1=0$
From this, we have $R^2=-R+1$ . Dividing both sides by $R$ , we get $R=-1+\frac{1}{R}\implies R^{-1}=R+1$ . Therefore, $R^2+R^{-1}=-R+1+R+1=2$ . Finally, we have \[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \boxed{2}.\]
| 2
|
1,386
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_2
| 1
|
One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$
|
The total number of cubes is $10^3$ or $1000$ . Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$ , since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\boxed{488}$ cubes that have at least one face painted.
| 488
|
1,387
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_2
| 2
|
One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$
|
Each face has $100$ cubes, so multiply by six to get $600$ . However, we overcounted each small cube on the edge but not on corner of the big cube once and each small cube on the corner of the big cube twice. Thus, there are $600 - (12 \cdot 8 + 2 \cdot 8) = \boxed{488}$ cubes that have at least one face painted.
| 488
|
1,388
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_3
| 1
|
The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is
$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80$
|
We can guess and check small primes, subtract it from $126$ , and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \cdot 3^2 \cdot 7$ , we can eliminate $2$ $3$ , and $7$ as an option because subtracting these would result in a composite number.
If we subtract $5$ , then the resulting number is $121$ , which is not prime. If we subtract $11$ , then the resulting number is $115$ , which is also not prime. But when we subtract $13$ , the resulting number is $113$ , a prime number. The largest possible difference is $113-13=\boxed{100}$
| 100
|
1,389
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_6
| 1
|
If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$ , when written in base 10, equals
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$
|
Write out the numbers using the definition of base numbers \[554_b = 5b^2 + 5b + 4\] \[24_b = 2b+4\] Since $554_b = (24_b)^2$ , we can write an equation. \[5b^2 + 5b + 4 = (2b+4)^2\] \[5b^2 + 5b + 4 = 4b^2 + 16b + 16\] \[b^2 - 11b - 12 = 0\] \[(b-12)(b+1) = 0\] Since base numbers must be positive, $b$ in base 10 equals $\boxed{12}$
| 12
|
1,390
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_7
| 1
|
The sum of all integers between 50 and 350 which end in 1 is
$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$
|
The numbers that we are adding are $51,61,71 \cdots 341$ . The numbers are part of an arithmetic series with first term $51$ , last term $341$ , common difference $10$ , and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\tfrac{30 \cdot 392}{2} = \boxed{5880}$
| 880
|
1,391
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_8
| 1
|
If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high is
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$
|
The statues are similar, and since the height if the smaller statue is $\tfrac{1}{6}$ of the original statue, the surface area is $\tfrac{1}{36}$ of the original statue. Thus, $\tfrac{1}{36}$ pints of paint is needed for one 1 ft. statue, so painting 540 of these statues requires $\tfrac{540}{36} = \boxed{15}$ pints of paint.
| 15
|
1,392
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_9
| 1
|
In $\triangle ABC$ with right angle at $C$ , altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$ , then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$
|
[asy] pair A=(-6,0),B=(6,0),C=(-3,5.196),M=(0,0),H=(-3,0); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-3,5.196)--(0,0)); draw(C--H); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); dot(M); label("$M$",M,S); dot(H); label("$H$",H,S); markscalefactor=0.1; draw(anglemark((-6,0),C,(6,0))); draw((-3,0.5)--(-2.5,0.5)--(-2.5,0)); [/asy]
Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).
Note that because $CM$ is a median, $AM = BM$ . Also, by ASA Congruency, $\triangle CHA = \triangle CHM$ , so $AH = HM$ . That means $HM = \tfrac{1}{4} \cdot AB$ , and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{4}$
| 4
|
1,393
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_16
| 1
|
If the sum of all the angles except one of a convex polygon is $2190^{\circ}$ , then the number of sides of the polygon must be
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$
|
Let $n$ be the number of sides in the polygon. The number of interior angles in the polygon is $180(n-2)$ . We know that the sum of all but one of them is $2190^{\circ}$ , so the sum of all the angles is more than that. \[180(n-2) > 2190\] \[n-2 > 12 \tfrac{1}{6}\] \[n > 14 \tfrac{1}{6}\]
The sum of the angles in a 15-sided polygon is $2340^{\circ}$ , making the remaining angle $150^{\circ}$ . The angles of a convex polygon are all less than $180^{\circ}$ , and since adding one more side means adding $180^{\circ}$ to the measure of the remaining angle, we can confirm that there are $\boxed{15}$ sides in the polygon.
| 15
|
1,394
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_23
| 1
|
There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$
|
There are three red faces, and two are on the card that is completely red, so our answer is $\frac{2}{3}$ , which is $\boxed{23}$
| 23
|
1,395
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_26
| 1
|
The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$
|
Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$
We can write an equation on the difference between the last and first term based on the conditions. \[a+r(n-1)-a =10.5\] \[rn-r=10.5\] Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so \[24 + r\frac{n}{2} = 30\] \[nr = 12\] Substituting the value back to a previous equation, \[12-r=10.5\] \[r=1.5\] Substituting to a previous equation again, \[1.5n-1.5=10.5\] \[n=8\] Thus, there are $\boxed{8}$ terms in the arithmetic sequence.
| 8
|
1,396
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_29
| 1
|
Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ \text{infinity} \qquad \textbf{(E)}\ \text{none of these}$
|
Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\tfrac{dx}{5}$ . Since the time elapsed for both boys is equal, one boy ran $5(\tfrac{dx}{5})$ feet while the other boy ran $9(\tfrac{dx}{5})$ feet. Because both finished at the starting point, both ran an integral number of laps, so $5(\tfrac{dx}{5})$ and $9(\tfrac{dx}{5})$ are multiples of $d$ . Because both stopped when both met at the start for the first time, $x = 5$
Note that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other. When $0 < x \le 5$ and either $5(\tfrac{dx}{5})$ or $9(\tfrac{dx}{5})$ is a multiple of $d$ , one of the runners completed a lap. This is achieved when $x = \tfrac59, 1, \tfrac{10}{9}, \tfrac{15}{9}, 2, \tfrac{20}{9}, \tfrac{25}{9}, 3, \tfrac{30}{9}, \tfrac{35}{9}, 4, \tfrac{40}{9}, 5$ , so the two meet each other (excluding start and finish) a total of $\boxed{13}$ times.
| 13
|
1,397
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_31
| 1
|
In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$
|
The right side of the equation can be rewritten as $111T = 37 \cdot 3T$ . With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$
The only possibility that works is $37 \cdot 27 = 999$ . That means $E+M+T+Y = \boxed{21}$
| 21
|
1,398
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_32
| 1
|
The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$
|
[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label("6",(4.5,2.598,0),SW); label("$\sqrt{15}$",(4.5,0.866,0.866),N); currentprojection=orthographic(1/6,1/2,1/3); [/asy]
Draw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\sqrt{3}$
[asy] draw((0,1.732)--(0,0)--(3.464,0),dotted); draw((0,1.732)--(3.464,0)); label("$2\sqrt{3}$",(1.732,0),S); label("$\sqrt{15}$",(1.732,0.866),NE); [/asy]
Using the Pythagorean Theorem , the length of the altitude is $\sqrt{3}$ , so the volume of the triangular pyramid is $\tfrac13 \cdot \tfrac{6^2 \cdot \sqrt{3}}{4} \cdot \sqrt{3} = \boxed{9}$
| 9
|
1,399
|
https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_33
| 1
|
When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is
$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textbf{(D)}\ 30\% \qquad \textbf{(E)}\ 33\frac13 \%$
|
Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water. We can write two equations since we know the percentage of acid after some water and acid. \[\frac{a}{a+w+1} = \frac{1}{5}\] \[\frac{a+1}{a+w+2} = \frac{1}{3}\] Cross-multiply to get rid of the fractions. \[5a = a+w+1\] \[3a+3=a+w+1\] Solve the system to get $a=1$ and $w=3$ . The percentage of acid in the original mixture is $\tfrac{1}{1+3} = \boxed{25}$
| 25
|
1,400
|
https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_3
| 1
|
If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$ , then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad \textbf{(E) }2$
|
Using DeMoivre's theorem, we can calculate $x^2=\frac{1+i\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer \[\boxed{1}.\] ~lopkiloinm
| 1
|
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