problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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48a22df7344d0edab73b0c1c78d26849 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_8 | If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$ | Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$
Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{300}$ | C | 300 |
bab24288566c327e0ab1154aa5d9f923 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_9 | The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$ | The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \... | A | 2 |
81be900c6c9c29a8647ad9b2fbd94a91 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_16 | The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{5}$ | B | 5 |
97e603f0b9a637b72c5b661d2693bbcd | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_18 | Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ ... | The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$ , so $\boxed{1}$ | E | 1 |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Using synthetic division, we get that the remainder is $\boxed{2}$ | D | 2 |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{2.}$ | D | 2. |
1593d8af2008d5782cf0f6c75a39905e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20 | When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$ , so $x^{13} - 1$ is divisible by $x-1$ , meaning $(x^{13} - 1) + 2$ leaves a remainder of $\boxed{2.}$ | D | 2. |
cc5521e4762c2a50f82e6ef088e90849 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_21 | The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ ... | If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2... | B | 24 |
5f22b1079465e17c89abf6fd894c88f6 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_22 | Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$ | Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{28}$ | D | 28 |
15cba6f5de754971777da9e876c0222b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_23 | A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textb... | $12\frac{1}{2}\%$ is the same as $\frac{1}{8}$ , so the man sets one eighth of each month's rent aside, so he only gains $\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\%$ , or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the w... | B | 83.33 |
78a66efd519363e3ceadc13a0fd88533 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24 | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | $x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 ... | E | 1 |
78a66efd519363e3ceadc13a0fd88533 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24 | The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$ | We can create symmetry in the equation: \[x+\sqrt{x-2} = 4\] \[x-2+\sqrt{x-2} = 2.\] Let $y = \sqrt{x-2}$ , then we have \[y^2+y-2 = 0\] \[(y+2)(y-1) = 0\] The two roots are $\sqrt{x-2} = -2, 1$
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remem... | E | 1 |
b783b3a52ff9a4137359a82df06e9845 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | $\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{5}$ | D | 5 |
b783b3a52ff9a4137359a82df06e9845 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25 | The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | $\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{5}$ | D | 5 |
367ee23bcd67eaf2f0f2db1b666d64b5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ $m=\boxed{10}$ | E | 10 |
367ee23bcd67eaf2f0f2db1b666d64b5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26 | If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$ | adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom | E | 10 |
7e95454dc1fc51d206821c757babb9c7 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_27 | A car travels $120$ miles from $A$ to $B$ at $30$ miles per hour but returns the same distance at $40$ miles per hour. The average speed for the round trip is closest to:
$\textbf{(A)}\ 33\text{ mph}\qquad\textbf{(B)}\ 34\text{ mph}\qquad\textbf{(C)}\ 35\text{ mph}\qquad\textbf{(D)}\ 36\text{ mph}\qquad\textbf{(E)}\ 37... | The car takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{30 \text{ miles }}=4 \text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \text{ miles }\cdot\dfrac{1 \text{ hr }}{40 \text{ miles }}=3 \text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\dfrac{240\text{ miles }}{7 \text{ hr}}=34\dfrac{2... | B | 34 |
6b8497181c28b13d625df7adeff4a048 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_28 | Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$ $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph}... | Let the speed of boy $A$ be $a$ , and the speed of boy $B$ be $b$ . Notice that $A$ travels $4$ miles per hour slower than boy $B$ , so we can replace $b$ with $a+4$
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to ... | B | 8 |
e33582148f9921d5e7bce3838ab5ab9e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_30 | From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:
$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{... | Let us represent the number of boys $b$ , and the number of girls $g$
From the first sentence, we get that $2(g-15)=b$
From the second sentence, we get $5(b-45)=g-15$
Expanding both equations and simplifying, we get $2g-30 = b$ and $5b = g+210$
Substituting $b$ for $2g-30$ , we get $5(2g-30)=g+210$ . Solving for $g$ , ... | A | 40 |
bcd47560365e468133a540eb0e79784b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textb... | By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$ | D | 8 |
bcd47560365e468133a540eb0e79784b | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_32 | $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textb... | We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has mov... | D | 8 |
2403f18fe6adb2ad7dd64fcbe0f7c1d6 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_33 | The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:
$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$ | It must be assumed that the pipes have an equal height.
We can represent the amount of water carried per unit time by cross sectional area.
Cross sectional of Pipe with diameter $6 in$ \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]
Cross sectional area of pipe with diameter $1 in$
\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4... | D | 36 |
4eafea8883767b672c63c70b8a6fd8b0 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$ . Therefore the inradius is $\boxed{4}$ | B | 4 |
4eafea8883767b672c63c70b8a6fd8b0 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_35 | In triangle $ABC$ $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\] | null | 4 |
72bae22d13b2c82284fa2e47ce9e654e | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_36 | A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)... | Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.
We then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\boxed{125}$ | null | 125 |
6dbba58c38cdfe2363eea836904d69a1 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_38 | If the expression $\begin{pmatrix}a & c\\ d & b\end{pmatrix}$ has the value $ab-cd$ for all values of $a, b, c$ and $d$ , then the equation $\begin{pmatrix}2x & 1\\ x & x\end{pmatrix}= 3$
$\textbf{(A)}\ \text{Is satisfied for only 1 value of }x\qquad\\ \textbf{(B)}\ \text{Is satisified for only 2 values of }x\qquad\\ \... | By $\begin{pmatrix}a & c\\ d & b\end{pmatrix}=ab-cd$ , we have $2x^2-x=3$ . Subtracting $3$ from both sides, giving $2x^2-x-3=0$ . This factors to $(2x-3)(x+1)=0$ . Thus, $x=\dfrac{3}{2},-1$ , so the equation is $\boxed{2}$ | B | 2 |
baa96ad69c523059b0a7eb9e75654ed2 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$ , using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$ .
Thus, the answer is $\boxed{2}$ | D | 2 |
baa96ad69c523059b0a7eb9e75654ed2 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_40 | The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | The numerator of $\frac {x^2-1}{x-1}$ can be factored as $(x+1)(x-1)$ . The $x-1$ terms in the numerator and denominator cancel, so the expression is equal to $x+1$ so long as $x$ does not equal $1$ . Looking at the function's behavior near 1, we see that as $x$ approaches one, the expression approaches $\boxed{2}$ | D | 2 |
a3ca15d83e3806c6c621114b3714a4e5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | Each diagonal has its two endpoints as vertices of the 100-gon. Each pair of vertices determines exactly one diagonal. Therefore the answer should be $\binom{100}{2}=4950$ . However this also counts the 100 sides of the polygon, so the actual answer is $4950-100=\boxed{4850}$ | A | 4850 |
a3ca15d83e3806c6c621114b3714a4e5 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_45 | The number of diagonals that can be drawn in a polygon of 100 sides is:
$\textbf{(A)}\ 4850 \qquad \textbf{(B)}\ 4950\qquad \textbf{(C)}\ 9900 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 8800$ | The formula for the number of diagonals of a polygon with $n$ sides is $n(n-3)/2$ . Taking $n=100$ , we see that the number of diagonals that may be drawn in this polygon is $100(97)/2$ or $\boxed{4850}$ | A | 4850 |
a2d7d27379eee5911205046c472a46e9 | https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_46 | In triangle $ABC$ $AB=12$ $AC=7$ , and $BC=10$ . If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:
$\textbf{(A)}\ \text{The area is doubled} \qquad\\ \textbf{(B)}\ \text{The altitude is doubled} \qquad\\ \textbf{(C)}\ \text{The area is four times the original area} \qquad\\ \textbf{(D)}\ \text{The m... | If you double sides $AB$ and $AC$ , they become $24$ and $14$ respectively. If $BC$ remains $10$ , then this triangle has area $0$ because ${14} + {10} = {24}$ , so two sides overlap the third side. Therefore the answer is $\boxed{0}$ | E | 0 |
90f8e608d5698ebf11846079d189ecfe | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ... | Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$
We know this must be true: \[|a_1b... | null | 197 |
90f8e608d5698ebf11846079d189ecfe | https://artofproblemsolving.com/wiki/index.php/2020_USOMO_Problems/Problem_4 | Suppose that $(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i, j)$ satisfying $1 \le i < j \le 100$ and $|a_ib_j - a_j b_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ ... | We claim the answer is $197$
Study the points $(0, 0), (a_i, b_i), (a_j, b_j)$ . If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\frac{1}{2}|0\times{b_i}+{a_i}\times{b_j}+{b_i}\times{0}-0\times{a_i}-{b_i}\times{a_j}-{b_j}\times{0} = \frac{1}{2}|a_ib_j - a_j b_i| = \frac{1}{... | null | 197 |
4eddc86a2b76bcb0824003bb66e85760 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_1 | Let $a,b,c,d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take. | Using the hint we turn the equation into $\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies (b-d-1)^2 + (a-c)^2 \implies \boxed{16}$ . This minimum is achieved when all the $x_i$ are equal to $1$ | null | 16 |
f25c446de036ffaadaf5676a0b0756b1 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | null | 6 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to s... | null | 13 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the larges... | null | 13 |
d89b02c176f387e6b9c6a679272db18b | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_4 | Find the smallest positive integer $n$ such that if $n$ squares of a $1000 \times 1000$ chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | We claim that $n = 1999$ is the smallest such number. For $n \le 1998$ , we can simply color any of the $1998$ squares forming the top row and the left column, but excluding the top left corner square.
We now show that no configuration with no colored right triangles exists for $n = 1999$ . We call a row or column fill... | null | 1999 |
2dec4c965bc88a3697fd054e8ba36687 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | [asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label("$\mathsf{A}$", A, W); label("$\mathsf{B}$", B, E);... | null | 77 |
2dec4c965bc88a3697fd054e8ba36687 | https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1 | In triangle $ABC$ , angle $A$ is twice angle $B$ , angle $C$ is obtuse , and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter | In $\triangle ABC$ let $\angle B = \beta, \angle A = 2\beta, \angle C = 180^{\circ} - 3\beta$ . From the law of sines, we have \[\frac{a}{\sin 2\beta} = \frac{b}{\sin \beta} = \frac{c} {\sin (180^{\circ} - 3\beta)} = \frac{c}{\sin 3\beta}\] Thus the ratio \[b : a : c = \sin\beta : \sin 2\beta : \sin 3\beta\] We can sim... | null | 77 |
bc9fb059b7f4baf332344340e897b028 | https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_3 | What is the smallest integer $n$ , greater than one, for which the root-mean-square of the first $n$ positive integers is an integer?
$\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[\left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2}\] | Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \[1^3 + 3\cdot 1^2 + 3 \cdot 1 + 1 = 2^3,\] \[1^3 + 3 \cdot(1^2 + 2^2) + 3 \cdot (1 + 2) + 1 + 1 = 3^3,\] and \[1^3 + 3\... | null | 337 |
982766bb6c57b0aec7d8441644d076c0 | https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5 | Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$ | We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$ , we... | null | 1700 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Using Vieta's formulas, we have:
\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end{align*}
From the last of these equations, we see that $cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62$ . Thus, the second equation becomes $-32+ac+ad+bc+bd+62=k$ , and so $ac+ad+bc+bd=... | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | We start as before: $ab=-32$ and $cd=62$ . We now observe that a and b must be the roots of a quadratic, $x^2+rx-32$ , where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic $x^2+sx+62$
Now
\begin{align*}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+... | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Let the roots of the equation be $a,b,c,$ and $d$ . By Vieta's, \begin{align*} a+b+c+d &= 18\\ ab+ac+ad+bc+bd+cd &= k\\ abc+abd+acd+bcd &=-200\\ abcd &=-1984.\\ \end{align*} Since $abcd=-1984$ and $ab=-32$ , then, $cd=62$ . Notice that \[abc + abd + acd + bcd = -200\] can be factored into \[ab(c+d)+cd(a+b)=-32(c+d)+... | null | 86 |
adbce7149b71691685e6ff8865546ef9 | https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1 | In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$ , the product of $2$ of its roots is $- 32$ . Find $k$ | Since two of the roots have product $-32,$ the equation can be factored in the form \[x^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\] Expanding, we get \[x^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\] Matching coefficients, we get
\begin{align*}
... | null | 86 |
ee0e8c72a8aeb1e1cbd508babe7d886e | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1 | In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? | We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $... | null | 1979 |
c4b2a2c596bc5c3b88aec10541b75586 | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2 | Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$
$(*)$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$
for $(m,n)=(2,3),(3,2),(2,5)$ , or $(5,2)$ . Determine all other pairs of integers $(m,n)$ if any, so that $(*)$ holds for all real numbers $x,y,z$ such that $x+y+z=0$ | Claim Both $m,n$ can not be even.
Proof $x+y+z=0$ $\implies x=-(y+z)$
Since $\frac{S_{m+n}}{m+n} = \frac{S_m S_n}{mn}$
by equating cofficient of $y^{m+n}$ on LHS and RHS ,get
$\frac{2}{m+n}=\frac{4}{mn}$
$\implies \frac{m}{2} + \frac {n}{2} = \frac{m\cdot n}{2\cdot2}$
So we have, $\frac{m}{2} \biggm{|} \frac{n}{2}$ an... | null | 2 |
12319197a4824f2c1f825b93a283e3bb | https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4 | Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$ | Indeed, $\boxed{2935363331541925531}$ has the requisite property. | null | 2935363331541925531 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then \[0=P(x)+P(y)+P(z)=3-3a+3b-3c\] using our system of equations, so $P(1)=0$ . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$ . Then we can use the system of equations to find that $y=z=1$ as well, and so $\boxed{1,1,1}$ is the ... | null | 1,1,1 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \[a+b+c=0,\] \[a^2+b^2+c^2=0,\] \[a^3+b^3+c^3=0.\] We have \begin{align*} 0&=(a+b+c)^3\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\ &=0-3a^3-3b^3-3c^3+6abc\\ &=6abc. \end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\box... | null | 1,1,1 |
02c5b310832b5c75b71a805a7f2ad412 | https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4 | Determine all the roots real or complex , of the system of simultaneous equations | We are going to use Intermediate Algebra Techniques to solve this equation.
Let's start with the first one: $x+y+z=3$ . This will be referred as the FIRST equation.
We are going to use the first equation to relate to the SECOND one ( $x^2+y^2+z^2=3$ ) and the THIRD one ( $x^3+y^3+z^3=3)$
Squaring this equation: $x^2+y^... | null | 1,1,1 |
91064cf7717d0f5be97285821d1c166e | https://artofproblemsolving.com/wiki/index.php/2021_USAJMO_Problems/Problem_1 | Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\] | The answer is $\boxed{1}$ , which works.
To show it is necessary, we first get $f(1)=f(1)^2$ , so $f(1)=1$ .
Then, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$ | null | 1 |
5878a9d27bdabf85aea862589abc464f | https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_3 | An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
What is the smallest positive number of beam... | Place the cube in the xyz-coordinate, with the positive x-axis pointing forward, the positive y-axis pointing right, and the positive z-axis pointing up. Let the position of a unit cube be $(x, y, z)$ if it is centered at $(x, y, z)$ . Place the $2020 \times 2020 \times 2020$ cube so that the edges are parallel to the ... | null | 3030 |
da6969b6e725522504576dde9744f9cc | https://artofproblemsolving.com/wiki/index.php/2020_USOJMO_Problems/Problem_5 | Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\leq i<j\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Determine the largest possible value of $N$ over all possible choices of the $100$ order... | Call the pair $(i, j)$ good if $1\leq i < j \leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\leq a_2\leq\ldots\leq a_{100}$ . Furthermore, reorder them so that if ... | null | 197 |
8f163c68b8bf11a7dbb1d1ef054ef7cc | https://artofproblemsolving.com/wiki/index.php/2016_USAJMO_Problems/Problem_4 | Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ | Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392... | null | 6097392 |
f25c446de036ffaadaf5676a0b0756b1 | https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_4 | problem_id
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1 Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object | We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$ | null | 6 |
b8db09096d7162ee8404b842152eba4b | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | First of all, note that $f(n)$ $\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \le n$ . We let $f(0) = 1$ for convenience.
From here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \in \mathbb{Z}$
We in... | null | 2047 |
b8db09096d7162ee8404b842152eba4b | https://artofproblemsolving.com/wiki/index.php/2013_USAJMO_Problems/Problem_4 | Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$ , where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$ $2+2$ $2+1+1$ $1+2+1$ $1+1+2$ , and $1+1+1+1$ . Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd. | Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\sum_{\textrm{power}=0}^{\textrm{pow}_{\textrm{larg}}} f(n-2^{\textrm{power}})$ , where $\textrm{pow}_{\textrm{larg}}$ represents the larg... | null | 2047 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to s... | null | 13 |
a45915bc778c3b967b94f7cee4faa46d | https://artofproblemsolving.com/wiki/index.php/2012_USAMO_Problems/Problem_1 | problem_id
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object | Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the larges... | null | 13 |
981346a8c4f46a2e3ea8c0d9d00c7ec5 | https://artofproblemsolving.com/wiki/index.php/2012_USAJMO_Problems/Problem_5 | For distinct positive integers $a$ $b < 2012$ , define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$ , where $a$ and $b$ range over all pairs of distinct positive i... | Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$ . Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$ . Thus, for every value of $k$ where $r_{a} > r_{b}$ , there is a value of $k$ where $r_{b} > r_{a}$ . Therefore, we merely have to... | null | 502 |
f628ab1e2295d16ab604e15d621af50b | https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_1 | Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square. | We will first take the expression modulo $3$ . We get $2^n+12^n+2011^n \equiv -1^n+1^n \pmod 3$
Lemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$ .
We can prove this by testing the residues modulo $3$ . We have $0^2 \equiv 0 \pmod 3$ $1^2 \equiv 1 \pmod 3$ , and $2^2 \equiv 1 \pmod 3$ , so the lemma is tr... | null | 1 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | We claim that the smallest $n$ is $67^2 = \boxed{4489}$ | null | 4489 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers $n$ in the form $p\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Ob... | null | 4489 |
f9c3533133e74a99960a349b5036838f | https://artofproblemsolving.com/wiki/index.php/2010_USAJMO_Problems/Problem_1 | A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $... | It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \cdot g(n)$ , no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.
We prove first: If $f(k) = f(a_k)$ $k \cdot a_k$ is a perfect square.
$k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)$ , ... | null | 4489 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\] | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbar... | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$ | E | 27 |
ae97e4a746f937b4142fe0a22411657d | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_1 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\q... | Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total d... | E | 27 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{8}$ | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{8}$ perfect squares less than 2023. | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{8}$ positive perfect squares no greater than $80$ | A | 8 |
a2658f3c6493f664125b0c5940db82a4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_3 | How many positive perfect squares less than $2023$ are divisible by $5$
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$ | We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . Th... | A | 8 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the sh... | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Say the chosen side is $a$ and the other sides are $b,c,d$
By the Generalised Polygon Inequality, $a<b+c+d$ . We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$
Combining these two, we get $a<26-a\Rightarrow a<13$
The largest length that satisfies this is $a=\boxed{12}$ | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing... | D | 12 |
613300bbad713a1527d2802dbf19ad74 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_4 | A quadrilateral has all integer sides lengths, a perimeter of $26$ , and one side of length $4$ . What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$ . Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$ . So ... | D | 12 |
a0794e532ec989d50607630576c1ba1c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$
$10^{15}$ gives us $15$ digits and $243$ gives us $3$ digits. $15+3=\text{\boxed{18}$ | E | 18 |
a0794e532ec989d50607630576c1ba1c | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_5 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$
$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$ | Multiplying it out, we get that $8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000$ . Counting, we have the answer is $\text{\boxed{18}$ ~andliu766 | E | 18 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Each of the vertices is counted $3$ times because each vertex is shared by three different edges.
Each of the edges is counted $2$ times because each edge is shared by two different faces.
Since the sum of the integers assigned to all vertices is $21$ , the final answer is $21\times3\times2=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Note that each vertex is counted $2\times 3=6$ times. Thus, the answer is $21\times6=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \cdot 3=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Since there are 8 vertices in a cube, there are $\dfrac{21}4$ vertices for two edges. There are $4$ edges per face, and $6$ faces in a cube, so the value of the cube is $\dfrac{21}4 \cdot 24 = \boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{126}$ | D | 126 |
b0a92d1ae5d9dd3d4d763c0be62c5d76 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_6 | An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppos... | The wording of the problem implies that the answer should hold for any valid combination of integers. Thus, we choose the numbers $21, 0, 0, 0, 0, 0, 0, 0$ , which are indeed $8$ integers that add to $21$ . Doing this, we find three edges that have a value of $21$ , and from there, we get three faces with a value of $4... | D | 126 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | To solve this question, you can use $y = mx + b$ where the $x$ is Fahrenheit and the $y$ is Breadus. We have $(110,0)$ and $(350,100)$ . We want to find the value of $y$ in $(200,y)$ that falls on this line. The slope for these two points is $\frac{5}{12}$ $y = \frac{5}{12}x + b$ . Solving for $b$ using $(110, 0)$ $\fr... | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | Let $^\circ B$ denote degrees Breadus. We notice that $200^\circ F$ is $90^\circ F$ degrees to $0^\circ B$ , and $150^\circ F$ to $100^\circ B$ . This ratio is $90:150=3:5$ ; therefore, $200^\circ F$ will be $\dfrac3{3+5}=\dfrac38$ of the way from $0$ to $100$ , which is $\boxed{37.5.}$ | D | 37.5. |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | From $110$ to $350$ degrees Fahrenheit, the Breadus scale goes from $1$ to $100$ $110$ to $350$ degrees is a span of $240$ , and we can use this to determine how many Fahrenheit each Breadus unit is worth. $240$ divided by $100$ is $2.4$ , so each Breadus unit is $2.4$ Fahrenheit, starting at $110$ Fahrenheit. For exam... | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | We note that the range of F temperatures that $0-100$ $\text{Br}^\circ$ represents is $350-110 = 240$ $\text{F}^\circ$ $200$ $\text{F}^\circ$ is $(200-110) = 90$ $\text{F}^\circ$ along the way to getting to $240$ $\text{F}^\circ$ , the end of this range, or $90/240 = 9/24 = 3/8 = 0.375$ of the way. Therefore if we swit... | D | 37.5 |
dcf55bfafb3d3e10c1e41b1657dc0ab0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_8 | Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at $110$ degrees Fahrenheit, which is $0$ degrees on the Breadus scale. Bread is baked at $350$ degrees Fahrenheit, which is $100$ degrees on the Breadus scale. B... | We have the points $(0, 110)$ and $(100, 350)$ . We want to find $(x, 200)$ . The equation of the line is $y=\frac{12}{5}x+110$ . We use this to find $x=\frac{75}{2}=37.5$ , or $\boxed{37.5}$ .
~MC413551 | D | 37.5 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the... | Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{9}$ .
For curious readers, the numbers (in chronological order) are: | E | 9 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the... | There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two... | E | 9 |
77fb83402afdf5d51537d2f43b0ce047 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_9 | A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the... | We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month ... | E | 9 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\] \[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying $(x+1)$ by $(a+1)$ and solving, we get: \[ax+11=ax+a+x+1\] \[11=a+x+1\] \[a+x=10\qquad (3)\]
M... | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Suppose Maureen took $n$ tests with an average of $m$
If she takes another test, her new average is $\frac{(nm+11)}{(n+1)}=m+1$
Cross-multiplying: $nm+11=nm+n+m+1$ , so $n+m=10$
If she takes $3$ more tests, her new average is $\frac{(nm+33)}{(n+3)}=m+2$
Cross-multiplying: $nm+33=nm+2n+3m+6$ , so $2n+3m=27$
But $2n+3m$ ... | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Let $s$ represent the sum of Maureen's test scores previously and $t$ be the number of scores taken previously.
So, $\frac{s+11}{t+1} = \frac{s}{t}+1$ and $\frac{s+33}{t+3} = \frac{s}{t}+2$
We can use the first equation to write $s$ in terms of $t$
We then substitute this into the second equation: $\frac{-t^2+10t+33}{t... | D | 7 |
95e0b26ee26a8b0a1e053bf6decdc7b0 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10A_Problems/Problem_10 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$ . If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$ . What is the mean of her quiz scores currently? $\textbf{(A) }4\qquad\textbf{(B) }5\q... | Let's consider all the answer choices. If the average is $8$ , then, we can assume that all her test choices were $8$ . We can see that she must have gotten $8$ twice, in order for another score of $11$ to bring her average up by one. However, adding three $11$ 's will not bring her score up to 10. Continuing this proc... | D | 7 |
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