Answer: 840

Small Hint:

Write the digits as $a,$ $ar,$ $ar^2;$ the ratio $r$ may be a fraction as long as all three digits are distinct integers

Big Hint:

For the largest, start with hundreds digit $9$ and try ratios with denominator $3;$ for the smallest, start with $1$ and an integer ratio

Solution:

Write the digits as $a,$ $ar,$ $ar^2.$ For the largest geometric number, take $a = 9.$ An integer ratio at least $2$ would push the next digit past $9,$ and $r = 1$ repeats digits, so $r$ is a fraction whose denominator squares into $9:$ the choices $r = \frac{2}{3}$ and $r = \frac{1}{3}$ give $964$ and $931.$ The largest is $964.$

For the smallest, take hundreds digit $1.$ Then the tens digit $r$ must be an integer at least $2$ (the digits are distinct), and $r = 2$ gives $124,$ which beats $r = 3$’s $139.$

The difference is $964 - 124 = 840.$

Answer: 697

Small Hint:

Write $z = a + 164i$ and clear the denominator: $z = 4i(z + n)$

Big Hint:

Expand $4i(z + n)$ and match real parts, then imaginary parts, to find $a$ and then $n$

Solution:

Write $z = a + 164i.$ Clearing the denominator gives $z = 4i(z + n),$ that is, $$\begin{aligned} a + 164i &= 4i\,(a + n + 164i) \\ &= -656 + 4(a + n)i. \end{aligned}$$

Real parts give $a = -656,$ and imaginary parts give $164 = 4(a + n),$ so $a + n = 41$ and $n = 41 + 656 = 697.$

Answer: 11

Small Hint:

Set up $\binom{8}{3}p^3(1-p)^5 = \frac{1}{25}\binom{8}{5}p^5(1-p)^3$

Big Hint:

The binomial coefficients cancel, leaving $(1-p)^2 = \frac{p^2}{25};$ both sides are positive, so take square roots

Solution:

The condition says $$\begin{aligned} &\binom{8}{3} p^3 (1-p)^5 \\ &= \frac{1}{25} \binom{8}{5} p^5 (1-p)^3. \end{aligned}$$ Since $\binom{8}{3} = \binom{8}{5}$ and both $p$ and $1 - p$ are positive, dividing by $p^3(1-p)^3$ leaves $(1-p)^2 = \frac{p^2}{25},$ so $1 - p = \frac{p}{5}.$

Hence $p = \frac{5}{6},$ and $m + n = 5 + 6 = 11.$

Answer: 177

Small Hint:

Set $A$ as the origin and express $M,$ $N,$ and $C$ in terms of the vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$

Big Hint:

Points of line $MN$ have the form $tM + (1-t)N;$ also write $P = s\,(\overrightarrow{AB} + \overrightarrow{AD}),$ match coefficients, and add the two resulting equations

Solution:

Place $A$ at the origin and let $\mathbf{b} = \overrightarrow{AB}$ and $\mathbf{d} = \overrightarrow{AD},$ so that $C = \mathbf{b} + \mathbf{d},$ $M = \frac{17}{1000}\mathbf{b},$ and $N = \frac{17}{2009}\mathbf{d}.$ Since $P$ lies on $\overline{AC},$ write $P = s\,(\mathbf{b} + \mathbf{d})$ where $s = \frac{AP}{AC};$ since $P$ also lies on line $MN,$ write $P = tM + (1 - t)N$ for some $t.$

Because $\mathbf{b}$ and $\mathbf{d}$ are independent, the coefficients must agree: $$s = \frac{17t}{1000}$$ and $$s = \frac{17(1-t)}{2009}.$$ Thus $t = \frac{1000s}{17}$ and $1 - t = \frac{2009s}{17};$ adding gives $1 = \frac{3009s}{17}.$

Therefore $\frac{AC}{AP} = \frac{1}{s} = \frac{3009}{17} = 177.$

Answer: 72

Small Hint:

$AK = CK$ and $K$ the midpoint of $\overline{PM}$ mean the diagonals of quadrilateral $APCM$ bisect each other, so $APCM$ is a parallelogram

Big Hint:

$AM \parallel CP$ makes triangles $BLP$ and $BAM$ similar; get $\frac{BL}{BA}$ from the angle bisector theorem

Solution:

Because $AK = CK$ and $K$ is the midpoint of $\overline{PM},$ the diagonals of quadrilateral $APCM$ bisect each other, so $APCM$ is a parallelogram and $AM \parallel CP.$ Since $P$ lies on line $CL$ and $B,$ $P,$ $M$ all lie on line $BK,$ triangles $BLP$ and $BAM$ are similar.

Thus $\frac{LP}{AM} = \frac{BL}{BA}.$ The angle bisector theorem gives $\frac{AL}{LB} = \frac{AC}{BC} = \frac{450}{300} = \frac{3}{2},$ so $\frac{BL}{BA} = \frac{2}{2 + 3} = \frac{2}{5}.$

Therefore $LP = \frac{2}{5} \cdot AM = \frac{2}{5} \cdot 180 = 72.$

Answer: 412

Small Hint:

If $\lfloor x \rfloor = k,$ then as $x$ runs over $[k, k+1)$ the value $x^k$ covers exactly the integers from $k^k$ to $(k+1)^k - 1$

Big Hint:

Sum the counts $(k+1)^k - k^k$ for $k = 1,$ $2,$ $3,$ and $4;$ for $k = 5$ the values already exceed $1000$

Solution:

Suppose $\lfloor x \rfloor = k$ for a positive integer $k.$ As $x$ runs over $[k, k+1),$ the value $x^k$ increases continuously from $k^k$ toward $(k+1)^k,$ so the attainable integers $N$ are exactly those with $k^k \le N \le (k+1)^k - 1:$ there are $(k+1)^k - k^k$ of them, and these ranges are disjoint for different $k.$ If $0 \lt x \lt 1,$ then $x^{\lfloor x \rfloor} = x^0 = 1,$ which is already attained. If $x \lt 0,$ then $\lfloor x \rfloor$ is negative; an odd exponent gives a negative value, while an even exponent gives a value strictly between $0$ and $1.$ Thus no $x \lt 1$ produces a new positive integer.

For $k = 1,$ $2,$ $3,$ and $4,$ the counts are $2 - 1 = 1,$ $9 - 4 = 5,$ $64 - 27 = 37,$ and $625 - 256 = 369,$ and every such $N$ is at most $624 \lt 1000.$ For $k = 5$ the smallest value is $5^5 = 3125 \gt 1000.$

The total is $1 + 5 + 37 + 369 = 412.$

Answer: 41

Small Hint:

Rewrite the relation as $5^{a_{n+1} - a_n} = \frac{3n+5}{3n+2}$ and multiply successive instances to telescope

Big Hint:

The closed form is $a_k = \log_5(3k+2),$ so you need $3k + 2$ to be a power of $5$

Solution:

The relation says $5^{a_{n+1} - a_n} = 1 + \frac{3}{3n + 2} = \frac{3n+5}{3n+2}.$ Multiplying these equations for $n = 1,$ $2,$ $\ldots,$ $k - 1$ telescopes: $$5^{a_k - a_1} = \frac{3k + 2}{5},$$ so $$\begin{aligned} a_k &= 1 + \log_5 \frac{3k+2}{5} \\ &= \log_5 (3k + 2). \end{aligned}$$

Thus $a_k$ is an integer exactly when $3k + 2$ is a power of $5.$ Since $5^j \equiv 2^j \pmod 3,$ only odd exponents $j$ give numbers of the form $3k + 2.$ The power $5^1 = 5$ gives $k = 1,$ which is excluded, and the next, $5^3 = 125 = 3 \cdot 41 + 2,$ gives $k = 41.$

Answer: 398

Small Hint:

In the sum of all positive differences, count how many times each $2^k$ is added and how many times it is subtracted

Big Hint:

$N = \sum_{k=0}^{10}(2k - 10)\,2^k;$ evaluate with $\sum k\,2^k$ and a geometric series, then reduce mod $1000$

Solution:

In the sum of all positive differences, the element $2^k$ is added once for each smaller element ($k$ times) and subtracted once for each larger element ($10 - k$ times). Hence $$\begin{aligned} N &= \sum_{k=0}^{10} (2k - 10)\,2^k \\ &= 2\sum_{k=0}^{10} k\,2^k - 10\sum_{k=0}^{10} 2^k. \end{aligned}$$

The standard sums are $\sum_{k=0}^{10} k\,2^k = 9 \cdot 2^{11} + 2 = 18434$ and $\sum_{k=0}^{10} 2^k = 2^{11} - 1 = 2047,$ so $$\begin{aligned} N &= 2 \cdot 18434 - 10 \cdot 2047 \\ &= 36868 - 20470 = 16398. \end{aligned}$$

The remainder upon division by $1000$ is $398.$

Answer: 420

Small Hint:

Concatenating the three prices in order gives an arrangement of the seven digits; count arrangements, then the ways to cut one into three prices

Big Hint:

Each arrangement is cut into three nonempty blocks of at most $4$ digits: the block lengths are permutations of $1,2,4;$ $2,2,3;$ or $1,3,3$

Solution:

Concatenating the three guessed prices in order produces an arrangement of the seven given digits, and each guess is recovered uniquely from an arrangement together with a way to cut it into three consecutive nonempty blocks of at most four digits each (prices run from $\$1$ to $\$9999,$ and no price can start with $0$ here since every digit is $1$ or $3$). There are $\frac{7!}{4!\,3!} = 35$ arrangements of four $1$s and three $3$s.

The ordered block lengths are the ways to write $7$ as an ordered sum of three parts between $1$ and $4:$ the permutations of $(1, 2, 4),$ $(2, 2, 3),$ and $(1, 3, 3),$ giving $6 + 3 + 3 = 12$ cuts for each arrangement.

The total is $35 \cdot 12 = 420.$

Answer: 346

Small Hint:

Count planet-to-chair patterns first; the rules force blocks of Martians, Venusians, and Earthlings to repeat in that clockwise order

Big Hint:

If the cycle $MVE$ repeats $k$ times, each planet’s $5$ members split into $k$ ordered nonempty groups in $\binom{4}{k-1}$ ways; cube and sum over $k$

Solution:

First choose which planet sits in each chair; the individuals from each planet can then be assigned to their chairs in $5!$ ways apiece, so $N$ counts the planet patterns. The adjacency rules say exactly that, reading clockwise, each maximal block of Martians must be followed by a block of Venusians and then a block of Earthlings before Martians can appear again. Since chair $1$ holds a Martian and chair $15$ holds an Earthling, the chairs from $1$ to $15$ consist of the pattern (Martian block, Venusian block, Earthling block) repeated $k$ times, for some $1 \le k \le 5.$

For a given $k,$ each planet’s five members are distributed into $k$ nonempty blocks in order, and the number of ways to write $5$ as an ordered sum of $k$ positive integers is $\binom{4}{k-1}.$ The three planets’ block sizes are independent, so $$\begin{aligned} N &= \sum_{k=1}^{5} \binom{4}{k-1}^3 \\ &= 1^3 + 4^3 + 6^3 + 4^3 + 1^3 \\ &= 346. \end{aligned}$$

Answer: 600

Small Hint:

The lattice points on the line are $(i,\, 2009 - 41i)$ for $i = 0,$ $1,$ $\ldots,$ $49$

Big Hint:

The triangle on points $i$ and $j$ has area $\frac{2009}{2}\,|i - j|,$ an integer exactly when $i$ and $j$ have the same parity

Solution:

The points on the line with nonnegative integer coordinates are $P_i = (i,\, 2009 - 41i)$ for $i = 0,$ $1,$ $\ldots,$ $49$ — fifty points in all. For $P = P_i$ and $Q = P_j,$ the shoelace formula gives $$\begin{aligned} [OPQ] &= \small \frac{1}{2}\left|\,i(2009 - 41j) - j(2009 - 41i)\,\right| \\ &= \frac{2009}{2}\,|i - j|. \end{aligned}$$

This is automatically positive for distinct points, and since $2009$ is odd, it is an integer exactly when $i - j$ is even, that is, when $i$ and $j$ have the same parity. There are $25$ even and $25$ odd indices, so the number of triangles is $\binom{25}{2} + \binom{25}{2} = 300 + 300 = 600.$

Answer: 11

Small Hint:

$AB$ is tangent to $\omega$ at $D,$ so $\omega$ is the inscribed circle of triangle $ABI$

Big Hint:

Let $x$ be the tangent length from $I$ and compute the area of $ABI$ two ways: $rs$ with $r = \frac{1}{2}\sqrt{AD \cdot BD},$ and Heron’s formula

Solution:

Because $\overline{CD} \perp \overline{AB}$ and $D$ is an endpoint of the diameter, $AB$ is tangent to $\omega$ at $D.$ Together with the tangent lines $AI$ and $BI,$ this makes $\omega$ the inscribed circle of triangle $ABI.$ Write $AD = y,$ $BD = z,$ and let $x$ be the tangent length from $I.$ The right-triangle altitude satisfies $CD^2 = AD \cdot BD,$ so the inradius of $ABI$ is $r = \frac{1}{2}\sqrt{yz}.$

With semiperimeter $s = x + y + z,$ the tangent lengths are exactly $s - AB = x,$ $s - BI = y,$ and $s - AI = z,$ so the area of $ABI$ equals both $rs$ and, by Heron’s formula, $\sqrt{s \cdot xyz}.$ Equating and squaring, $$\frac{s^2\,yz}{4} = s\,xyz,$$ so $$s = 4x,$$ which gives $AB = y + z = s - x = 3x.$

The perimeter is $2s = 8x,$ so its ratio to $AB$ is $\frac{8x}{3x} = \frac{8}{3}$ (independent of the given legs), and $m + n = 8 + 3 = 11.$

Answer: 90

Small Hint:

Clear denominators: $a_{n+2}(1 + a_{n+1}) = a_n + 2009,$ then subtract consecutive instances

Big Hint:

Unless $a_{n+2} = a_n$ always, the differences $|a_{n+2} - a_n|$ strictly decrease forever, which is impossible; the resulting two-periodic case forces $a_1 a_2 = 2009$

Solution:

Clearing denominators, $a_{n+2}(1 + a_{n+1}) = a_n + 2009$ for all $n \ge 1.$ Subtracting each instance from the next gives $$\begin{aligned} &a_{n+2} - a_n \\ &= (a_{n+2} + 1)(a_{n+3} - a_{n+1}). \end{aligned}$$

If some difference $a_{n+2} - a_n$ were nonzero, then every later difference would be nonzero as well, and since each $a_{n+2} + 1 \ge 2,$ the identity would force $|a_3 - a_1|$ $\gt |a_4 - a_2|$ $\gt |a_5 - a_3|$ $\gt \cdots,$ an infinite strictly decreasing sequence of positive integers — impossible. Hence $a_{n+2} = a_n$ for all $n:$ the odd-indexed terms are all equal and the even-indexed terms are all equal.

The recursion then reads $a_1(1 + a_2) = a_1 + 2009,$ so $a_1 a_2 = 2009 = 7^2 \cdot 41.$ Among the factor pairs of $2009,$ the sum is smallest for $41 \cdot 49,$ giving $41 + 49 = 90.$

Answer: 905

Small Hint:

Let $m_j$ count how many $a_i$ equal $j;$ the given data are linear equations in $m_1,$ $m_2,$ $m_3,$ $m_4$

Big Hint:

Eliminating $m_1$ and then $m_2$ leaves $5m_3 + 21m_4 = 195,$ so $m_4$ is $0$ or $5;$ compare the two resulting values of $S_2$

Solution:

For $j = 1, 2, 3, 4,$ let $m_j$ be the number of $a_i$ equal to $j.$ Then $$\begin{aligned} &m_1 + m_2 + m_3 + m_4 = 350, \\ &m_1 + 2m_2 + 3m_3 + 4m_4 = 513, \\ &m_1 + 16m_2 + 81m_3 \\ &\quad {}+ 256m_4 = 4745. \end{aligned}$$