Answer: 372

Small Hint:

Split into cases by how many $0$’s the sequence uses; every other character can appear at most once

Big Hint:

With two $0$’s, place them in $\binom{5}{2}$ ways and fill three spots from the six other characters

Solution:

The available characters are the seven distinct symbols A, I, M, E, $2,$ $0,$ $7,$ where $0$ may be used up to twice and every other character at most once. Sequences using at most one $0$ consist of five distinct characters chosen from the seven, in order: $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = 2520.$

Sequences with two $0$’s: choose the two positions for the $0$’s in $\binom{5}{2} = 10$ ways, then fill the remaining three positions with distinct characters from the other six in $6 \cdot 5 \cdot 4 = 120$ ways, for $1200$ sequences.

Thus $N = 2520 + 1200 = 3720,$ and $\frac{N}{10} = 372.$

Answer: 200

Small Hint:

Since $a$ divides both $b$ and $c,$ it must divide $a + b + c = 100$

Big Hint:

Write $b = as$ and $c = at:$ then $s + t = \frac{100}{a} - 1,$ which has $\frac{100}{a} - 2$ positive solutions

Solution:

Since $a$ divides $b$ and $c,$ it divides $a + b + c = 100.$ Write $b = as$ and $c = at$ with $s, t \ge 1;$ then $a(1 + s + t) = 100,$ so $$s + t = \frac{100}{a} - 1.$$ For positive $s$ and $t$ we need $\frac{100}{a} \ge 3,$ so $a \in \{1, 2, 4, 5, 10, 20, 25\}.$

For each such $a,$ the equation $s + t = \frac{100}{a} - 1$ has $\frac{100}{a} - 2$ ordered positive solutions. Summing, $$\begin{aligned} &\small (100 + 50 + 25 + 20 + 10 + 5 + 4) \\ &\quad {}- 2 \cdot 7 \\ &= 214 - 14 = 200. \end{aligned}$$

Answer: 578

Small Hint:

Since $5^2 + 12^2 = 13^2,$ angles $AEB$ and $CFD$ are right angles

Big Hint:

Extend $EA$ beyond $A$ and $FD$ beyond $D$ to meet at $G,$ and look for right triangles congruent to $AEB$

Solution:

Since $5^2 + 12^2 = 13^2,$ triangles $AEB$ and $CFD$ are right-angled at $E$ and $F,$ and they are congruent (sides $5,$ $12,$ $13$). Extend $EA$ beyond $A$ and $FD$ beyond $D$ until the two lines meet at $G.$

Then $$\begin{aligned} \angle GAD &= 90^\circ - \angle EAB \\ &= \angle ABE, \\ \angle GDA &= 90^\circ - \angle FDC \\ &= \angle DCF = \angle BAE. \end{aligned}$$ These two angles sum to $90^\circ,$ so $\angle AGD = 90^\circ,$ and triangle $AGD$ is congruent to $BEA$ (equal angles and hypotenuse $AD = BA = 13$). Hence $GA = EB = 5$ and $GD = EA = 12.$

Therefore $GE = GA + AE = 5 + 12 = 17$ and $GF = GD + DF = 12 + 5 = 17,$ with a right angle between them at $G,$ so $EF^2 = 17^2 + 17^2 = 578.$

Answer: 450

Small Hint:

Let $a$ and $b$ be the worker-hours needed for one widget and one whoosit; each scenario gives one linear equation

Big Hint:

The first two scenarios give $3a + 2b = 1$ and $4a + 5b = 2;$ the third supplies $150$ worker-hours in total

Solution:

Let $a$ and $b$ be the worker-hours required to make one widget and one whoosit. The three scenarios supply $100,$ $120,$ and $150$ worker-hours, so $$\begin{aligned} 300a + 200b &= 100, \\ 240a + 300b &= 120, \\ 150a + mb &= 150. \end{aligned}$$

The first two equations simplify to $3a + 2b = 1$ and $4a + 5b = 2,$ giving $a = \frac{1}{7}$ and $b = \frac{2}{7}.$ Substituting into the third, $\frac{150}{7} + \frac{2m}{7} = 150,$ so $150 + 2m = 1050$ and $m = 450.$

Answer: 888

Small Hint:

The segment runs from $(0, 9)$ to $(223, 0);$ count how many unit squares it cuts through in the $223 \times 9$ rectangle of $2007$ squares

Big Hint:

Since $\gcd(9, 223) = 1$ the segment hits no interior lattice point, so it crosses $223 + 9 - 1$ squares; symmetry splits the untouched squares evenly

Solution:

The line meets the axes at $(223, 0)$ and $(0, 9),$ so all qualifying squares lie inside the $223 \times 9$ rectangle, which contains $223 \cdot 9 = 2007$ unit squares. Because $\gcd(9, 223) = 1,$ the segment passes through no interior lattice point; it crosses $222$ interior vertical lines and $8$ interior horizontal lines, entering a new square at each crossing, so it passes through the interiors of $223 + 9 - 1 = 231$ squares.

The other $2007 - 231 = 1776$ squares lie entirely above or entirely below the segment. The segment’s midpoint $\left(\frac{223}{2}, \frac{9}{2}\right)$ is the center of the rectangle, so rotating by $180^\circ$ about it swaps the two groups. Hence exactly half of them, $888,$ lie below the graph.

Answer: 640

Small Hint:

Build the number from right to left, counting how many digits may immediately precede a given digit

Big Hint:

For every digit $d,$ exactly $4$ digits can precede it: the odd digits below $d$ together with the even digits above $d.$ The digit $0$ never qualifies.

Solution:

A digit $a_i$ may immediately precede $a_{i+1} = d$ exactly when $a_i$ is odd and less than $d,$ or even and greater than $d.$ Checking each $d$ from $0$ to $9,$ this always allows exactly $4$ digits: for example, $d = 0$ allows $2,$ $4,$ $6,$ $8;$ $d = 4$ allows $1,$ $3,$ $6,$ $8;$ $d = 9$ allows $1,$ $3,$ $5,$ $7.$ (Raising $d$ by $1$ trades odd choices for even ones, keeping the total at $4.$) Note that $0$ is never an allowed predecessor, since $0$ is even but exceeds no digit.

So choose the last digit $a_4$ in $10$ ways, then each of $a_3,$ $a_2,$ $a_1$ in $4$ ways; the leading digit is automatically nonzero. The count is $4^3 \cdot 10 = 640.$

Answer: 553

Small Hint:

$\lfloor\sqrt[3]{n}\rfloor = k$ means $k^3 \le n \lt k^3 + 3k^2 + 3k + 1;$ count the multiples of $k$ in that range

Big Hint:

There are $3k + 4$ such multiples, so solve $3k + 4 = 70,$ then take the largest multiple

Solution:

The condition $\lfloor\sqrt[3]{n}\rfloor = k$ means $k^3 \le n \lt (k+1)^3$ $= k^3 + 3k^2 + 3k + 1.$ The multiples of $k$ in this range are $k \cdot k^2,$ $k(k^2 + 1),$ $\ldots,$ $k(k^2 + 3k + 3),$ so there are exactly $3k + 4$ of them.

Setting $3k + 4 = 70$ gives $k = 22.$ The maximum of $\frac{n_i}{k}$ is $k^2 + 3k + 3 = 484 + 66 + 3$ $= 553.$

Answer: 896

Small Hint:

With $h$ drawn lines of length $4$ and $v$ of length $5,$ we get $4h + 5v = 2007,$ and adjacent pairs form $(h-1)(v-1)$ basic rectangles

Big Hint:

Maximize $xy$ over $4x + 5y = 1998:$ take $x$ satisfying $x \equiv 2 \pmod 5$ and as close as possible to the vertex $\frac{999}{4}$

Solution:

Suppose $h$ of the drawn lines have length $4$ and $v$ have length $5,$ so $4h + 5v = 2007.$ A basic rectangle is bounded by two adjacent lines in each direction, so the lines determine $(h - 1)(v - 1)$ basic rectangles. Setting $x = h - 1$ and $y = v - 1,$ we must maximize $xy$ subject to $4x + 5y = 1998.$

As a function of $x,$ the product $xy = x \cdot \frac{1998 - 4x}{5}$ is a downward parabola with vertex at $x = \frac{999}{4} = 249.75.$ For $y$ to be an integer we need $4x \equiv 1998 \pmod 5,$ i.e. $x \equiv 2 \pmod 5.$ The nearest candidates are $x = 247$ (giving $y = 202$ and product $49894$) and $x = 252$ (giving $y = 198$ and product $49896$).

So $N = 49896,$ and the remainder upon division by $1000$ is $896.$

Answer: 259

Small Hint:

Compute $BE,$ $BF,$ and $EF$ with the Pythagorean theorem; triangles $BEF$ and $DFE$ turn out congruent

Big Hint:

The tangency point splits $EF$ by tangent lengths: $EP = s - BF$ and $FQ = s - DE,$ where $s$ is the common semiperimeter. Then $PQ = EF - EP - FQ.$

Solution:

Place $A = (0, 0),$ $B = (63, 0),$ $C = (63, 448),$ $D = (0, 448),$ so $E = (0, 84)$ and $F = (63, 364).$ Then $BE = DF = \sqrt{63^2 + 84^2}$ $= 21\sqrt{3^2 + 4^2} = 105,$ $BF = DE = 448 - 84 = 364,$ and $EF = \sqrt{63^2 + 280^2}$ $= 7\sqrt{9^2 + 40^2} = 287.$ In particular triangles $BEF$ and $DFE$ are congruent, with common semiperimeter $s = \frac{105 + 364 + 287}{2} = 378.$

In any triangle, the distance from a vertex to the incircle’s tangency points on its two sides is the semiperimeter minus the opposite side. In triangle $BEF,$ $EP = s - BF$ $= 378 - 364 = 14;$ in triangle $DEF,$ $FQ = s - DE$ $= 378 - 364 = 14.$

Therefore $PQ = EF - EP - FQ$ $= 287 - 14 - 14 = 259.$

Answer: 710

Small Hint:

Condition on $|A| = k$ and count the sets $B$ with $B \subseteq A$ or $B \subseteq S - A;$ only the empty set satisfies both

Big Hint:

Sum over $k$ the expression $\binom{6}{k}\left(2^k + 2^{6-k} - 1\right)$ using $\sum_k \binom{6}{k} 2^k = 3^6$

Solution:

Fix $A$ with $|A| = k.$ There are $2^k$ subsets $B \subseteq A$ and $2^{6-k}$ subsets $B \subseteq S - A,$ and only the empty set is counted twice, so $2^k + 2^{6-k} - 1$ choices of $B$ succeed. Since there are $\binom{6}{k}$ sets $A$ of size $k$ and $2^6$ choices for each of $A$ and $B,$ the probability is $$\begin{aligned} &\frac{1}{2^{12}}\sum_{k=0}^{6} \binom{6}{k}\left(2^k + 2^{6-k} - 1\right) \\ &= \frac{2 \cdot 3^6 - 2^6}{2^{12}}, \end{aligned}$$ using $\sum_k \binom{6}{k} 2^k = \sum_k \binom{6}{k} 2^{6-k}$ $= (1+2)^6 = 3^6.$

This simplifies to $\frac{3^6 - 2^5}{2^{11}} = \frac{697}{2^{11}}.$ Since $697 = 17 \cdot 41$ is odd, we take $m = 697,$ $n = 2,$ $r = 11,$ and $m + n + r = 710.$

Answer: 179

Small Hint:

When the tubes touch while both rest on the ground, the segment joining their centers has length $96$ and vertical component $48$

Big Hint:

Total turning is $360^\circ:$ while crossing the small tube, the big tube’s rotation equals the center’s $120^\circ$ sweep plus the angular contribution from the contact arc; the rest is flat rolling

Solution:

When the rolling tube touches both the ground and the small tube, the segment between centers has length $72 + 24 = 96$ and vertical component $72 - 24 = 48,$ so it makes a $30^\circ$ angle with the horizontal. As the big tube rolls over the small one, its center swings along an arc of radius $96$ about the small tube’s center, from $30^\circ$ above the horizontal on one side to $30^\circ$ on the other: a sweep of $120^\circ,$ advancing the center horizontally by $2 \cdot 96\cos 30^\circ = 96\sqrt{3}.$

During that sweep, the contact arc on the small tube is $120^\circ \cdot \frac{24}{72} = 40^\circ$ worth of the big tube’s circumference, and the sweep itself also rotates the big tube by $120^\circ,$ so crossing the small tube turns the big tube by $160^\circ$ in all. To complete exactly one revolution, the remaining $360^\circ - 160^\circ = 200^\circ$ of turning happens rolling on flat ground, where the center advances the rolled distance $\frac{200}{360} \cdot 2\pi \cdot 72 = 80\pi.$

Hence $x = 80\pi + 96\sqrt{3},$ and $a + b + c = 80 + 96 + 3 = 179.$

Answer: 91

Small Hint:

Write $x_n = 3^{a+bn}$ with integers $a$ and $b \ge 1;$ the log-sum condition gives $2a + 7b = 77$

Big Hint:

The sum $x_0 + \cdots + x_7$ lies strictly between $x_7$ and $3 x_7,$ so its base-$3$ log pins down $a + 7b$

Solution:

Every term is a power of $3$ and the ratio is a quotient of powers of $3,$ so $x_n = 3^{a + bn}$ for integers $a$ and $b,$ with $b \ge 1$ since the sequence increases. The first condition gives $$\sum_{n=0}^{7} (a + bn) = 8a + 28b = 308,$$ i.e. $$2a + 7b = 77.$$

For the second condition, $x_7 = 3^{a+7b}$ is the largest term, and $$\begin{aligned} x_7 &\lt \sum_{n=0}^{7} x_n \\ &\lt x_7\left(1 + \tfrac{1}{3} + \tfrac{1}{9} + \cdots\right) \\ &= \tfrac{3}{2}\,x_7 \lt 3x_7, \end{aligned}$$ so $\log_3\left(\sum x_n\right)$ lies strictly between $a + 7b$ and $a + 7b + 1.$ The given bounds then force $a + 7b = 56.$

Subtracting from $2a + 7b = 77$ yields $a = 21,$ then $b = 5.$ Therefore $\log_3(x_{14}) = a + 14b = 21 + 70$ $= 91.$

Answer: 640

Small Hint:

Each entry feeds upward like Pascal’s triangle: for bottom entries $x_0,$ $\ldots,$ $x_{10},$ the top square is $\sum_i \binom{10}{i} x_i$

Big Hint:

Modulo $3,$ except for $i = 0,$ $1,$ $9,$ $10,$ the coefficient $\binom{10}{i}$ vanishes, so only $x_0 + x_1 + x_9 + x_{10}$ matters

Solution:

Label the bottom-row entries $x_0,$ $x_1,$ $\ldots,$ $x_{10}.$ Since each square is the sum of the two below it, the contributions accumulate with Pascal’s-triangle weights: the top square equals $\sum_{i=0}^{10} \binom{10}{i} x_i.$

Modulo $3,$ direct checking (or Lucas’ theorem with $10 = 101_3$) shows that for $2 \le i \le 8,$ $\binom{10}{i} \equiv 0,$ while $\binom{10}{0} = \binom{10}{10} = 1$ and $\binom{10}{1} = \binom{10}{9} = 10 \equiv 1.$ So the top square is a multiple of $3$ exactly when $x_0 + x_1 + x_9 + x_{10}$ $\equiv 0 \pmod 3.$

Because the four entries are $0$ or $1,$ their sum is divisible by $3$ only when it is $0$ or $3:$ either all four are $0$ (one way) or exactly three are $1$ (four ways), for $5$ choices. The remaining seven entries $x_2,$ $\ldots,$ $x_8$ are free, so the count is $5 \cdot 2^7 = 640.$

Answer: 676