Answer: 839

Small Hint:

Compute the inner factorials: $3! = 6,$ so the expression is $\frac{720!}{6}$

Big Hint:

Write $720! = 720 \cdot 719!$ and note that $n = 720$ would make $k \cdot n!$ too large

Solution:

Since $3! = 6$ and $6! = 720,$ the expression is $$\begin{aligned} \frac{((3!)!)!}{3!} &= \frac{720!}{6} \\ &= \frac{720 \cdot 719!}{6} \\ &= 120 \cdot 719!. \end{aligned}$$

If $n$ were $720$ or more, then $k \cdot n! \ge 720!,$ which exceeds $\frac{720!}{6}.$ So the largest possible value of $n$ is $719,$ achieved with $k = 120,$ and $k + n = 120 + 719 = 839.$

Answer: 301

Small Hint:

The colors alternate, so the green regions are the annuli between radii $1$ and $2,$ $3$ and $4,$ and so on

Big Hint:

By difference of squares, the green areas total $\pi\,(1 + 2 + 3 + \cdots + 100)$

Solution:

The regions alternate red, green, red, green, $\ldots$ from the center outward, so the green regions are the annuli between radii $1$ and $2,$ between $3$ and $4,$ and so on up to the annulus between $99$ and $100.$ Their total area is $$\begin{aligned} &\scriptsize \pi\left[(2^2 - 1^2) + (4^2 - 3^2) + \cdots + (100^2 - 99^2)\right] \\ &\scriptsize = \pi\left[(2 + 1) + (4 + 3) + \cdots + (100 + 99)\right], \end{aligned}$$ which is $\pi\,(1 + 2 + \cdots + 100) = 5050\pi.$

The desired ratio is $\frac{5050\pi}{100^2 \pi} = \frac{101}{200},$ so $m + n = 101 + 200 = 301.$

Answer: 484

Small Hint:

Each element is written on the list once for every smaller element in the set

Big Hint:

Sort the set as $1, 2, 3, 5, 8, 13, 21, 34$ and multiply each element by the count of smaller elements

Solution:

An element $x$ is the greater element of a two-element subset exactly once for each smaller element of the set, so $x$ contributes to the sum once per element below it. Sorting the set as $1, 2, 3, 5, 8, 13, 21, 34,$ the sum of the list is $$\begin{aligned} &0(1) + 1(2) + 2(3) + 3(5) \\ &\quad {}+ 4(8) + 5(13) + 6(21) \\ &\quad {}+ 7(34) \\ &= 2 + 6 + 15 + 32 \\ &\quad {}+ 65 + 126 + 238 = 484. \end{aligned}$$

Answer: 12

Small Hint:

Add the logarithms: the first equation gives $\sin x \cos x = \frac{1}{10}$

Big Hint:

Square $\sin x + \cos x$ and use $\sin^2 x + \cos^2 x = 1$ to evaluate it

Solution:

The first equation says $\log_{10}(\sin x \cos x) = -1,$ so $\sin x \cos x = \frac{1}{10}.$ Then $$\begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x \\ &\quad {}+ 2 \sin x \cos x \\ &= 1 + \frac{2}{10} = \frac{12}{10}. \end{aligned}$$

Taking logarithms, $2\log_{10}(\sin x + \cos x)$ $= \log_{10} \frac{12}{10}$ $= \log_{10} 12 - 1,$ so $\log_{10}(\sin x + \cos x)$ $= \frac{1}{2}(\log_{10} 12 - 1)$ and $n = 12.$

Answer: 505

Small Hint:

The region is the box plus slabs over each face, quarter-cylinders along each edge, and sphere octants at each corner

Big Hint:

The twelve quarter-cylinders form three full cylinders with heights $3,$ $4,$ $5,$ and the eight octants form one unit sphere

Solution:

The region consists of the box itself, six slabs of thickness $1$ projecting outward from the faces, quarter-cylinders of radius $1$ along the twelve edges, and eighth-spheres of radius $1$ at the eight corners. The box has volume $3 \cdot 4 \cdot 5 = 60,$ and the slabs total $2(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94.$

The four quarter-cylinders along edges parallel to each dimension combine into a full cylinder, so the cylinders total $\pi \cdot 1^2 (3 + 4 + 5) = 12\pi.$ The eight octants combine into one unit sphere of volume $\frac{4\pi}{3}.$

The total volume is $$\begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3}, \end{aligned}$$ so $m + n + p$ $= 462 + 40 + 3$ $= 505.$

Answer: 348

Small Hint:

Each side is a cube edge, a face diagonal of length $\sqrt{2},$ or a space diagonal of length $\sqrt{3};$ classify the $\binom{8}{3} = 56$ triangles

Big Hint:

The three types are edge-edge-face diagonal, three face diagonals, and edge-face diagonal-space diagonal; there are $24,$ $8,$ and $24$ of them

Solution:

Every side of such a triangle is a cube edge, a face diagonal of length $\sqrt{2},$ or a space diagonal of length $\sqrt{3}.$ Only three shapes occur. A triangle of two adjacent edges and a face diagonal is right with area $\frac{1}{2};$ there are $4$ per face, or $24.$ A triangle of three face diagonals is equilateral with area $\frac{\sqrt{3}}{2};$ each is determined by the three vertices adjacent to one of the $8$ cube vertices, so there are $8.$ A triangle of an edge, a face diagonal, and a space diagonal is right with legs $1$ and $\sqrt{2},$ so its area is $\frac{\sqrt{2}}{2};$ each of the $4$ space diagonals forms one with each of the $6$ vertices off that diagonal, so there are $24.$ (Indeed $24 + 8 + 24 = \binom{8}{3} = 56.$)

The total area is $$\begin{aligned} &24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} \\ &= 12 + 4\sqrt{3} + 12\sqrt{2} \\ &= 12 + \sqrt{48} + \sqrt{288}, \end{aligned}$$ so $m + n + p$ $= 12 + 48 + 288$ $= 348.$

Answer: 380

Small Hint:

The foot of the perpendicular from $D$ to $\overline{AC}$ is the midpoint of $\overline{AC},$ which is $6$ units from $B$

Big Hint:

With $AD = a$ and $BD = b,$ the two right triangles give $a^2 - b^2 = 15^2 - 6^2 = 189;$ factor it

Solution:

Let $AD = CD = a$ and $BD = b,$ and let $E$ be the foot of the perpendicular from $D$ to $\overline{AC}.$ Since $AD = CD,$ point $E$ is the midpoint of $\overline{AC},$ so $AE = 15$ and $BE = 15 - 9 = 6.$ The right triangles $DEA$ and $DEB$ share leg $DE,$ so $$a^2 - 15^2 = DE^2 = b^2 - 6^2,$$ that is $$(a+b)(a-b) = 189.$$

The factorizations $189 = 189 \cdot 1$ $= 63 \cdot 3$ $= 27 \cdot 7 = 21 \cdot 9$ give $(a, b) = (95, 94),$ $(33, 30),$ $(17, 10),$ and $(15, 6).$ The last is rejected: $b = 6$ would put $D$ on $\overline{AC}.$ Each valid pair gives a triangle with perimeter $2a + 30.$

Therefore $s = (190 + 30)$ $+ (66 + 30)$ $+ (34 + 30)$ $= 220 + 96 + 64$ $= 380.$

Answer: 129

Small Hint:

Write the terms as $a,$ $a + d,$ $a + 2d,$ and $a + 30$

Big Hint:

The geometric condition becomes $3a(10 - d) = 2d(2d - 15),$ so $10 - d$ and $2d - 15$ must have the same sign

Solution:

Write the terms as $a,$ $a + d,$ $a + 2d,$ and $a + 30,$ where $a$ and $d$ are positive integers. The geometric condition on the last three terms says $(a + 30)(a + d) = (a + 2d)^2.$ Expanding both sides and simplifying, $$30a + 30d = 3ad + 4d^2,$$ that is $$3a(10 - d) = 2d(2d - 15).$$

Since $a, d \gt 0,$ the factors $10 - d$ and $2d - 15$ must have the same sign, forcing $7.5 \lt d \lt 10,$ so $d = 8$ or $d = 9.$ For $d = 8,$ we get $6a = 16,$ which has no integer solution. For $d = 9,$ we get $3a = 54,$ so $a = 18.$

The sequence is $18, 27, 36, 48$ (indeed $27, 36, 48$ has ratio $\frac{4}{3}$), and the sum is $18 + 27 + 36 + 48 = 129.$

Answer: 615

Small Hint:

Group by the common digit-pair sum $s;$ count leftmost pairs (first digit nonzero) and rightmost pairs separately

Big Hint:

For $s \le 9$ the counts are $s$ and $s + 1;$ for $s \ge 10$ both equal $19 - s.$ Sum the products.

Solution:

Group the balanced integers by the common sum $s$ of each digit pair, where $1 \le s \le 18.$ For $s \le 9,$ the leftmost pair (first digit at least $1$) can be formed in $s$ ways and the rightmost pair in $s + 1$ ways. For $s \ge 10,$ both digits of each pair must be at least $s - 9,$ giving $19 - s$ ways for each pair.

The total count is $$\begin{aligned} &\sum_{s=1}^{9} s(s+1) + \sum_{s=10}^{18} (19 - s)^2 \\ &= \sum_{s=1}^{9} (s^2 + s) + \sum_{k=1}^{9} k^2 \\ &= 2 \cdot 285 + 45 = 615. \end{aligned}$$

Answer: 83

Small Hint:

The base angles are $37^\circ.$ In triangle $AMC,$ the Law of Sines with $AC = 1$ gives $CM = 2\sin 7^\circ.$

Big Hint:

Apply the Law of Cosines in triangle $BMC,$ using the fact that the cosine of $\angle MCB$ equals $\sin 7^\circ,$ to show $MB = CB$

Solution:

Assume $AC = BC = 1.$ In triangle $AMC,$ the angles at $A$ and $C$ are $7^\circ$ and $23^\circ,$ so $\angle AMC = 150^\circ,$ and the Law of Sines gives $$CM = \frac{\sin 7^\circ}{\sin 150^\circ} = 2\sin 7^\circ.$$

Also $\angle MCB = 106^\circ - 23^\circ = 83^\circ,$ whose cosine is $\sin 7^\circ.$ The Law of Cosines in triangle $BMC$ then gives $$\begin{aligned} MB^2 &= CM^2 + CB^2 \\ &\quad {}- 2 \cdot CM \cdot CB \cos 83^\circ \\ &= 4\sin^2 7^\circ + 1 \\ &\quad {}- 4\sin^2 7^\circ = 1. \end{aligned}$$

So $MB = 1 = CB,$ making triangle $BMC$ isosceles with $\angle CMB = \angle MCB = 83^\circ.$ The answer is $83.$

Answer: 92

Small Hint:

For $0^\circ \lt x \le 45^\circ$ the largest of the three numbers is $\cos^2 x,$ so only one triangle inequality can fail

Big Hint:

$\cos^2 x \ge \sin^2 x + \sin x \cos x$ rearranges via double angles to $\tan 2x \le 2$

Solution:

Replacing $x$ by $90^\circ - x$ swaps $\sin x$ and $\cos x,$ so the failure probability on $(45^\circ, 90^\circ)$ matches that on $(0^\circ, 45^\circ),$ and it suffices to consider $0^\circ \lt x \le 45^\circ.$ There $\cos^2 x \ge \sin x \cos x \ge \sin^2 x,$ so the three numbers fail to form a triangle exactly when $$\cos^2 x \ge \sin^2 x + \sin x \cos x.$$

Since $\cos^2 x - \sin^2 x = \cos 2x$ and $\sin x \cos x = \frac{1}{2}\sin 2x,$ this says $\cos 2x \ge \frac{1}{2} \sin 2x,$ i.e. $\tan 2x \le 2.$ Because tangent increases on this range, that happens exactly for $x \le \frac{1}{2}\arctan 2.$

Therefore $$p = \frac{\frac{1}{2}\arctan 2}{45^\circ} = \frac{\arctan 2}{90^\circ},$$ so $m = 2$ and $n = 90,$ with $m + n = 92 \lt 1000,$ and the answer is $92.$

Answer: 777

Small Hint:

Compute $BD^2$ by the Law of Cosines in triangles $ABD$ and $CDB,$ and set the results equal

Big Hint:

Because $AD \ne BC,$ dividing by $AD - BC$ leaves $\cos A = \frac{AD + BC}{360},$ and $AD + BC = 640 - 360$

Solution:

Let $\angle A = \angle C = \alpha,$ $AD = x,$ and $BC = y.$ Applying the Law of Cosines to diagonal $BD$ in triangles $ABD$ and $CDB,$ $$\begin{aligned} BD^2 &= x^2 + 180^2 \\ &\quad {}- 2 \cdot 180x\cos\alpha \\ &= y^2 + 180^2 \\ &\quad {}- 2 \cdot 180y\cos\alpha. \end{aligned}$$

Rearranging gives $x^2 - y^2 = 2 \cdot 180(x - y)\cos\alpha,$ and since $x \ne y$ we may divide by $x - y:$ $$\begin{aligned} \cos\alpha &= \frac{x + y}{360} \\ &= \frac{640 - 2 \cdot 180}{360} \\ &= \frac{280}{360} = \frac{7}{9}. \end{aligned}$$

Then $1000\cos A = \frac{7000}{9} = 777.7\ldots,$ so $\lfloor 1000\cos A \rfloor = 777.$

Answer: 155

Small Hint:

Every such integer has at most $11$ binary digits; for $(d+1)$-digit numbers, count by the number of $1$’s using row $d$ of Pascal’s triangle

Big Hint:

Rows $0$ through $10$ contribute $\frac{2047 + 351}{2}$ numbers by symmetry; then discard the qualifying integers from $2004$ to $2047.$

Solution:

Since $2003 \lt 2^{11} = 2048,$ every integer in question has at most $11$ binary digits. A $(d+1)$-digit binary number starts with $1,$ and choosing $k$ more $1$’s among the remaining $d$ digits gives $\binom{d}{k}$ numbers with $k + 1$ ones; the $1$’s outnumber the $0$’s exactly when $k \ge \frac{d}{2}.$ So the count over all numbers up to $2047$ is the total of the entries on or to the right of the center of rows $0$ through $10$ of Pascal’s triangle.

Those rows sum to $1 + 2 + \cdots + 2^{10} = 2047,$ and the central entries sum to $\sum_{i=0}^{5}\binom{2i}{i}$ $= 1 + 2 + 6 + 20 + 70 + 252$ $= 351,$ so by symmetry the count is $\frac{2047 + 351}{2} = 1199.$

The $44$ integers from $2004$ to $2047$ all exceed $1984 = 11111000000_2,$ so each has the prefix $11111$ plus at least one more $1,$ hence at least six $1$’s among eleven digits — all $44$ were counted. Therefore $N = 1199 - 44 = 1155,$ whose remainder upon division by $1000$ is $155.$

Answer: 127

Small Hint:

Show it suffices for $251$ to appear immediately after the decimal point, so $\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000}$

Big Hint:

Use the neighboring fractions $\frac14 \lt \frac{32}{127} \lt \frac{63}{250}:$ each adjacent pair has cross-difference $1,$ which bounds the denominators between them

Solution:

It suffices to make $251$ appear immediately after the decimal point: if $\frac{m}{n} = 0.A251\ldots$ with $A$ a block of $k \ge 1$ digits, then $10^k \frac{m}{n} - A = 0.251\ldots$ is a fraction between $0$ and $1$ whose reduced denominator is at most $n.$ So we need the smallest $n$ admitting an $m$ with $$\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000},$$ that is $$0 \le 1000m - 251n \lt n.$$

The fraction $\frac{32}{127}$ lies in this interval because $$\frac{251}{1000} \lt \frac{32}{127} \lt \frac{63}{250} = \frac{252}{1000}.$$ It remains to prove that no smaller denominator works. We use the following elementary fact: if $\frac{a}{b} \lt \frac{u}{v} \lt \frac{c}{d}$ and $bc-ad=1,$ then $$\begin{aligned} v &= b(cv-du)+d(bu-av) \\ &\ge b+d, \end{aligned}$$ because both parenthesized quantities are positive integers.

Now $4\cdot32-1\cdot127=1$ and $127\cdot63-32\cdot250=1.$ Therefore every fraction strictly between $\frac14$ and $\frac{32}{127}$ has denominator at least $4+127=131,$ while every fraction strictly between $\frac{32}{127}$ and $\frac{63}{250}$ has denominator at least $127+250=377.$ Since our target interval lies inside $\left(\frac14,\frac{63}{250}\right)$ and contains $\frac{32}{127},$ no fraction in it has denominator below $127.$

The smallest possible value of $n$ is $127.$