Answer: 93

Small Hint:

Pair the odd-indexed terms with the even-indexed terms

Big Hint:

Each even-indexed term is $1$ greater than the odd-indexed term immediately before it

Solution:

Let $E=a_2+a_4+\cdots+a_{98}$ and $O=a_1+a_3+\cdots+a_{97}.$ There are $49$ pairs, and $a_{2k}-a_{2k-1}=1,$ so $E-O=49.$ Also $E+O=137.$ Adding the two equations gives $2E=186,$ and hence $E=93.$

Answer: 592

Small Hint:

Divisibility by $5$ determines the last digit

Big Hint:

Divisibility by $3$ restricts the number of digits equal to $8$

Solution:

A multiple of $15$ must end in $0$ and have digit sum divisible by $3.$ Because $8\equiv2\pmod3,$ the number of $8$’s must be a positive multiple of $3.$ The smallest possible number therefore has three $8$’s followed by $0,$ namely $n=8880.$ Thus $\frac{n}{15}=592.$

Answer: 144

Small Hint:

Each of the three smaller triangles is similar to $\triangle ABC$

Big Hint:

Convert each area ratio into a linear ratio and add the three linear ratios

Solution:

Let the area of $\triangle ABC$ be $K.$ The three small triangles are similar to $\triangle ABC,$ so their corresponding linear ratios are $$\frac{2}{\sqrt K},\qquad \frac{3}{\sqrt K},\qquad \frac{7}{\sqrt K}.$$ Each such scale factor is also the perpendicular distance from $P$ to one side divided by the altitude to that side. These are the three barycentric area ratios $\frac{[PBC]}{K},$ $\frac{[PCA]}{K},$ and $\frac{[PAB]}{K},$ which add to $1.$ Hence $$\frac{2+3+7}{\sqrt K}=1.$$ Therefore $\sqrt K=12$ and $K=144.$

Answer: 649

Small Hint:

Let $N$ be the number of entries in the original list

Big Hint:

To maximize one entry, make every unrestricted positive integer as small as possible

Solution:

If the original list has $N$ entries, then $$56N-68=55(N-1),$$ so $N=13$ and the total of the entries is $56\cdot13=728.$ Besides the entry $68,$ make eleven entries equal to the least positive integer, $1.$ The remaining entry is then $$728-68-11=649.$$ This construction is valid, and no larger entry is possible.

Answer: 512

Small Hint:

Express every logarithm using base $2$

Big Hint:

Set $A=\log_2a$ and $B=\log_2b$ to obtain a linear system

Solution:

Let $A=\log_2a$ and $B=\log_2b.$ The two equations become $$\frac A3+B=5,\qquad \frac B3+A=7.$$ Equivalently, $A+3B=15$ and $3A+B=21,$ which give $A=6$ and $B=3.$ Therefore $ab=2^{A+B}=2^9=512.$

Answer: 24

Small Hint:

The line already bisects the circle centered at $(17,76)$

Big Hint:

For the other two equal circles, their signed distances from the line must be opposites

Solution:

For a circle of radius $3,$ let $G(d)$ be the signed difference between the areas on the two sides of a line when the center’s signed distance from the line is $d.$ The function $G$ is odd and strictly increasing for $-3<d<3,$ and is constant only after the line no longer cuts the circle. The circle centered at $(17,76)$ contributes zero, so the line must separate the other two centers.

Let $M=(\frac{33}{2},88)$ be the midpoint of those two centers. Among the lines through $(17,76)$ that separate them, the smaller of their two distances to the line is largest when the distances are equal, namely for the line through $M.$ Even then, the distance is $\frac{56}{\sqrt{577}}<3,$ so the line must cut at least one of the two circles. Balance then forces it to cut both. The strict increase of $G$ therefore forces the two signed distances to be opposites, so the required line passes through $M.$ Its slope is $$\frac{88-76}{\frac{33}{2}-17}=-24.$$ Its absolute value is $24.$

Answer: 997

Small Hint:

Start by evaluating $f(995),$ $f(996),$ $\ldots,$ $f(999)$

Big Hint:

Use downward induction to find a parity pattern for every integer below $1000$

Solution:

Directly from the definition, $$\begin{aligned} f(999)&=f(f(1004))\\ &=f(1001)=998,\\ f(998)&=f(f(1003))\\ &=f(1000)=997. \end{aligned}$$ Continuing gives $f(997)=998,$ $f(996)=997,$ and $f(995)=998.$

We now use downward induction. If $n<995$ is even, then $n+5$ is odd, so the established pattern above $n$ gives $f(n)=f(f(n+5)).$ Thus $f(n)=f(998)=997.$ If $n$ is odd, the same argument gives $f(n)=f(997)=998.$ Thus every even $n<1000$ has value $997.$ Since $84$ is even, $f(84)=997.$

Answer: 160

Small Hint:

Substitute $u=z^3$

Big Hint:

The two possible arguments of $z^3$ are $120^\circ$ and $240^\circ$

Solution:

Let $u=z^3.$ Then $u^2+u+1=0,$ so $u$ has argument $120^\circ$ or $240^\circ.$ Taking cube roots gives possible arguments $$40^\circ,160^\circ,280^\circ,80^\circ,200^\circ,320^\circ.$$ The only one strictly between $90^\circ$ and $180^\circ$ is $160^\circ.$

Answer: 20

Small Hint:

Find the altitudes from $C$ and $D$ to the common edge $AB$

Big Hint:

Express the tetrahedron’s volume using the common edge, the two altitudes, and the sine of the dihedral angle

Solution:

Let $h_C$ and $h_D$ be the perpendicular distances from $C$ and $D$ to $AB.$ From the two face areas, $$\begin{aligned} h_C&=\frac{2\cdot15}{3}=10,\\ h_D&=\frac{2\cdot12}{3}=8. \end{aligned}$$ The angle between these two perpendicular directions is the $30^\circ$ dihedral angle. Hence the scalar triple product gives $$\begin{aligned} V&=\frac16(AB)h_Ch_D\sin30^\circ\\ &=\frac16\cdot3\cdot10\cdot8\cdot\frac12\\ &=20. \end{aligned}$$

Answer: 119

Small Hint:

For a fixed score $s,$ express the number wrong in terms of $s$ and $c$

Big Hint:

Use $w\geq0$ and $c+w\leq30$ to bound the possible integer values of $c$

Solution:

From $s=30+4c-w,$ we have $w=30+4c-s.$ The conditions $w\geq0$ and $30-c-w\geq0$ give $$\left\lceil\frac{s-30}{4}\right\rceil \leq c\leq \left\lfloor\frac{s}{5}\right\rfloor.$$ Evaluating these integer endpoints for scores $81$ through $118$ always leaves at least two possible values of $c.$ At $s=119,$ both endpoints equal $23,$ so John can determine $c=23.$ Thus the first score over $80$ with the required property is $119.$

Answer: 106

Small Hint:

First choose the five positions occupied by birch trees

Big Hint:

Place the seven non-birch trees first and use the eight gaps around them

Solution:

The five birch positions form a uniformly chosen $5$-element subset of the $12$ positions, so there are $\binom{12}{5}$ possibilities. After the seven non-birch trees are placed, there are eight gaps, including the two end gaps. Choosing five distinct gaps gives $\binom85$ arrangements with no adjacent birches. Therefore $$\frac{m}{n}=\frac{\binom85}{\binom{12}{5}} =\frac{56}{792}=\frac7{99},$$ and $m+n=106.$

Answer: 401

Small Hint:

Interpret the two identities as reflection symmetries about $2$ and $7$

Big Hint:

Composing the two reflections produces a translation by $10$

Solution:

The identities make the root set invariant under reflection about $2$ and about $7.$ Their composition is translation by $10.$ Starting from the root $0,$ these operations force every number in $$\begin{gathered} \{10k:k\in\mathbb Z\}\\ {}\cup\{4+10k:k\in\mathbb Z\} \end{gathered}$$ to be a root. In the interval, the first set contributes $201$ roots and the second contributes $200,$ for a total of $401.$

This bound is attainable: define $f$ to be $0$ on this invariant set and $1$ elsewhere. It has both required reflection symmetries. Hence the least possible number is $401.$

Answer: 15

Small Hint:

Use $\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}$

Big Hint:

Combine the four inverse-cotangent angles two at a time from left to right

Solution:

Let $\alpha=\cot^{-1}3+\cot^{-1}7.$ The cotangent addition formula gives $$\begin{aligned} \cot\alpha&=\frac{3\cdot7-1}{3+7}\\ &=2. \end{aligned}$$ Combining the next angle gives $$\frac{2\cdot13-1}{2+13}=\frac53,$$ and combining the last gives $$\frac{(\frac{5}{3})\cdot21-1}{\frac{5}{3}+21}=\frac32.$$ Multiplying by $10$ yields $15.$

Answer: 38

Small Hint:

Test candidate even integers by listing the odd composites no greater than half the candidate

Big Hint:

For sufficiently large even $N,$ work modulo $6$ and try subtracting $9,$ $25,$ or $35$

Solution:

In a representation of $38,$ the smaller odd composite would be at most $19.$ The only possibilities are $9$ and $15,$ whose complements $29$ and $23$ are prime. Thus $38$ is not representable.

Now let $N>38$ be even. If $N\equiv0\pmod6,$ write $N=9+(N-9).$ If $N\equiv2\pmod6,$ write $N=35+(N-35).$ If $N\equiv4\pmod6,$ write $N=25+(N-25).$ In each case the second summand is an odd multiple of $3$ greater than $3,$ hence is composite; the fixed first summand is also odd and composite. Therefore every even integer greater than $38$ is representable, so the largest exception is $38.$

Answer: 36

Small Hint:

Regard the four left sides as values of one rational function in $T$

Big Hint:

Compare the coefficient of $\frac{1}{T}$ as $T$ tends to infinity

Solution:

Define $$\begin{aligned} R(T)&=\frac{x^2}{T-1}+\frac{y^2}{T-9}\\ &\quad{}+\frac{z^2}{T-25}+\frac{w^2}{T-49}\\ &\quad{}-1. \end{aligned}$$ Put $$\begin{aligned} N(T)&=(T-4)(T-16)\\ &\quad{}\cdot(T-36)(T-64),\\ D(T)&=(T-1)(T-9)\\ &\quad{}\cdot(T-25)(T-49). \end{aligned}$$ With common denominator $D(T),$ the numerator of $R(T)$ has leading coefficient $-1.$ The four equations say its zeros are $4,$ $16,$ $36,$ and $64.$ Therefore $R(T)=-\frac{N(T)}{D(T)}.$

Let $S=w^2+x^2+y^2+z^2.$ As $T$ tends to infinity, the defining expression gives $$R(T)=-1+\frac{S}{T}+O(T^{-2}).$$ On the other hand, the sum of the four numerator roots is $120,$ while the sum of the four denominator roots is $84,$ so the factored expression gives $$\begin{aligned} R(T)&=-1+\frac{120-84}{T}\\ &\quad{}+O(T^{-2}). \end{aligned}$$ Hence $w^2+x^2+y^2+z^2=36.$