Small Hint:

Count all nonempty proper subsets of the ten buttons

Big Hint:

Subtract the five-button combinations that the original design already allowed

Solution:

The redesigned lock permits every nonempty subset except the full set, giving $2^{10}-2=1022$ combinations. The original lock permits $\binom{10}{5}=252.$ Thus the number of additional combinations is $1022-252=770.$

Small Hint:

Compute the first several iterates until a value repeats

Big Hint:

After the transient values, the sequence alternates between two numbers

Solution:

The iterates begin $4,$ $16,$ $49,$ $169,$ $256,$ followed by $169,$ $256,$ $\ldots.$ Thus from the fourth iterate onward, even indices give $169$ and odd indices give $256.$ Since $1988$ is even, $f_{1988}(11)=169.$

Small Hint:

Set $y=\log_2x$ and rewrite every base-$8$ logarithm in base $2$

Big Hint:

Solve the resulting linear equation for $\log_2 y$

Solution:

Put $y=\log_2x\gt0.$ Since $\log_8x=\frac{y}{3},$ the equation becomes $$\log_2(\frac{y}{3})=\frac13\log_2y.$$ Hence $\frac23\log_2y=\log_23,$ so $y=3^{\frac{3}{2}}=3\sqrt3.$ Therefore $(\log_2x)^2=y^2=27.$

Small Hint:

Let $P$ be the sum of the positive terms and $N$ the sum of the absolute values of the negative terms

Big Hint:

The left side minus the final absolute value equals $2\min(P,N)$

Solution:

Let $P$ be the sum of the positive $x_i$ and $N$ the sum of the absolute values of the negative $x_i.$ Then $$\begin{aligned}P+N-|P-N|&=2\min(P,N)\\&=19,\end{aligned}$$ so both $P$ and $N$ are at least $9.5.$ Because every $|x_i|\lt1,$ each sign requires at least $10$ terms, giving $n\geq20.$ Equality is attainable with ten terms equal to $0.95$ and ten equal to $-0.95,$ so the minimum is $20.$

Small Hint:

Write every divisor as $2^a5^b$

Big Hint:

Count exponent pairs with $a\geq88$ and $b\geq88$ and reduce the resulting probability

Solution:

Every divisor is $2^a5^b$ with $0\leq a\leq99$ and $0\leq b\leq99,$ so there are $100^2=10000$ divisors. A multiple of $10^{88}$ requires $88\leq a\leq99$ and $88\leq b\leq99,$ giving $12^2=144$ divisors. The probability is $\frac{144}{10000}=\frac{9}{625},$ and $m+n=9+625=634.$

Small Hint:

Index the rows and columns from $0$ through $4$

Big Hint:

A grid whose rows and columns are arithmetic has the form $A+Bi+Cj+Dij$

Solution:

Index rows and columns from $0$ to $4.$ The row and column conditions give the general entry $A+Bi+Cj+Dij.$ The four shown values yield $$\begin{aligned}A+4B&=0,\\A+B+C+D&=74,\\A+2B+4C+8D&=186,\\A+3B+2C+6D&=103.\end{aligned}$$ Solving gives $A=52,$ $B=-13,$ $C=30,$ and $D=5.$ The asterisk is at $(i,j)=(0,3),$ so its value is $A+3C=52+90=142.$

Small Hint:

Let the altitude have length $h$ and place its foot at the origin

Big Hint:

Use cross product over dot product to express the tangent of the angle between the two side vectors

Solution:

Let the altitude length be $h.$ From $A,$ vectors to the endpoints of $BC$ may be taken as $(-3,-h)$ and $(17,-h).$ Therefore $$\tan\angle CAB=\frac{20h}{h^2-51}=\frac{22}{7}.$$ Thus $11h^2-70h-561=0,$ whose positive root is $h=11.$ Since $BC=3+17=20,$ the area is $\frac12(20)(11)=110.$

Small Hint:

Apply the third property to $(x,y-x)$ when $y\gt x$

Big Hint:

Compare the resulting subtraction rule with the same rule for $\operatorname{lcm}(x,y)$

Solution:

For $y\gt x,$ the third property applied to $(x,y-x)$ gives $$f(x,y)=\frac{y}{y-x}f(x,y-x).$$ The least common multiple obeys the identical relation, because $\gcd(x,y)=\gcd(x,y-x).$ Hence the quotient $\frac{f(x,y)}{\operatorname{lcm}(x,y)}$ is unchanged by each subtraction step of the Euclidean algorithm. On the diagonal it equals $\frac{f(g,g)}{g}=1,$ so $f(x,y)=\operatorname{lcm}(x,y).$ Therefore $$\begin{aligned}f(14,52)&=\frac{14\cdot52}{\gcd(14,52)}\\&=364.\end{aligned}$$

Small Hint:

The units digit forces the number to have the form $10a+2$

Big Hint:

Reduce $(10a+2)^3\equiv888\pmod{1000}$ first modulo $125$

Solution:

Write the number as $10a+2.$ Expanding the required congruence modulo $125$ and dividing by $5$ gives $$20a^2-a-1\equiv0\pmod{25}.$$ Reducing first modulo $5$ gives $a\equiv4\pmod5;$ substituting $a=4+5b$ then gives $b\equiv3\pmod5.$ Thus $a\equiv19\pmod{25},$ and the smallest candidate is $10(19)+2=192.$ It is even, so its cube is also $0\pmod8,$ and indeed $192^3=7{,}077{,}888.$

Small Hint:

Double-count face-vertex and face-edge incidences to find $V$ and $E$

Big Hint:

Count all vertex pairs, then subtract pairs lying together on a face, correcting the double count of edges

Solution:

The total number of face-vertex incidences is $12(4)+8(6)+6(8)=144.$ Three faces meet at each vertex, so $V=48.$ The same sum counts each edge twice, so $E=72.$

There are $\binom{48}{2}=1128$ vertex pairs. Summing pairs on faces gives $12\binom42+8\binom62+6\binom82=360.$ Every edge was counted twice in this sum and every other same-face pair once, so the number of distinct boundary pairs is $360-E=288.$ Hence $1128-288=840$ joining segments lie in the interior.

Small Hint:

Average the equation $\sum(z_k-w_k)=0$

Big Hint:

A mean line is exactly a line through the centroid of the given points

Solution:

The condition says that the average of the $z_k$ equals the average of the $w_k.$ Since all $z_k$ lie on $L,$ their average lies on $L;$ conversely, any line through the average works by taking all $z_k$ equal to that point. The centroid satisfies $$\begin{aligned}\bar x&=\frac{32-7-9+1-14}{5}=\frac35,\\\bar y&=\frac{170+64+200+27+43}{5}\\&=\frac{504}{5}.\end{aligned}$$ The line through this point and $(0,3)$ has slope $\frac{\bigl(\frac{504}{5}-3\bigr)}{\frac35}=163.$

Small Hint:

Express the three barycentric coordinates of $P$ using the ratios $a:d,$ $b:d,$ $c:d$

Big Hint:

Use $\frac3{a+3}+\frac3{b+3}+\frac3{c+3}=1$ and expand symmetrically

Solution:

Along the cevian from $A,$ the barycentric coordinate at $A$ is $\frac{d}{a+d},$ and similarly at the other vertices. Since the three coordinates sum to $1$ and $d=3,$ $$\frac3{a+3}+\frac3{b+3}+\frac3{c+3}=1.$$ Put $s_1=a+b+c=43,$ $s_2=ab+bc+ca,$ and $s_3=abc.$ Clearing denominators gives $$\begin{aligned}3(s_2+6s_1+27)&=s_3+3s_2\\&\quad+9s_1+27.\end{aligned}$$ Therefore $s_3=9s_1+54,$ so $s_3=9(43)+54=441.$

Small Hint:

Modulo $x^2-x-1,$ powers satisfy $x^n=F_nx+F_{n-1}$

Big Hint:

Set both coefficients of the linear remainder equal to zero and use Cassini’s identity

Solution:

Modulo $x^2-x-1,$ the relation $x^2=x+1$ gives $x^n=F_nx+F_{n-1}.$ Hence the remainder is $$\begin{aligned}r(x)={}&(1597a+987b)x\\&+(987a+610b+1).\end{aligned}$$ Both coefficients must vanish. The determinant of the coefficient matrix is $$1597(610)-987^2=1$$ by Cassini’s identity. Cramer’s rule then gives $a=987.$

Small Hint:

Find the reflection matrix for the line spanned by $(1,2)$

Big Hint:

Because reflection is its own inverse, substitute the reflected coordinates into $uv=1$

Solution:

Reflection in the line spanned by $(1,2)$ has matrix $$\frac15\begin{pmatrix}-3&4\\4&3\end{pmatrix}.$$ Thus a point $(x,y)$ on $C^*$ came from $u=\frac{-3x+4y}{5}$ and $v=\frac{4x+3y}{5}$ on $C.$ Substituting $uv=1$ gives $$-12x^2+7xy+12y^2-25=0,$$ or $12x^2-7xy-12y^2+25=0.$ Hence $b=-7,$ $c=-12,$ and $bc=84.$

Small Hint:

After $8$ has been typed, any remaining letters among $1,\ldots,7$ must later be typed in decreasing order

Big Hint:

Separate the cases according to whether $9$ was typed before lunch or remains to be inserted into the later order

Solution:

Any subset of $1,\ldots,7$ can remain in the stack after $8$ has been typed, and those remaining letters must later be typed in decreasing order. If $9$ was already typed, choosing that subset gives $2^7=128$ possible orders.

If $9$ remains, choose $k$ of the seven smaller letters and insert $9$ into any of the $k+1$ positions in their decreasing order. This gives $$\begin{aligned}\sum_{k=0}^7\binom7k(k+1)&=7\cdot2^6+2^7\\&=576.\end{aligned}$$ Every such order can be realized by suitable morning choices, so the total is $128+576=704.$