Small Hint:

Treat the four decimal places independently

Big Hint:

A target digit $d$ can be split into an ordered pair of digits in $d+1$ ways without carrying

Solution:

For a target digit $d,$ there are $d+1$ ordered pairs of nonnegative digits with sum $d.$ The four digits $1,$ $4,$ $9,$ $2$ therefore give $2,$ $5,$ $10,$ $3$ choices independently. The answer is $2\cdot5\cdot10\cdot3=300.$

Small Hint:

First find the distance between the two centers

Big Hint:

The maximum occurs along the line of centers, on the two outward sides

Solution:

The squared distance between the centers is $14^2+18^2+(-21)^2=961,$ so they are $\sqrt{961}=31$ apart. By the triangle inequality, the greatest point-to-point distance is obtained on the line of centers and equals $19+31+87=137.$

Small Hint:

Express the product of all positive divisors in terms of the number and its divisor count

Big Hint:

The nice numbers are exactly those having four positive divisors

Solution:

If $N$ has $d$ positive divisors, their product is $N^{\frac{d}{2}}.$ Removing $1$ and $N$ leaves product $N^{\frac{d}{2}-1},$ which equals $N$ exactly when $d=4.$ Thus nice numbers are precisely $p^3$ and $pq$ for distinct primes. The first ten are $6,$ $8,$ $10,$ $14,$ $15,$ $21,$ $22,$ $26,$ $27,$ $33,$ whose sum is $182.$

Small Hint:

Solve for $|y|$ and determine where its right-hand side is nonnegative

Big Hint:

The boundary is a kite whose vertices occur at the breakpoints of the absolute values

Solution:

We need $|y|=\frac{|x|}{4}-|x-60|\geq0.$ This forces $48\leq x\leq80.$ At $x=48,$ $x=60,$ $x=80,$ the boundary values are respectively $y=0,$ $y=\pm15,$ $y=0.$ Hence the region is a kite with perpendicular diagonals $80-48=32$ and $30,$ so its area is $\frac12(32)(30)=480.$

Small Hint:

Move a suitable multiple of $3x^2+1$ to create a product

Big Hint:

Factor $507$ and use that $3x^2+1\equiv1\pmod3$

Solution:

Rearranging gives $(3x^2+1)(y^2-10)=507,$ where $507=3\cdot13^2.$ The positive divisors of $507$ congruent to $1\pmod3$ are $1,$ $13,$ $169.$ If $3x^2+1=1,$ then $x^2=0$ but $y^2=517,$ which is not a square. If $3x^2+1=13,$ then $x^2=4$ and $y^2=49.$ Finally, $3x^2+1=169$ gives $x^2=56,$ not a square. Therefore $3x^2y^2=3\cdot4\cdot49=588.$

Small Hint:

Call the common boundary length $t$ and the rectangle width $L$

Big Hint:

Equal areas above and below $PQ$ place $PQ$ halfway up the rectangle

Solution:

Put $AB=L$ and let the four equal boundary lengths be $t.$ Writing $AX=x$ and $DW=w,$ the left boundary condition gives $x+w=t-19.$ Substituting $XY=ZW=t$ into the right condition gives $2L+38=4t,$ so $t=\frac{L+19}{2}.$

The upper and lower central regions are trapezoids with the same bases $t$ and $87.$ Since their areas are equal, $PQ$ lies halfway up the $19$-cm rectangle. Each central region therefore has area $\frac{19(t+87)}{4}.$ This is one quarter of the rectangle, $\frac{19L}{4},$ so $t+87=L.$ Combining with $t=\frac{L+19}{2}$ yields $L=193.$

Small Hint:

Treat the exponents of $2$ and $5$ independently

Big Hint:

For each prime, translate every least common multiple into a condition on pairwise maxima

Solution:

For the exponent of $5,$ all three pairwise maxima equal $3.$ Thus at least two exponents are $3:$ there is one all-$3$ triple and $3\cdot3$ triples with exactly two $3$’s, for $10$ choices.

For the exponent of $2,$ the first pair has maximum $3$ while the other two have maximum $4.$ The exponent of $c$ must be $4,$ and the exponents of $a$ and $b$ lie in $\{0,1,2,3\}$ with maximum $3,$ giving $4^2-3^2=7$ choices. Independence gives $10\cdot7=70.$

Small Hint:

Solve both inequalities for $k$

Big Hint:

Study the integers in the open interval $(\frac{6n}{7},\frac{7n}{8})$

Solution:

The inequalities are equivalent to $\frac{6n}{7}<k<\frac{7n}{8}.$ This interval has length $\frac{n}{56}.$ At $n=112$ it is $(96,98),$ containing only $97.$ For $n>112$ its length exceeds $2,$ so it contains at least two integers. Thus the largest possible $n$ is $112.$

Small Hint:

The three equal angles around $P$ are each $120^\circ$

Big Hint:

Use vectors from $P$ and translate the right angle at $B$ into a dot product

Solution:

Let $\mathbf a,$ $\mathbf b,$ and $\mathbf c$ be the vectors from $P$ to $A,$ $B,$ $C,$ and put $|\mathbf c|=x.$ Their pairwise angles are $120^\circ,$ so $\mathbf a\cdot\mathbf b=-30,$ $\mathbf b\cdot\mathbf c=-3x,$ and $\mathbf a\cdot\mathbf c=-5x.$ Since $AB\perp BC,$

$(\mathbf a-\mathbf b)\cdot(\mathbf c-\mathbf b)=0.$ Expanding gives $-5x+30+3x+36=0,$ so $x=33.$

Small Hint:

Let Bob’s speed be $b,$ the escalator’s upward speed be $e,$ and the visible count be $N$

Big Hint:

Write $N$ as net speed times travel time for each person

Solution:

Bob’s time is $\frac{75}{b},$ so $N=(b+e)(\frac{75}{b}),$ or $N=75(1+\frac{e}{b}).$ Al’s time is $\frac{150}{3b}=\frac{50}{b},$ so $N=(3b-e)(\frac{50}{b}),$ or $N=150-\frac{50e}{b}.$ Equating gives $\frac{125e}{b}=75,$ hence $\frac{e}{b}=\frac{3}{5}$ and $N=75(1+\frac{3}{5})=120.$

Small Hint:

Write the sum as $\frac{k(2a+k-1)}{2}$

Big Hint:

The only possible lengths divide $2\cdot3^{11}$; then enforce a positive first term

Solution:

If the first term is $a,$ then $2\cdot3^{11}=k(2a+k-1).$ Thus $k$ is $3^j$ or $2\cdot3^j.$ The largest viable even choice is $k=2\cdot3^5=486,$ for which $2a+k-1=3^6=729$ and $a=122>0.$ The next candidates, $3^6$ and $2\cdot3^6,$ force a nonpositive first term, as do all larger choices. Hence $k=486.$

Small Hint:

For fixed $n,$ the smallest possible integer is $m=n^3+1$

Big Hint:

Compare $n^3+1$ with $(n+0.001)^3$

Solution:

For a given $n,$ the closest integer cube-root candidate above $n$ is $m=n^3+1.$ We need $n^3+1<(n+0.001)^3,$ or $1<0.003n^2+0.000003n+10^{-9}.$ This fails at $n=18$ and holds at $n=19.$ The right-hand side is increasing for positive $n,$ so every smaller $n$ fails; every larger $n$ has a larger least candidate $m=n^3+1.$ Hence the smallest $m$ occurs with $n=19.$

Small Hint:

After the comparison reaching position $j,$ that position holds the maximum of the first $j$ original terms

Big Hint:

Characterize the relative ranks of $r_{20}$ and $r_{31}$ among the first $31$ terms

Solution:

For $r_{20}$ to move right to position $30,$ it must exceed every other term among $r_1,\ldots,r_{30}.$ It stops at position $30$ exactly when $r_{31}>r_{20}.$ Thus among the first $31$ terms, $r_{31}$ must be greatest and $r_{20}$ second greatest. These two ordered rank assignments have probability $\frac1{31}\cdot\frac1{30}=\frac1{930}.$ Therefore $p+q=1+930=931.$

Small Hint:

Use Sophie Germain’s identity with $324=4\cdot3^4$

Big Hint:

If $g(x)=x^2+6x+18,$ rewrite $x^4+324$ as $g(x-6)g(x)$

Solution:

Let $g(x)=x^2+6x+18.$ Sophie Germain’s identity gives $x^4+324=g(x-6)g(x).$ The numerator therefore supplies $g(4),g(10),\ldots,g(58),$ while the denominator supplies $g(-2),g(4),\ldots,g(52).$ Everything cancels except $$\begin{aligned}\frac{g(58)}{g(-2)}&=\frac{58^2+6(58)+18}{4-12+18}\\&=\frac{3730}{10}=373.\end{aligned}$$

Small Hint:

Let the legs be $a$ and $b$ and use the first square to relate $ab$ to $a+b$

Big Hint:

For the second square, use the altitude to the hypotenuse and similar cross-sections

Solution:

Put $p=a+b,$ $q=ab,$ and let the hypotenuse be $c.$ Since $S_1$ has side $21,$ the standard leg-aligned-square relation gives $21=\frac{ab}{a+b},$ so $q=21p.$ Hence $c^2=p^2-2q=p(p-42).$

The altitude to the hypotenuse is $h=\frac{q}{c}.$ If the side of $S_2$ is $t,$ similarity gives $t=\frac{ch}{c+h}.$ Substitution simplifies this to $t=\frac{21c}{p-21}.$ Therefore $$440=\frac{441p(p-42)}{(p-21)^2}.$$ Since $p(p-42)=(p-21)^2-441,$ this becomes $$440=441\left(1-\frac{441}{(p-21)^2}\right).$$ Thus $(p-21)^2=441^2,$ and $p>42$ gives $p=462.$