AMC 10 practice test 1

Expanding the two specified products gives $E(n)=9n+28$⁠.

If $E(n)=k^2$⁠, the given $n$ bounds imply $7\le k\le 42$ and $k^2$ is $1$ modulo $9$⁠.

The complete roots of $1$ modulo $9$ are $1$ and $8$⁠, so the possible positive $k$ in this range are $8,10,17,19,26,28,35,37$⁠.

Divisibility of $k^2$ by $25$ but not $625$ means $k$ is divisible by $5$ but not $25$⁠.

This leaves $k=10$ and $35$⁠.

Recovering $n=\frac{k^2-28}{9}$ gives $n=8$ and $133$⁠, both within the domain.

The answer is $2$⁠.

Bounds on the square root and the complete residue classes exclude every other $n$⁠.

Write the seven even integers as $2a,2(a+1)$⁠, through $2(a+6)$⁠, with $aa$ positive integer.

Their sum is $14a+42$⁠, while the first $m$ odd integers sum to $m^2$⁠.

Hence $$m^2=14a+28$$

Divisibility by the square-free integer $14$ forces $m$ to be a multiple of $14$⁠.

Within $1$ through $60$ these are $14,28,42$ and $56$⁠.

For each, $a=\frac{m^2-28}{14}$ is positive and integral.

Thus exactly $4$ indices work.

Rearrange the inequality as $(k-2)n\le k$⁠.

If $k$ is one or two, its left side is nonpositive for every positive $n$⁠, so the condition holds.

If $k\ge 3$⁠, choose the permitted positive integer $n=k+1$⁠.

Then $(k-2)n\ge k+1>k$⁠, so the inequality fails.

The universal condition therefore holds exactly for $k=1$ and $k=2$⁠.

If $W$ and $L$ are the team scores, $W+L=67$ and $W-L=17$ give $W=42$ and $L=25$⁠.

For the winner, $2w_2+5w_5=42$ forces $w_5$ to be even.

Its possible values are $0,2,4,6,8$⁠, each giving one nonnegative $w_2$⁠.

For the loser, $2l_2+5l_5=25$ forces $l_5$ to be odd, with values $1,3,5$⁠.

The choices for the two teams are independent, so there are $5\times 3=15$ ordered quadruples.

Volume is $30t^3$ and surface area is $2(6+10+15)t^2=62t^2$⁠.

The scales $1,2,3,4$ are feasible, while $t=5$ already gives volume $3750>2000$ and every later scale is larger.

Surface area increases with positive $t$⁠, so $t=4$ gives the maximum $62\times 16=992$⁠.

Modulo $7$ the ordered pair transition is $(a,b)\to (b,a+b)$⁠.

It is invertible because the preceding pair is $(b-a,a)$⁠, so the first repeated pair closes the cycle from the initial pair.

Starting from $(2,5)$⁠, the successive first coordinates in one full cycle are $2,5,0,5,5,3,1,4,5,2,0,2,2,4,6,3$⁠, after which the pair $(2,5)$ returns.

Zero occurs at indices $2$ and $10$ modulo $16$⁠.

Equivalently the qualifying indices in the given interval are $2,10,\dots ,98$⁠, so there are $13$⁠.

Substitute the two given points:$$b+c=6$$$$5b+c=-18.$$

Subtracting gives $b=-6$ and $c=12$⁠, so $y=(x-3{)}^2+3$⁠.

Two distinct integral intersections with a horizontal line have $x=3-d$ and $3+d$ for an integer $d\ge 1$⁠.

The coordinate interval gives $d\le 5$⁠.

The corresponding heights are $4,7,12,19,28$⁠, all within the allowed $k$ interval.

These five values are distinct.

No other integral intersection pair is possible by the completed-square formula, so the answer is $5$⁠.

Represent an unlabelled partition by the unique team containing $J$ $1$⁠.

If $J$ $2$ is also on that team, choose its third junior in four ways.

$S$ $1$ and $S$ $2$ must both be outside, so that team's seniors must be $S$ $3$⁠, $S$ $4$⁠: four partitions.

If $J$ $2$ is outside $J$ $1$'$s$ team, choose the other two juniors from $J$ $3,\dots ,J$ $6$ in $C(4,2)=6$ ways.

$S$ $1$ must be outside and $S$ $2$ inside, so choose the second senior as $S$ $3$ or $S$ $4$⁠: twelve partitions.

The two cases are disjoint and exhaustive, giving $4+12=16$⁠.

There is no further division by two, because the $J$ 1-team convention has already removed team labels.

Every positive divisor of $72=2^3\times 3^2$ has the form $2^i\times 3^j$ with $0\le i\le 3$ and $0\le j\le 2$⁠, giving $12$ divisors.

There are $66$ unordered distinct pairs.

For divisibility in the smaller-to-larger direction, choose exponent intervals $i\le i$' and $j\le j$'.

There are $10$ choices of the first interval and $6$ of the second, giving $60$ pairs including $12$ equal-divisor cases.

Removing equality leaves $48$ comparable distinct pairs.

The required probability is $\frac{48}{66}=\frac{8}{11}$⁠.

Let the long and short sides have $a$⁠, $b$ spacing intervals.

The closed perimeter has $24$ gaps, giving $2a+2b=24$ and $a+b=12$⁠.

Side post counts including corners are $a+1$ and $b+1$⁠; hence $a+1\ge 2(b+1)$⁠, or $a\ge 2b+1$⁠.

Substitution gives $b\le 3$⁠.

The areas for $b=1,2,3$ are $4b(12-b)=44,80,108$ square metres.

The maximum is $108$⁠, attained with $a=9,b=3$⁠, and every positive case has been considered.

Since all variables are nonzero, $$x^2=\frac{(xy)(xz)}{yz}=36$$

In the same way $y^2=4$ and $z^2=9$⁠.

Expanding the requested invariant yields $$\begin{array}{rl}(x+y+z{)}^2& =x^2+y^2+z^2+2(xy+xz+yz)\\ & =36+4+9+2(12+18+6)\\ & =121\end{array}$$

Both $(6,2,3)$ and $(-6,-2,-3)$ satisfy the given products; their sums differ in sign, but the requested square is the same.

Thus no positivity assumption is needed.

Column $1$ has four cells and must contain all four letters, with its first $A$ already fixed.

Each later row must contain $B$⁠, because three $B$ tokens must occupy three distinct rows and row $1$ is $A$⁠.

Now consider the six permutations of $B,C,D$ in the remaining column-1 cells, and fill each remaining row with distinct letters while observing the column bans and remaining inventory.

Writing the three later rows as words, the full possibilities are $\frac{\frac{BA}{CBA}}{DCB}$⁠, $\frac{\frac{BA}{DCB}}{CBA}$⁠, $\frac{\frac{BC}{CAB}}{DBA}$⁠, $\frac{\frac{BC}{CBA}}{DAB}$⁠, $\frac{\frac{BC}{DAB}}{CBA}$⁠, $\frac{\frac{BC}{DBA}}{CAB}$⁠, $\frac{\frac{CB}{BCA}}{DAB}$⁠, $\frac{\frac{CB}{DAB}}{BCA}$⁠, $\frac{\frac{DB}{BCA}}{CAB}$⁠, $\frac{\frac{DB}{CAB}}{BCA}$⁠.

Each has the required inventory and no repeated row or column letter.

This six-column-order classification yields $10$ fillings.

Letters are indistinguishable, so no extra token factorial appears.

The unordered pairs of distinct single-digit primes are $\{2,3\},\{2,5\},\{2,7\},\{3,5\},\{3,7\},\{5,7\}$⁠.

Test both palindromes for each pair.

The first two pairs have neither palindrome prime.

The other four have exactly one: $\{2,7\}$ gives $272$ and $727$⁠; $\{3,5\}$ gives $353$ and $535$⁠; $\{3,7\}$ gives $373$ and $737$⁠; $\{5,7\}$ gives $575$ and $757$⁠.

Exact trial division through each square root establishes the indicated primality statuses; even endings, a final $5$⁠, or a divisor certificate reject the others.

There are $4$ admissible unordered pairs.

Perpendicular directions $(-s,p)$ and $(q-s,-s)$ give $p+q=s$⁠.

Parametrise $$R=B+t(P-B)$$

Solving its intersection with $CQ$ gives $t=\frac{sp}{s^2+p^2}$⁠.

The ratio $BR$⁠: $$PR=12:13$$ means $$t=\frac{12}{25}$$

For $u=\frac{p}{s}$ this gives $12u^2-25u+12=0$⁠, with roots $\frac{3}{4}$ and $\frac{4}{3}$⁠.

The latter is outside the side, so $p=\frac{3s}{4}$ and $q=\frac{s}{4}$⁠.

Then $$BP=\frac{5s}{4}$$ and $$BR=\frac{3s}{5}$$

Integral side coordinates and these lengths require both $4$ and $5$ to divide $s$⁠.

Thus $s$ is a multiple of $20$⁠; the given range permits $20,40,60$⁠, three values.

Each gives an intersection inside both segments and the required ratio.

$$S(n)=\frac{d{n}^2}{2}+(a-\frac{d}{2})n$$

Expanding $S(2n+1)-4S(n)$ gives $(3d-2a)n+a$⁠.

It is a constant $K$ precisely when $3d=2a$⁠, and then $K=a$⁠.

A positive square at most $100$ divisible by $3$ must be $9$⁠, $36$ or $81$⁠.

These give $(a,d)=(9,6),(36,24),(81,54)$⁠, each within the stated bounds.

Therefore $3$ pairs work.

A few matching numerical indices would not by themselves prove the required identity.

For each side length $k$⁠, count positioned squares containing each mark and subtract twice the count containing both.

This gives the exclusive-or count.

The counts for $k=1,2,3,4,5,6$ are $2,8,13,9,2,0$⁠.

For example at $k=4$ there are $4$ containing the first mark, $9$ containing the second and $2$ containing both, giving $4+9-4=9$⁠.

The total across all side lengths is $34$⁠.

Positions use the lower-left cell indices within the actual $7$ by $6$ grid.

The three exchanges change the total value by $15,60$ and $-12$⁠, all multiples of three.

The total therefore remains congruent to one modulo three.

This excludes $102$ and $104$⁠.

To reach $100$⁠, exchange the initial $X$ for four $Y$⁠; exchange two of those $Y$ for eight $Z$⁠; then exchange three of those $Z$ for twelve $X$⁠.

The final inventory is twelve $X$⁠, two $Y$ and five $Z$⁠, of total value $12+8+80=100$⁠.

Only $100$ is reachable among the listed values.

The two end layers contain eight corner dice, each exposing three mutually adjacent faces.

The three smallest labels $1,2$ and $3$ lie on different opposite-face pairs, so they meet at a corner and can all face outward; their sum six is the smallest possible.

The four dice in the middle layer expose two adjacent faces each, whose smallest possible sum is $1+2=3$⁠.

These position types exhaust the twelve dice.

Independent rotations attain every local minimum simultaneously, giving total $8\times 6+4\times 3=60$⁠.

For $A$ to make the final safe output, its input $u$ must satisfy $760\le 3u<800$⁠, since the following addition of $40$ must fail.

Thus $254\le u\le 266$⁠.

One full round sends an input $x$ for $A$ to $3x+40$⁠.

Before $A$'$s$ turn numbered $k+1$⁠, that input is $3^k(x+20)-20$⁠, as follows by adding $20$ to the full-round recurrence.

Therefore a winning turn requires $274\le 3^k(x+20)\le 286$⁠.

For $k=0$⁠, the smallest $x$ is $254$⁠.

For $k=1$⁠, the integer starts are $72$ through $75$⁠.

For $k=2$⁠, only $x=11$ works.

For $k\ge 3$⁠, even $x=0$ gives $3^k\times 20\ge 540$⁠, above the permitted range.

Hence $11$ is the smallest start.

Its outputs are $33,73,219,259,777$ and $817$⁠; the final safe one, $777$⁠, belongs to $A$⁠.

Fix $A$ to remove rotations.

Its unordered neighbour pair is chosen from $D,E,F,G$⁠.

If the pair is {D, E}, both orders allow all $24$ interior permutations, giving $48$⁠.

If it is {F, G}, D and $E$ are both interior: of $24$ permutations, $2\times 3!=12$ put them adjacent, so the two neighbour orders give $24$ valid arrangements.

For each of the four pairs containing one of $D,E$ and one of $F,G$⁠, the other member of $D$⁠, $E$ is interior.

Exactly $3!=6$ of its $24$ permutations put it next to its forbidden neighbour; both neighbour orders therefore give $36$⁠.

The total oriented count is $48+24+4\times 36=216$⁠.

Reflection fixes no anchored arrangement with seven distinct labels, since it would equate paired seat labels.

Thus all reflection orbits have size $2$ and there are $108$ seating classes.

Across the four readings, each chosen digit occurs once as a hundreds digit, once as a tens digit and once as a units digit.

Thus the sum is $111$ times the sum $S$ of the chosen digits.

Divisibility by $999=111\times 9$ is equivalent to $S$ being divisible by $9$⁠.

Four distinct digits from $1$ through $8$ have sum from $10$ through $26$⁠, so the only possible multiple is $18$⁠.

The eight qualifying digit sets are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},\{1,4,6,7\},\{2,3,5,8\},\{2,3,6,7\},\{2,4,5,7\},\{3,4,5,6\}$⁠.

Each has $4!/4=6$ directed cyclic orders because all digits are distinct, and no reflection identification is requested.

Therefore the total is $8\times 6=48$⁠.

Divide vertical coordinates by $\sqrt{3}$⁠, leaving hexagon area $6$ and band half-height $k$⁠.

Its six vertices are $(-2+k,-k),(2-k,-k),(2,0),(2-k,k),(-2+k,k),(-2,0)$⁠.

Decomposition or shoelace gives band area $8k-2k^2$⁠.

If this is twice the remaining caps, it must equal $4$⁠.

Thus $$k^2-4k+2=0$$ and $$k=2\pm \sqrt{2}$$

Only $2-\sqrt{2}$ lies between $0$ and $1$⁠.

Squaring gives $$k^2=6-4\sqrt{2}$$ so $$u+v=6+4=10$$

Substituting $k=2-\sqrt{2}$ into $8k-2k^2$ gives band area $4$ and combined caps area $2$ in the scaled coordinates.

The threshold condition at $j=2$ forces at least one $1$ among the two remaining digits; the $j=1$ and $j=3$ thresholds already follow from the stocks.

Thus the remaining stock is either one $1$ and one $2$⁠, or two $1$s.

Nonadjacent cyclic zero positions have $9$ choices among the $15$ position pairs.

In the first stock class, the other four digits have $12$ arrangements, giving $108$ labelled words, all with rotation orbit size $6$ and thus $18$ classes.

In the second, there are $6$ arrangements of the remaining two $1$s and two $3$s, giving $54$ labelled words.

Six of these have period $3$⁠: they form two orbits represented by $013013$ and $031031$⁠.

The other $48$ form eight size $6$ orbits.

Total classes are $$18+2+8=28$$

Dividing all $162$ labelled words by $6$ would miss the smaller orbits.

Number cart travel intervals by $j=0,1,2,\dots$ .

Each lasts $24$ seconds and begins at time $72j$⁠, since a travel-plus-pause cycle lasts $72$ seconds.

At departure the cart is $120(j+1)$ metres from the robot’s starting marker, while the robot is at $144j$⁠, giving signed gap $120-24j$⁠.

At arrival the gap is $192-24j$⁠, and at the end of the pause it is $96-24j$⁠.

During travel the gap increases linearly, so a zero occurs exactly for $j=5,6,7,8$⁠, at times $360,440,520,600$⁠.

During pauses it decreases linearly, so a zero occurs exactly for $j=4,5,6,7,8$⁠, at times $360,420,480,540,600$⁠.

The shared boundary times $360$ and $600$ count once each.

The union has seven times: $360,420,440,480,520,540,600$ seconds.

Let $h$ be the positive integer height and $x$⁠, $y$ the positive integer horizontal components of the sides $10$ and $17$⁠.

Then $$h^2+x^2=100$$ and $$h^2+y^2=289$$

Because $x$ is positive, $h\le 9$⁠.

The positive integer possibilities for the 10-side are $(h,x)=(6,8)$ or $(8,6)$⁠.

Only $h=8$ also gives an integer 17-side component, $y=15$⁠.

The shorter base lies between the endpoint perpendiculars, so the base difference is $x+y=21$⁠.

The base sum from perimeter is $68-10-17=41$⁠.

The bases are therefore $10$ and $31$⁠, both positive integers.

The area is $$(\frac{41}{2})\times 8=164$$