AMC 10 practice test 2

Let $N$ be the marker count, $C$ the number of singles and packs, and $K$ the dollar cost.

Each available container satisfies $3$ times its cost $=2$ times its marker count $+1$⁠, so $$3K=2N+C=46+C$$

Three containers contain at most $21$ markers, so $C\ge 4$⁠.

Integral $K$ forces $C$ congruent to $2$ modulo $3$⁠, hence $C\ge 5$ and $K\ge 17$⁠.

A purchase with three 7-packs and two singles attains $C=5$ and $K=17$⁠.

For every optimum, write $s+f+q=5$ and $s+4f+7q=23$⁠.

Subtraction gives $$f+2q=6$$

Nonnegative solutions are $$\begin{array}{rl}& (s,f,q)\\ & =(0,4,1),(1,2,2),(2,0,3)\end{array}$$

Thus exactly $3$ distinct purchase mixes attain the minimum; permutations of packs do not create new mixes.

A binary repunit is $$R_k=2^k-1$$

Summing the geometric part gives $S_n=2^{n+1}-2-n$⁠.

Modulo $7$⁠, the power has period $3$ and the index term period $7$⁠, giving a joint period $21$⁠.

If $n\equiv 0(\mathrm{mod}3)$⁠, divisibility requires $n\equiv 0(\mathrm{mod}7)$⁠; if $n\equiv 1(\mathrm{mod}3)$⁠, it requires $n\equiv 2(\mathrm{mod}7)$⁠; if $n\equiv 2(\mathrm{mod}3)$⁠, it requires $n\equiv 6(\mathrm{mod}7)$⁠.

The resulting residue classes modulo $21$ are $0,16,20$⁠.

In $1$ through $42$ the qualifying indices are $16,20,21,37,41,42$⁠.

There are $6$⁠.

On either horizontal edge $y=\pm 3$⁠, an intersection satisfies $x^2=r^2-9$ with $|x|\le 6$⁠.

Each horizontal edge therefore contributes no points for $r<3$⁠, one at $r=3$ and two for $3<r\le \sqrt{45}$⁠.

On a vertical edge $x=\pm 6$⁠, use $y^2=r^2-36$ and $|y|\le 3$⁠.

For integer $r=1,\dots ,10$⁠, the distinct intersection counts are respectively $0,0,2,4,4,6,0,0,0,0$⁠.

No integer radius has $r^2=45$⁠, so no corner is counted twice.

Only $r=4$ and $r=5$ qualify.

The answer is $2$⁠.

At $k=3$⁠, rewrite the expression as $(x^2-1)+3(y+2)$⁠.

Both terms are strictly positive on the stated open intervals, so $k=3$ works even though their infimum is zero.

For $k=4$ take $x=\frac{5}{4}$ and $y=-\frac{7}{4}$⁠, which are admissible.

The expression is $\frac{25}{16}-7+5=-\frac{7}{16}$⁠, so $k=4$ fails.

The same pair gives an even smaller value for every larger integer $k$ because $y$ is negative.

Hence the greatest integer is $3$⁠.

Every four-digit palindrome is $1001a+110b$⁠, where $1\le a\le 9$ and $0\le b\le 9$⁠.

Within a fixed a block, consecutive $b$ values have gap $110$⁠; those pairs have smaller $b$ from $0$ through $8$⁠.

Across an a-block boundary the gap is $11$⁠, so those eight pairs are excluded.

Modulo $7$ the smaller palindrome is $5b$⁠, because $1001$ is divisible by $7$⁠.

Hence $b$ must be $0$ or $7$⁠.

Each of nine a values supplies two pairs, giving $18$⁠.

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 $62\times 16=992$⁠.

The calibration gives $f(x)=(\frac{5}{2})x+\frac{41}{2}$ and $g(x)=(\frac{1}{2})x-3$⁠.

Therefore $$f(g(n))=(\frac{5}{4})n+13$$ whereas $$g(f(n))=(\frac{5}{4})n+\frac{29}{4}$$

The first inequality becomes $n\ge -52$⁠; the second becomes $n<-29$⁠.

Intersecting with the stated integer domain leaves $-52$⁠, $-51,\dots$⁠, $-30$⁠, which is $23$ integers.

The endpoint $-29$ fails the strict second inequality.

Although both composite slopes agree, their intercepts differ, so reversing the composition order loses information.

Reindex the layer sum as the sum of $i(b-a+i)$ for $i$ from $1$ through $a$⁠.

It equals $\frac{a(a+1)(3b-a+1)}{6}$⁠.

Since $b$ is at least $a$⁠, the minimum for a fixed $a$ is the sum of the first a squares.

At $a=8$ that is $204$⁠, already above $180$⁠, so only $a=2$ through $7$ need checking.

Solving for $b$ gives respectively $\frac{181}{3}$⁠, $\frac{92}{3},19,\frac{40}{3}$⁠, $\frac{215}{21}$⁠, $\frac{59}{7}$⁠.

Only $b=19$ is integral and within the permitted range.

Hence the unique pair is $(4,19)$⁠, and the count is $1$⁠.

Because $t\le 50$⁠, only $x=0$⁠, plus or minus $1$⁠, or plus or minus $2$ can occur.

Put $g(u)=1$ for $u=0$⁠, $2$ for a positive square, and $0$ otherwise, including negative $u$⁠.

The number of ordered solutions at $t$ is $g(t)+2g(t-1)+2g(t-16)$⁠.

The coefficients $2$ count the two nonzero $x$ signs; $g$ already accounts for the two nonzero $y-2$ signs.

Testing the square lists through $50$ gives count $4$ precisely for $t=1,2,5,10,16,20,26,32,37,41,50$⁠.

There are $11$⁠.

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.

A prime product has exactly one nonunit factor, and that factor must be one of the four primes $2,3,5,7$⁠.

The other four dice must all be $1$⁠.

There are five choices of exceptional position and four choices of prime face, making $20$ equally likely outcomes under the conditioning.

Both last dice are odd unless the exceptional face is $2$ and occurs at one of those two positions.

Exactly two conditioned outcomes fail, so the probability is $\frac{18}{20}=\frac{9}{10}$⁠.

Repeated prime factors cannot occur because their product would be composite.

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.

For a three-digit number its first digit sum is between $1$ and $27$⁠.

Applying $S$ again gives $4$ exactly for first sums $4$⁠, $13$ or $22$⁠, equivalently $N$ congruent to $4$ modulo $9$ in this domain.

Along with divisibility by $7$ this gives $N$ congruent to $49$ modulo $63$⁠.

Check the resulting three-digit progression and reject repeated digits.

The survivors are $175,238,301,364,427,490,679,742,805,931$⁠, so the count is $10$⁠.

Every altitude equals twice the same area divided by its corresponding side length.

The longest altitude is therefore $\frac{168}{13}\text{cm}$ and the shortest is $\frac{168}{15}\text{cm}$⁠.

Their ratio is $$\frac{\frac{168}{13}}{\frac{168}{15}}=\frac{15}{13}$$

The swap increases value by $$\begin{array}{rl}& (19a+7b)-(7a+19b)\\ & =12(a-b)\\ & =72\end{array}$$ so $a=b+6$⁠.

The count cap gives $2b+6\le 45$⁠, hence $b\le 19$⁠.

The value cap gives $26b+42\le 450$⁠, hence $b\le 15$⁠.

With $b$ positive, precisely $b=1$ through $15$ work and $a=b+6$ is positive.

The answer is $15$⁠; the value cap is the tighter one.

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 possible $A$ hands containing $1$ and summing $14$ are$\{1,2,11\},\{1,3,10\},\{1,4,9\},\{1,5,8\},\{1,6,7\}$⁠.

For each, choose a disjoint three-card $B$ hand summing $22$⁠, then a disjoint three-card $C$ hand summing $18$⁠; the remaining $D$ hand must have three cards and sum $24$⁠.

The respective legal completion counts are $11,11,7,7,11$⁠, and successful counts with $12$ in $B$ are $4,4,4,2,3$⁠.

Thus the conditioned sample space has $47$ equiprobable deals, $17$ successful, and probability $\frac{17}{47}$⁠.

Uniformity is over complete deals, not over the five possible $A$ hands.

If opposite face pairs have sums $s,t,u$⁠, expanding stu selects one face from each pair and equals the eight-vertex total.

There are $15$ matchings: choose the partner of $1$⁠, then one of the three pairings of the other four labels.

Grouped by partner $2,3,4,5,6$⁠, the possible products are respectively $(231,240,243),(264,280,288),(275,315,320),(300,324,336),(315,336,343)$⁠.

The unique minimizing opposite pairs are $(1,2),(3,4),(5,6)$⁠, with product $3\cdot 7\cdot 11=231$⁠.

Assign these three pairs to the three opposite-face pairs in $3!$ ways and orient each in $2$ ways, giving $48$ minimizing face assignments.

Proper cube rotations form $24$ orientations.

Distinct face labels make the action free, so each orbit has $24$ members and there are $\frac{48}{24}=2$ classes.

Reflections are not identified.

Let $p=3^a$ be the largest power of $3$ not exceeding $N$⁠, and write $L_N=pM$ with $M$ not divisible by $3$⁠.

Because $N<3p$⁠, the only denominators having maximal 3-adic exponent are $p$⁠, and also $2p$ when $2p\le N$⁠.

Every other term $\frac{L_N}{k}$ is divisible by $3$⁠.

The term at $p$ has positive sign because $p$ is odd; at $2p$ the sign is negative.

Thus $h_N$ is congruent to $M$ if $N<2p$⁠, or $M(1-\frac{1}{2})\equiv -M(\mathrm{mod}3)$ otherwise, using the inverse $2$ modulo $3$⁠.

Both residues are nonzero.

This includes $N=1$⁠, $2$ with $p=1$⁠.

No $h_N$ is divisible by $3$⁠, so the requested count is $0$⁠.

Place the tether midpoint at $(0,0)$⁠, with the shed occupying $-3\le x\le 3$ and $0\le y\le 4$⁠.

Direct access gives an outward half-disk of radius $5$⁠, area $25\times \frac{\pi }{2}$⁠.

Each near corner is $3$ metres along the wall from the tether, leaving $2$ metres of rope.

Wrapping around either gives an outward quarter-disk of radius $2$⁠, area $\pi$ each.

These quarter-disks have $y\ge 0$ and lie respectively to the left of $x=-3$ and right of $x=3$⁠, so their interiors are disjoint from each other and from the direct half-disk.

The next corners require $3+4=7$ metres of rope and cannot be reached.

The total area divided by $\pi$ is $\frac{25}{2}+1+1=\frac{29}{2}$⁠.

After seeing first value $a$⁠, choosing a gives expected payoff $\frac{4+5\cdot 1}{6}=\frac{3}{2}$⁠.

Choosing any different value gives $\frac{1+5\cdot (-2)}{6}=-\frac{3}{2}$⁠.

Thus choosing the observed value is optimal for every first roll, with adaptive expectation $\frac{3}{2}$⁠.

A preset number matches both dice in $1$ of $36$ outcomes, exactly one in $10$ and neither in $25$⁠.

Its expectation is $\frac{4+10-50}{36}=-1$⁠, independent of the preset number.

The requested difference is $\frac{3}{2}-(-1)=\frac{5}{2}$⁠.

Randomising a choice cannot exceed the largest of its possible expected payoffs.

Every array with arithmetic rows and columns has form $A_{i,j}=2+4(i-1)+3(j-1)+d(i-1)(j-1)$⁠: the first column and row determine the linear parts, and the row differences themselves form an arithmetic progression.

The entry at $(2,2)$ shows $d$ is integral.

The last-row step is $3+3d>0$⁠, giving $d\ge 0$⁠.

The final corner is $23+9d\le 100$⁠, giving $d\le 8$⁠.

All nine values $0$ through $8$ give increasing bounded rows/columns.

Substitution reveals collisions for $d=1$ (values $10,14,20$ repeat), $d=2$ ($11$ repeats), and $d=5$ ($14$ repeats).

For $d=0,3,4,6,7,8$ all sixteen values are distinct.

This complete finite check gives $6$ arrays.

Integrality of the mixed coefficient is stronger than merely requiring an integer final corner.

Two-digit strict decrease implies $a\ge b\ge c$⁠: if $a<b,ab<bc$⁠, and if $b<c,bc<cd$⁠.

Distinctness therefore forces $a>b>c$⁠.

The first difference is $10(a-b)+(b-c)\ge 11$⁠.

Equality requires $$a-b=b-c=1$$ and the progression then forces $c-d=1$⁠.

The minimum strings are $3210,4321,5432,6543,7654,8765,9876$⁠: seven.

All have distinct digits and valid decreasing two-digit windows.

Thus the lower bound is attained and every minimum case is classified.

The horizontal distance from the square centre to each support centre is $\sqrt{8}$⁠.

Tangency makes the centre distance $R+2$⁠, so the upper centre is at height $2+\sqrt{(R+2{)}^2-8}$⁠.

The top height is $H=R+2+\sqrt{(R+2{)}^2-8}$⁠.

It is integral exactly when the square root is an integer $t$⁠.

Then $$(R+2-t)(R+2+t)=8$$

The positive factors must have the same parity, leaving only $2$ and $4$⁠.

Hence $$R+2=3,R=1$$ and $H=4$⁠.

This sphere is above the support centre plane and ground, so it is valid.

There is $1$ radius.

Put $$u_n=\frac{x_n-1}{x_n+1}$$

Since all $x_n$ are positive, every denominator in both recurrences is nonzero.

Substitution gives $$u_{n+1}=\frac{u_n}{3}$$ with $$u_0=\frac{1}{2}$$

Therefore $$x_n=\frac{2\times 3^n+1}{2\times 3^n-1}$$

The numerator and denominator are odd and differ by $2$⁠, so their gcd is $1$⁠.

The reduced denominator is divisible by $5$ precisely when $3^n\equiv 3(\mathrm{mod}5)$⁠.

Powers of $3$ have period $4$⁠, giving $n\equiv 1(\mathrm{mod}4)$⁠.

Between $1$ and $8$⁠, $n=1$ and $5$ qualify, so the count is $2$⁠.