ACT Math challenge

The insight is to eliminate small drinks to expose a small integer equation, then push to the extreme. Let $s$, $m$, $l$ be the numbers of small, medium, and large drinks. Then $s+m+l=20$ and $2s+3m+5l=65$. Subtracting twice the first from the second gives $m+3l=25$. Then $s=20-m-l=2l-5$, so maximizing $s$ means maximizing $l$. With $m=25-3l$ nonnegative, $l$ is at most 8, giving $m=1$ and $s=11$, which indeed uses $11+1+8=20$ drinks and $22+3+40=65$ dollars. Enumerating all triples would take far too long.

The insight is that the logs must be integers, turning Pythagoras into a triple search. Write $a={\log }_{2}m=r$ and $b={\log }_{2}n=s$ with integers $a,b\ge 0$ and $a^2+b^2=25$. Positive legs for a genuine triangle force $\{a,b\}=\{3,4\}$, since $(0,5)$ gives a zero leg and degenerate area. Hence $\{m,n\}=\{8,16\}$ and $mn=2^{3+4}=2^7=128$.

The insight is to factor instead of searching triples. Let legs be $11,b$ and hypotenuse $c$ with $11^2+b^2=c^2$, so $121=c^2-b^2=(c-b)(c+b)$. Both factors are positive integers with the same parity and product $121=1\times 121$ or $11\times 11$. The second gives $b=0$, degenerate and not a triangle, so $c-b=1$ and $c+b=121$. Adding gives $2c=122$, so $c=61$ and $b=60$.

Rationalize every term. Multiply top and bottom by the conjugate: $1\mathrm{/}(\sqrt{k+2}+\sqrt{k})=(\sqrt{k+2}-\sqrt{k})\mathrm{/}((k+2)-k)=(\sqrt{k+2}-\sqrt{k})\mathrm{/}2$. With $k=2,4,\dots ,48$ the sum becomes $((\sqrt{4}-\sqrt{2})+(\sqrt{6}-\sqrt{4})+\cdots +(\sqrt{50}-\sqrt{48}))\mathrm{/}2$, which telescopes to $(\sqrt{50}-\sqrt{2})\mathrm{/}2$. Since $\sqrt{50}=5\sqrt{2}$, this is $(5\sqrt{2}-\sqrt{2})\mathrm{/}2=2\sqrt{2}$. Adding twenty-four decimal reciprocals directly would exceed two minutes.

The shortcut is the parallelogram law obtained by adding the squared-distance expansions so the cross terms cancel, avoiding four unknown coordinates. Writing the numbers as a plus b i and c plus d i, the squared magnitude of the sum is a plus c squared plus b plus d squared and that of the difference is a minus c squared plus b minus d squared, whose sum is twice a squared plus b squared plus c squared plus d squared. Hence the sum of the two squared magnitudes equals twice the sum of the individual squared magnitudes. With magnitudes 3, 4, and 5, this gives 25 plus the unknown square equal to 50, so the unknown square is 25 and the magnitude is 5. Solving the coordinate system directly is underdetermined and long, but the invariant is short.

The insight is to use Vieta relations to turn unknown intercepts into factor pairs of 36 and then use the axis inequality as a bound. Let the intercepts be integers $r$ and $s$ with $r\mathrm{\ne }s$. Then $rs=36$ and $r+s=-b$, and the axis is $h=(r+s)\mathrm{/}2$. The bound $5<h<7$ gives $10<r+s<14$. Positive factor pairs of 36 give sums 37, 20, 15, 13, and 12 for $(1,36)$, $(2,18)$, $(3,12)$, $(4,9)$, and $(6,6)$. Only 13 lies strictly between 10 and 14 once the double root $(6,6)$ is excluded by distinctness, so $r+s=13$ from $(4,9)$ and $b=-(r+s)=-13$. The arithmetic is short once the enumeration is framed this way.

The insight is central symmetry plus two-point form. A $180^{\circ }$ rotation about the rectangle center $(5,3)$ swaps the two halves, so every equal-area line must pass through that center; otherwise one half would rotate strictly inside the other. Hence the desired line joins $(3,1)$ to $(5,3)$. The slope is $(3-1)\mathrm{/}(5-3)=1$, and with point $(3,1)$ the equation is $y-1=1(x-3)$, namely $y=x-2$. Checking areas polygon by polygon for each candidate would be the long route.

The insight is that five sevenths has a short repetend whose digit sum repeats, so counting complete cycles plus a remainder replaces one hundred divisions. Long division gives $5\mathrm{/}7=\mathrm{0.714285714285...}$ with six-digit block 714285 summing to 27. Since $100=16\times 6+4$, there are sixteen full blocks contributing 432 plus the next four digits $7+1+4+2=14$, for 446 total. A calculator cannot display one hundred places, but the six-place cycle makes the total immediate.

The insight is to use cosine squared equals one minus sine squared to get a quadratic in one function, then use boundedness to discard. Let $s$ denote sine of $x$, so cosine squared is $1-s^2$. Substituting gives $3(1-s^2)+5s-1=0$, or $3s^2-5s-2=0$, which factors as $(3s+1)(s-2)=0$. So sine is $-1\mathrm{/}3$ or 2. Since sine has absolute value at most one for real $x$, the value 2 is impossible. The remaining value $-1\mathrm{/}3$ lies strictly inside $(-1,1)$, so it occurs twice in the given interval of length two pi, once with positive cosine and once with negative cosine. Hence there are 2 solutions. Without the identity the equation mixes two functions and resists factoring.

The key is to see the quartic as a square in disguise. Set $y=x^2-7x$. Then $(x^2-7x{)}^2=x^4-14x^3+49x^2$, and subtracting from the radicand leaves $22x^2-154x+121=22y+121$. So the radicand is $y^2+22y+121=(y+11{)}^2=(x^2-7x+11{)}^2$. Hence the square root is $\mathrm{\mid }{x}^2-7x+11\mathrm{\mid }$. On $3<x<4$ that quadratic is negative, since its vertex at 3.5 gives $-1.25$ and its endpoint values are $-1$, so the absolute value flips the sign to $-x^2+7x-11$.

The insight is to pair each red ring with the white disk inside it through differences of squares. Red area equals $\pi$ times $1^2+(3^2-2^2)+(5^2-4^2)+\cdots +(99^2-98^2)$. Each difference $n^2-(n-1{)}^2=2n-1$ gives the odd numbers $1,5,9,\dots ,197$, an arithmetic progression with 50 terms. Its sum is $50(1+197)\mathrm{/}2=4950$, so the red area is $4950\pi$. Adding 50 separate ring areas directly is long, while the pairing makes it one series sum.

Expanding gives $x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so $a+b+c=15$ and $abc=90$ by Vieta matching from multiplying binomials. The insight is arithmetic-progression symmetry plus coefficient matching: since $a$, $b$, $c$ are in order, $b-a=c-b$, so $a+c=2b$ and $3b=15$, giving $b=5$. Then $a+c=10$ and $5ac=90$, so $ac=18$. Thus $a$ and $c$ are the two roots of $t^2-10t+18=0$, namely $5-\sqrt{7}$ and $5+\sqrt{7}$. The largest is $5+\sqrt{7}$. Trying integer triples with product 90 and sum 15 fails, which forces the irrational pair and blocks calculator guessing.

The insight is the tangent-radius right triangle plus point-slope to standard form. The distance from $(5,0)$ to the center is 5 and the radius is 4, perpendicular to the tangent, so the tangent segment is $\sqrt{25-16}=3$. Thus opposite 4 over adjacent 3 gives slope $4\mathrm{/}3$, and through $(5,0)$ the line is $y={\displaystyle \frac{4}{3}}(x-5)$, namely $4x-3y=20$. Solving a quadratic discriminant for each candidate line against the circle would be the long route.

Group as $A=x^4+x^2+1$ and $B=x^3+x$, so the product is $(A+B)(A-B)=A^2-B^2$ by difference of squares. The insight is difference-of-powers recognition plus rational cancellation: also $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$ and $(x+1)(x^4-x^3+x^2-x+1)=x^5+1$, so for $x^2\mathrm{\ne }1$ the product equals $(x^{10}-1)\mathrm{/}(x^2-1)=x^8+x^6+x^4+x^2+1$ by division, and the polynomial identity extends to all $x$. Expanding 25 signed terms directly is long and sign-prone, while the geometric-series view gives five even-degree terms at once.

The insight is periodicity: powers of a quarter turn repeat every four, so huge exponents collapse to remainders mod 4. Direct multiplication gives $R^2=-I$ with rows $(-1,0)$ and $(0,-1)$, $R^3=-R$ with rows $(0,1)$ and $(-1,0)$, and $R^4=I$. Since $2026=4\cdot 506+2$ and $2027=4\cdot 506+3$, we get $R^{2026}=R^2$ and $R^{2027}=R^3$. Thus the expression is $[[-1,0],[0,-1]]+[[0,2],[-2,0]]=[[-1,2],[-2,-1]]$, whose entries total $-2$. Multiplying 2026 times by hand or calculator is infeasible.

The insight is that factorials beyond a point are all multiples of 4, so all but three terms of the sum are 1 and the rest is short. Powers of i repeat every four as i, -1, -i, 1, so only the exponent mod 4 matters. Here 1 factorial is 1 giving i, 2 factorial is 2 giving -1, and 3 factorial is 6 giving -1 since 6 leaves remainder 2. For k at least 4, k factorial contains 4 as a factor, as seen in 24, so every later factorial is 0 mod 4 and contributes 1. There are 37 such ones from 4 factorial through 40 factorial. Adding i minus 1 minus 1 plus 37 gives 35 plus i, which is immediate once the stable tail is seen.

The insight is to identify airspeed as the hypotenuse of a right triangle and convert ground speed before applying Pythagoras. Since 22 ft/s equals 15 mi/h, 30 mi/h equals 44 ft/s. The ground velocity east and the wind north are perpendicular, so the air velocity must supply both, making airspeed the longest side. Hence airspeed is the square root of $44^2$ plus $33^2$, which is the square root of 3025, equal to 55 ft/s. Treating ground speed as the hypotenuse would subtract instead of add, and skipping conversion would mix 30 with 33.

The insight is that the difference quotient factors so only the sum of the times matters, which Vieta gives without solving. Since $s(b)-s(a)=(b^2-a^2)+3(b-a)=(b-a)(a+b+3)$, dividing by $b-a$ leaves $a+b+3$. From $t^2-10t+19=0$ the sum is 10, so the quotient is 13, which is change in position over change in time, hence average velocity in meters per second. Solving for $a=5\pm \sqrt{6}$ and squaring decimals is long, while factoring plus the sum finishes without a calculator.

The insight is to form right triangles from the center to the tangency points. Let $O$ be the origin and $P=(6,0)$. Each radius to a tangency point is perpendicular to its tangent, so each triangle $OPT$ is right with hypotenuse $OP=6$ and leg $OT=3$. Thus the sine of the half angle at $P$ is $3\mathrm{/}6=1\mathrm{/}2$, so that half angle is 30 degrees and the full angle between tangents is 60 degrees. A quadrilateral with two right angles is supplementary, so the minor central angle is $180-60=120$ degrees. Its arc is $(120\mathrm{/}360)2\pi (3)=2\pi$.

Reduce modulo 5 where 3 behaves like $-2$. Since $3\equiv -2(\mathrm{m}\mathrm{o}\mathrm{d}5)$, we have $2^n+3^n\equiv 2^n+(-2{)}^n=2^n(1+(-1{)}^n)$ modulo 5. The factor $2^n$ is never divisible by 5, so the sum is divisible by 5 exactly when $1+(-1{)}^n$ is, namely when $n$ is odd. In the inclusive interval from 0 to 100 there are 101 integers, and the odds $1,3,\dots ,99$ are exactly 50 values. Trying values directly would require one hundred modular exponentiations, while the congruence collapses the whole range at once.

The insight is that the total fixes the added sum while the range forces which original endpoint can survive. The original sum is $60+70+80+90=300$ and six numbers averaging 78 total 468, so the two added numbers sum to 168. Write them $a\le b$; the new minimum is $\min (60,a)$ and the new maximum $\max (90,b)$ with difference 40. If $a\ge 60$ then the minimum stays 60, so $b=100$ and $a=68$, which fits. If the minimum drops to 50 then $a=50$ and $b=118$, but then the maximum is 118 and the range is 68, impossible. If both extremes are new then $b-a=40$ with $a+b=168$ gives 104 and 64, but 64 is not below 60, contradicting both-new. Hence only 68 and 100 work, with greater 100.

The insight is to count ordered pairs by their maximum and then take a weighted average, which collapses 36 cases to 6. For $M=k$, one pair is $(k,k)$ with $k-1$ pairs having $k$ first and a smaller second and $k-1$ having $k$ second and a smaller first, so $2k-1$ ordered outcomes give $M=k$. Hence the total of $M$ is $1\times 1+2\times 3+3\times 5+4\times 7+5\times 9+6\times 11=161$ over 36 outcomes, so the mean is ${\displaystyle \frac{161}{36}}$, about 4.47. Enumerating all 36 pairs by calculator is long, while the $2k-1$ reframe finishes in six terms without a calculator.

The remainder theorem turns remainders into values: $P(0)=4$, $P(3)=10$, and the unknown remainder is $R=P(8)$. The non-obvious step uses integer coefficients: for any integers $u,v$, $u-v$ divides $P(u)-P(v)$, since each $u^k-v^k$ carries the factor $u-v$. Hence 8 divides $R-4$ and 5 divides $R-10$, so $R$ is 4 modulo 8 and 0 modulo 5. Checking the numbers, 15 fails the first divisibility, 28 fails the second, and 16 fails both, while 20 gives 16 divisible by 8 and 10 divisible by 5. The polynomial $2x+4$ shows 20 occurs. A student who tries to recover $P$ faces infinitely many possibilities.

The idea is to reduce each huge exponent modulo 4 and then group four consecutive terms that cancel to zero. Since i to a power depends only on the exponent mod 4, compute k times k plus 1 mod 4 from k mod 4, giving residues 2, 2, 0, 0 for k congruent to 1, 2, 3, 0, hence values -1, -1, 1, 1 whose sum is 0. Thus each block of four consecutive k contributes 0. With 99 terms there are 24 full blocks covering the first 96 terms, leaving k equal to 97, 98, 99 with values -1, -1, 1 totaling -1. Direct evaluation of powers like i to the 9900th is hopeless on a calculator, but residues finish it quickly.

The insight is to divide differences so the unknown angle cancels. Let the radii be $R>r$ with common angle $\theta$. Then $(R-r)\theta =4\pi$ for arc length and ${\displaystyle \frac{1}{2}}(R^2-r^2)\theta =48\pi$ for area. Factoring gives ${\displaystyle \frac{1}{2}}(R-r)(R+r)\theta =48\pi$. Dividing the area equation by the arc equation leaves ${\displaystyle \frac{1}{2}}(R+r)=48\pi \mathrm{/}4\pi =12$, so $R+r=24$. Attempting to find $\theta$, $R$, and $r$ separately stalls with three unknowns and two equations, and omitting the half yields half the sum.

The idea is to square to reveal $\sin 2\theta$ while retaining the size and sign tests that expose the phantom solutions. Squaring gives $1+\sin 2\theta =49\mathrm{/}25$, so $\sin 2\theta =24\mathrm{/}25$. With $0\le \theta <2\pi$, $2\theta$ ranges over $[0,4\pi )$, where $\sin 2\theta =24\mathrm{/}25$ has four preimages, but two make $\sin \theta +\cos \theta =-7\mathrm{/}5$ and must be discarded since $7\mathrm{/}5>0$. Also $(7\mathrm{/}5{)}^2=49\mathrm{/}25<2$ confirms feasibility, leaving exactly two angles near $37^{\circ }$ and $53^{\circ }$.

The insight is to convert the current to common units and solve the two-equation downstream-upstream system by eliminating distance. Since 22 ft/s equals 15 mi/h, 11 ft/s equals 7.5 mi/h. Let still-water speed be $s$ in mi/h and one-way distance be $D$ in miles. Downstream gives $D$ equals 2 times $s$ plus 7.5, and upstream gives $D$ equals 4 times $s$ minus 7.5. Equating yields $2s$ plus 15 equals $4s$ minus 30, so $s$ equals 22.5 and $D$ equals 60. The conversion is essential because hours and miles must match the current.

The insight is to combine the arithmetic-series sum with the polygon interior-sum formula and then use convexity as a bound to reject an extraneous root. For $n$ sides the series sums to $n\mathrm{/}2\cdot (2\cdot 120+5(n-1))$, while any $n$-gon sums to $(n-2)\cdot 180$. Equating gives $n(235+5n)\mathrm{/}2=180n-360$, which simplifies to $n^2-25n+144=0$ and factors as $(n-9)(n-16)=0$. The root $n=16$ would make the largest angle $120+15\cdot 5=195$, at least 180, impossible for a convex polygon, so it is rejected. The root $n=9$ gives largest angle $120+8\cdot 5=160$, strictly below 180, so the polygon has 9 sides.

The insight is detecting the hidden right angle by perpendicular slopes, after which the hypotenuse midpoint is the circumcenter. Slopes from $(4,3)$ are $3\mathrm{/}4$ to $(0,0)$ and $-4\mathrm{/}3$ to $(7,-1)$, whose product is $-1$, so the right angle is at $(4,3)$. The hypotenuse joins $(0,0)$ to $(7,-1)$, with midpoint $(7\mathrm{/}2,-1\mathrm{/}2)$. By Thales that midpoint is equidistant from all three vertices, so it is the circumcenter. Intersecting two perpendicular bisectors gives the same point with much more fraction arithmetic.

The point is to combine sign parity with the location of zeros. For $x=0$, each of the nine factors is negative, and nine negatives multiply to a negative, so 0 works. For $x=1$ through 9, one factor is zero, so $P(x)=0$, which is not less than zero. For $x$ at least 10, every factor is positive, so the product is positive. Hence no other nonnegative integer works, leaving exactly one value. A search without this misses that beyond the largest root all signs stay positive and that integer roots give zero rather than a sign change.

The observation is that adding $90^{\circ }$ swaps cosine-squared with sine-squared, so the long list collapses into complementary pairs. Since $\cos (\theta +90^{\circ })=-\sin \theta$, we have ${\cos }^2(\theta +90^{\circ })={\sin }^2\theta$, and ${\cos }^2\theta +{\sin }^2\theta =1$. The 36 angles split into 18 such $90^{\circ }$-separated pairs, for example $0^{\circ }$ with $90^{\circ }$ and $10^{\circ }$ with $100^{\circ }$, each summing to 1, so $S=18$.

The insight is to combine same-side supplementary angles with a logarithmic substitution that turns the condition into a quadratic plus an angle-range bound. Let $t={\log }_{2}x$. Parallelism makes the same-side interiors sum to $180^{\circ }$, so $t^2+3t=180$, which is $t^2+3t-180=0$ and factors as $(t+15)(t-12)=0$. Thus $t=12$ or $t=-15$. The value $t=-15$ would give angles $225^{\circ }$ and $-45^{\circ }$, impossible for interior angles that must lie strictly between $0^{\circ }$ and $180^{\circ }$, so it is rejected. Hence $t=12$ and $x=2^{12}=4096$, giving angles $144^{\circ }$ and $36^{\circ }$ that indeed sum to $180^{\circ }$.

The insight is that opposite sides of a square give area as squared perpendicular distance, so no vertices are needed. The direction from $(2,1)$ to $(6,4)$ is $(4,3)$ with slope $3\mathrm{/}4$, so the sides lie on $3x-4y=2$ and the parallel $3x-4y=17$ through $(7,1)$. The perpendicular distance is $\mathrm{\mid }17-2\mathrm{\mid }\mathrm{/}\sqrt{3^2+4^2}=15\mathrm{/}5=3$, using the 3-4-5 scale. That distance is the side length, so area is $3^2=9$. Finding all four corners by intersecting perpendiculars works but is much longer.

The idea is to count the grand total two ways and then build the extreme. Let the column sums be added to get the grand total. Adding by rows gives five odd numbers, so the grand total is odd. Adding by columns gives the same total, whose parity matches the number of odd columns, since even columns contribute nothing to parity. Hence the number of odd columns is odd, so with five columns the number of even columns is even and at most 4. The bound is attainable by putting 0 in the first four columns and 1 in the last column, giving each row sum 1 and four even column sums.

The insight is to add symmetric sums instead of solving for messy roots. By Vieta $p+q=11$ and $pq=23$, so $p^2+q^2=(p+q{)}^2-2pq=121-46=75$. Then $f(p)+f(q)=(p^2+q^2)+3(p+q)+14=75+33+14=122$. Solving $x^2-11x+23=0$ gives $(11\pm \sqrt{29})\mathrm{/}2$, and squaring those decimals and adding is long and error prone, while the symmetric computation is short without a calculator.

The insight is that $k$ stays constant on doubling intervals. The condition means $n$ lies in $[1,1]$, $[2,3]$, $[4,7]$, $[8,15]$, $[16,31]$, $[32,63]$, and $[64,127]$, truncated to 100 for the last block. Thus the block sizes are 1,2,4,8,16,32,64, with the last cut to 37 numbers 64 through 100. Even $k=0,2,4,6$ contributes $1+4+16+37=58$ integers. With 100 equally likely choices, the probability is $58\mathrm{/}100=29\mathrm{/}50$. Computing logarithms one by one would require one hundred evaluations, while grouping by powers of two needs only the seven block sizes.

The insight is to restrict to the diagonal and use the symmetry of the product. Given sum 101, the ordered pairs are $(x,101-x)$ for $x=1$ through 100, 100 equally likely pairs. The product $x(101-x)$ is a downward parabola with vertex at 50.5, increasing up to the middle and then decreasing symmetrically. Checking near the threshold, $43\ast 58=2494<2500$ while $44\ast 57=2508>2500$, so by monotonicity on each side the inequality holds exactly for $x=44$ through 57, 14 values. Thus the conditional probability is $14\mathrm{/}100=7\mathrm{/}50$. Evaluating all one hundred products would be long, while the vertex plus two multiplications settles the interval.

The insight is spotting hidden collinearity by slope and midpoint and then correcting the pair count. Any two points determine a line, so six points give 15 pairs. But $(1,1)$, $(3,5)$, and $(5,9)$ have slope 2 between each pair, and $(3,5)$ is the average of the other two, so all three lie on $y=2x-1$. Those three pairs produce only one line, overcounting by 2. Checking the remaining slopes shows no other three are collinear, so distinct lines equal $15-2=13$. Checking every slope pair by pair is long, while grouping the triple makes it short.

Write every factor with natural logs to reveal telescoping cancellation. Since ${\log }_a(a+1)=\ln (a+1)\mathrm{/}\ln a$, the product $P(k)$ becomes $(\ln 3\mathrm{/}\ln 2)(\ln 4\mathrm{/}\ln 3)\cdots (\ln (k+1)\mathrm{/}\ln k)$, where every interior logarithm cancels and only $\ln (k+1)\mathrm{/}\ln 2={\log }_2(k+1)$ remains. For integer $k+1$, this is an integer exactly when $k+1$ is a power of 2. Between 3 and 101 the powers are 4,8,16,32,64, corresponding to $k=3,7,15,31,63$, so there are 5 such $k$. Direct multiplication of up to 99 logarithms would be hopeless on a calculator, while cancellation finishes the work.

The insight is to combine the definitions of mean and median with ordering case analysis. The mean is $(36+x)\mathrm{/}5$ and the median is the third smallest among $1,2,3,30,x$, which depends on where $x$ falls. If the median is 2 then $(36+x)\mathrm{/}5=2$ gives $x=-26$, which indeed lies below 1 so the median is 2. If the median is $x$ then $x=9$, which is not between 2 and 3, and if the median is 3 then $x=-21$, which is not at least 3, so those cases fail the ordering check. Hence only $x=-26$ works, while assuming a median position without checking leads to failure.

The insight is to split on the definition of absolute value, which produces two different factorable quadratics whose roots must be checked against the split. For $x$ at least 1, absolute value $\mathrm{\mid }x-1\mathrm{\mid }$ equals $x-1$ and $x^2-7x-30=0$, so $(x-10)(x+3)=0$ giving 10 and $-3$, of which only 10 satisfies the case. For $x$ less than 1, $\mathrm{\mid }x-1\mathrm{\mid }$ equals $1-x$ and $x^2+7x-44=0$, so $(x+11)(x-4)=0$ giving $-11$ and 4, of which only $-11$ satisfies the case. Hence the real solutions are 10 and $-11$, with sum $-1$. Squaring to remove the absolute value would give a quartic, which is why the split is the short route.