ACT · Competition-style problems · Part 1 of 2

The insight is to track a point rather than compare coefficients. A dilation centered at the origin sends $(x,y)$ to $(mx,my)$ for scale factor $m$. The point $(1,3)$ lies on $y=3x^2$, so its image $(m,3m)$ must lie on $y=27x^2$, giving $3m=27m^2$. Since $m$ is nonzero, $m=3\mathrm{/}27=1\mathrm{/}9$. Directly forming $27\mathrm{/}3$ inverts dilation with vertical stretch, and square roots confuse linear scaling with area scaling.

The insight is to divide the quadratic by the linear factor and count shifts. Since $n^2+9n+50=(n+4)(n+5)+30$, any common divisor of the two expressions divides 30, and any divisor of $n+4$ and 30 divides the quadratic, so $g$ is the greatest common factor of $n+4$ and 30. As $n$ runs from 1 to 100, the shift $n+4$ runs from 5 to 104. Hence $g>1$ exactly when that shift is divisible by 2, 3, or 5. Counting those multiples gives 50 multiples of 2, 33 of 3, 20 of 5, minus 17, 10, 6 pairwise overlaps, plus 3 triple overlaps, for 73.

The insight is to trap the discriminant between consecutive squares so only finitely many $k$ can work. For integer roots the discriminant $k^2-4k-4$ must be a nonnegative perfect square. Completing the square gives $(k-2{)}^2=k^2-4k+4$, so the discriminant equals $(k-2{)}^2-8$. For $k>=7$, $(k-3{)}^2$ is below and $(k-2{)}^2$ is above the discriminant with no square strictly between, so no square. For $k<=-3$, writing $m=-k$ gives $(m+1{)}^2$ below and $(m+2{)}^2$ above, again no square. Hence only $-2<=k<=6$ need checking. Direct substitution shows the discriminant is $8,1,-4,-7,-8,-7,-4,1,8$, so only $k=-1$ with $x^2-x=0$ and $k=5$ with $x^2+5x+6=0$ have integer roots. Their sum is 4. Without the trap there are infinitely many $k$ to try.

The insight is to use relative speed for the meeting rate and modular periodicity for distinct points. Convert the $6:10$ ratio to $3:5$ after dividing by 2, so relative speed is 8 parts while Nadia contributes 3 parts. Meetings occur each time the combined distance covers one lap, giving 33 events including both endpoints for 32 laps combined, at times indexed by $n$ from 0 to 32. Nadia has covered $3n\mathrm{/}8$ laps at meeting $n$, so positions are multiples of $3\mathrm{/}8$ around the circle. Since 3 is invertible modulo 8, the sequence cycles every 8 meetings, producing 8 distinct points including the start.

Divisibility gives two conditions, but the second is hidden. First, the factor theorem gives $P(1)=0$, so $1+a+b=0$. The clever step deflates the known factor: $P(x)-P(1)=(x^{101}-1)+a(x-1)=(x-1)[x^{100}+x^{99}+\mathrm{.}\mathrm{.}\mathrm{.}+1+a]$. Since $(x-1{)}^2$ divides $P$, the bracket $Q(x)$ must still vanish at $x=1$. Evaluating the geometric sum at $x=1$ gives 101 ones, so $Q(1)=101+a=0$ and $a=-101$ with $b=100$. Long division by a degree 101 divisor is hopeless, while deflation needs only two evaluations.

Values are remainders, but three values do not determine a cubic, so solving for coefficients is an underdetermined failing route. The insight subtracts the line through the points: $(1,4)$, $(2,7)$, and $(3,10)$ lie on $y=3x+1$. Let $Q(x)=P(x)-(3x+1)$; $Q$ is cubic with the same leading coefficient $a$ and roots 1, 2, and 3, so $Q(x)=a(x-1)(x-2)(x-3)$. Hence $P(4)=6a+13$ with $a$ a positive integer, since integer coefficients give integer $a$ and cubic gives $a$ nonzero. The values 7, 13, and 18 correspond to $a=-1$, $a=0$, and noninteger $a=5\mathrm{/}6$, each violating one hypothesis, while $a=1$ gives 19, realized by $x^3-6x^2+14x-5$. The arithmetic is then one substitution.

The insight is to split 1 through 10 by remainder modulo 3. The chains are 3,6,9 and 1,4,7,10 and 2,5,8, and avoiding a difference of 3 means choosing no consecutive entries inside any chain. The longest gap-free choice takes 2 from each chain, for 6 total, realized only by 3 and 9, 2 and 8, and one of 1 and 7, 1 and 10, 4 and 10, giving 3 gap-free six-sets, including 1,2,3,8,9,10. The total number of six-sets is 210, so 3 have no suitable pair and 207 do, giving $207\mathrm{/}210=69\mathrm{/}70$. Enumerating all sets would be very long, while the residue split isolates the only exceptions.

The insight is to factor out the ratio scale and use parity of prime exponents. Write the integers as $7k$, $11k$, $13k$ with $k$ a positive integer. The product is 1001 times $k^3$, where 1001 equals 7 times 11 times 13 with three distinct primes. For a square every prime exponent must be even, but $k^3$ contributes exponents that are multiples of three, so 1 plus a multiple of three must be even, which forces each of the three primes to divide $k$ to an odd power. Hence $k$ is a multiple of 7 times 11 times 13, which is 1001. Writing $k$ as 1001 times $t$ makes the product a fourth power times $t^3$, so $t$ itself must be a square and the smallest choice is $t$ equals 1. The triple is 7007, 11011, 13013 with sum 31031.

The insight is to see the quartic as a quadratic in $y=x^2+2x$ so the given relation makes a factor vanish. Let $y=x^2+2x$; then $y^2=x^4+4x^3+4x^2$, so $f(x)=y^2+6y+9=(y+5)(y+1)+4$. For the given $z$, $z^2+2z=-5$, so $y=-5$ and the product $(y+5)(y+1)$ is zero, leaving $25-30+9=4$. Solving for $z=-1\pm 2i$ and raising each to the fourth power with complex arithmetic is long and error prone, while the substitution finishes without a calculator.

The insight is that integer length sharply limits possibilities, and the northeast condition forces swapped coordinates. From $x^2+y^2=65$ with positive integers, checking $x=1,\dots ,8$ leaves only $65=64+1=49+16$, so in the first quadrant the vectors are $(8,1)$, $(1,8)$, $(7,4)$ and $(4,7)$. Two distinct ones sum to equal components only when coordinates are swapped: $(8,1)+(1,8)=(9,9)$ with magnitude $9\sqrt{2}$ and $(7,4)+(4,7)=(11,11)$ with magnitude $11\sqrt{2}$. No other distinct pair works, so the distinct diagonal magnitudes total $9\sqrt{2}+11\sqrt{2}=20\sqrt{2}$. Treating the vectors as continuous leaves infinitely many possibilities and fails.

The insight is the angle-bisector proportion plus a weighted average. Here $AB=8$ and $BC=\sqrt{8^2+6^2}=10$, so the bisector meets $AC$ at $D$ with $AD\mathrm{/}DC=8\mathrm{/}10=4\mathrm{/}5$. Since $AC$ runs from $(0,0)$ to $(0,6)$, the point is $D=(0,8\mathrm{/}3)$ because $8\mathrm{/}3$ is $4\mathrm{/}9$ of the way from $A$ to $C$. The line through $B=(8,0)$ and $D=(0,8\mathrm{/}3)$ has slope $(8\mathrm{/}3-0)\mathrm{/}(0-8)=-1\mathrm{/}3$, so $y=(-1\mathrm{/}3)(x-8)$, namely $x+3y=8$. Measuring angles with inverse tangent for each candidate would be the long route.

The insight is to count by apex using equal arcs. Equally spaced points cut the circle into equal arcs, so symmetric points about the diameter through a vertex give equal chords and hence an isosceles triangle. Fix one vertex as the apex. Of the other 7 points, the opposite point cannot pair with the apex to make equal sides, while the remaining 6 form 3 symmetric pairs, giving 3 isosceles triangles with that apex. With 8 choices of apex this gives 24, and no triangle is counted twice because with 8 points none has two apices. Listing all 56 triples is long.

The insight is to reframe the total over the inner index and bound it, then build matrices attaining it. Let $a_k$ be column sums of $A$ and $b_k$ row sums of $B$; the sum of all entries of $AB$ equals $a_1{b}_1+a_2{b}_2+a_3{b}_3$. Each $a_k$ and $b_k$ is 0, 1 or 2 with total 3. Since $a_k\le 2$, the total is at most $2(b_1+b_2+b_3)=6$, and 6 would need $a_k=2$ wherever $b_k>0$, forcing either two full columns in $A$ or three ones in one row of $B$, both impossible with the counts. Hence at most 5, attained for example by $A$ with rows $(1,1,0)$ and $(1,0,0)$ and $B$ with rows $(1,1)$, $(1,0)$ and $(0,0)$, giving $AB$ with rows $(2,1)$ and $(1,1)$. Enumerating all 4096 pairs is infeasible.

Convert the mixed-base sum to a single variable and bound it with a square. For $x>1$ put $t={\log }_{2}x$, which is positive, and by change of base ${\log }_{x}8=\ln 8\mathrm{/}\ln x=3\ln 2\mathrm{/}\ln x=3\mathrm{/}t$. Then $f=t+3\mathrm{/}t$ with $t>0$, and $t+3\mathrm{/}t-2\sqrt{3}=(t-\sqrt{3}{)}^2\mathrm{/}t\ge 0$ because the numerator is a square and the denominator is positive. Thus $f\ge 2\sqrt{3}$, with equality when $t=\sqrt{3}$, that is $x=2^{\sqrt{3}}$, which indeed exceeds 1. Sampling convenient powers like 2 or 4 gives larger values and never proves optimality.

The insight is to combine triangle inequality bounds with a division re-framing. Triangle inequality gives $22<n<50$ from $36-14<n<36+14$. Rewriting $5n+55=5(n-1)+60$ shows the fraction equals $5+60\mathrm{/}(n-1)$, so $n-1$ must divide 60. Divisors of 60 that lie in 21 to 49 for $n-1$ are only 30, giving $n=31$ since 20 is too small and 60 is too large for the triangle range. Then $2n-15=62-15=47$, which is short once the bounds and divisibility are seen, while trying values across the range is long.

The insight is that a unique mode controls how many times the top value may repeat, and the total then fixes the cheapest layout. The total is $6\times 8=48$ and the middle two sum to 14 since the median is 7. Because the mode is 5, the minimum is at most 5, so with range $R$ the maximum is at most $5+R$. If $R=6$ then the minimum must be 5 and the maximum 11, otherwise the sum cannot reach 48, and the middle pair summing to 14 leaves 24 for the top two, but two numbers at most 11 with no value repeated twice can sum to at most 21, and even allowing a repeat gives at most 22, short of 24. Thus $R\ge 7$, achieved by 5,5,5,9,12,12 whose middle two are 5 and 9, mean 8, unique mode 5, and range 7.

The insight is to pair divisors. For $6^6$, write $6^6=2^6\ast 3^6$, so the number of positive divisors is $(6+1)(6+1)=49$ by the exponent rule. Each divisor $d$ pairs with $6^6\mathrm{/}d$, and the product of the pair is $6^6$. The pair collapses to a single divisor exactly when $d=6^3$, since $(6^3{)}^2=6^6$. Thus one divisor equals $6^3$ and the remaining 48 form 24 pairs with one member below $6^3$ and one above. Hence 24 of the 49 equally likely slips satisfy the condition, giving $24\mathrm{/}49$. Listing all divisors would take far too long, while the pairing finishes the count at once.

The insight is to factor each factor as a difference of squares and then see telescoping cancellation across numerators and denominators. Write $1-1\mathrm{/}{n}^2=(n^2-1)\mathrm{/}{n}^2=(n-1)(n+1)\mathrm{/}{n}^2$. Hence the product splits into $\prod (n-1)\mathrm{/}n$ times $\prod (n+1)\mathrm{/}n$ for $n=2$ to 50. The first product telescopes to $1\mathrm{/}50$ and the second telescopes to $51\mathrm{/}2$. Multiplying gives $51\mathrm{/}100$ in lowest terms, which is obtained without multiplying forty-nine decimals.

The insight is the reciprocal symmetry $t=x+1\mathrm{/}x$, which turns a quartic into two quadratics. Since $(x+1\mathrm{/}x{)}^2=x^2+2+1\mathrm{/}{x}^2$, we have $x^2+1\mathrm{/}{x}^2=t^2-2$, so the equation becomes $t^2-2-5t+6=0$, or $(t-1)(t-4)=0$. Thus $t=1$ or 4. Now $x+1\mathrm{/}x=1$ gives $x^2-x+1=0$ with negative discriminant, so no real $x$. While $x+1\mathrm{/}x=4$ gives $x^2-4x+1=0$ with two distinct real roots whose sum is 4 by Vieta. Hence the sum of all real $x$ is 4. Multiplying by $x^2$ without the substitution leaves a quartic that is much longer to factor.

The breakthrough is to clear denominators and use positivity to fix signs. From $1\mathrm{/}x+1\mathrm{/}y=1\mathrm{/}24$ we get $24x+24y=xy$, so $xy-24x-24y+576=576$, or $(x-24)(y-24)=576$. If $x$ were at most 24, then $1\mathrm{/}x$ would be at least $1\mathrm{/}24$, forcing $1\mathrm{/}y$ to be nonpositive, impossible for positive $y$, so both shifted factors are positive. With $576=24^2$ having 21 positive divisors, the unordered factor pairs with first less than second are $(21-1)$ divided by 2, or 10, each giving $x<y$. The pair with both 24 gives $x=y=48$ and is excluded by the strict inequality.

The insight is to convert integrality of the interior angle into a divisibility condition on the exterior angle and then count divisors. A regular $n$-gon has exterior $360\mathrm{/}n$ and interior $180-360\mathrm{/}n$. Since 180 is an integer, the interior is an integer exactly when $360\mathrm{/}n$ is an integer, so $n$ must be a positive divisor of 360. Writing $360=2^3\cdot 3^2\cdot 5$ gives $(3+1)(2+1)(1+1)=24$ positive divisors. The values 1 and 2 do not form polygons, and $n\ge 3$ excludes them, leaving $24-2=22$ values.

The insight is to take the square root of the area ratio to get the linear scale and then enforce integer divisibility with distinctness. Similar figures scale areas by the square of the linear factor, so $9:16$ gives linear $3:4$. Let the smaller sides be $a$ and $b$, so the larger are $4a\mathrm{/}3$ and $4b\mathrm{/}3$, which forces $a$ and $b$ to be multiples of 3. Write $a$ as $3m$ and $b$ as $3n$ so the four sides are $3m$, $3n$, $4m$, $4n$. Distinctness rules out $m$ equals $n$, since 3,3,4,4 repeats. The smallest distinct pair is $m$ equals 1 and $n$ equals 2, giving sides 3, 6, 4, 8. The perimeters are 18 and 24 for a total of 42.

The insight is that the median forces a straddle while the mode forces a duplicate, and the total then leaves only two duplicate layouts. Five numbers averaging 83 total 415, so with $70+80+90=240$ the two added $a\le b$ sum to 175. To keep median 80 with $70<80<90$, the added pair must straddle 80, i.e. $a\le 80\le b$, otherwise the middle shifts away from 80. With $a+b=175$ and straddle, many pairs work, e.g. 75 and 100 with all distinct and no mode. Requiring a unique mode forces a duplicate: either $a=80,b=95$ giving 70,80,80,90,95 with mode 80, or $a=70,b=105$ giving 70,70,80,90,105 with mode 70. Both satisfy mean 83 and median 80, so the greatest possible added value is 105.

Subtract the equations: $x^2-y^2+2y-2x=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)=0$ by GCF and difference of squares. The insight is factored subtraction plus case analysis with Vieta sums: either $x=y$ or $x+y=2$. If $x=y$ then $x^2+2x-7=0$, whose roots sum to $-2$ without solving the radicals. If $x+y=2$ then $y=2-x$ gives $x^2-2x-3=0$, whose roots sum to 2. Adding both cases gives 0. Solving all four pairs explicitly with square roots is long, while Vieta gives each sum at once.

The insight is that a rational terminates exactly when its reduced denominator has only prime factors two and five, so the factor seven must either be absent or cancel, forcing a divisibility split. If seven does not divide $n$, then $n$ itself must be of the form $2^a{5}^b$, giving fifteen values up to 100: 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, and 100. If seven divides $n$, write $n=7m$ so $7\mathrm{/}n=1\mathrm{/}m$, and $m$ must be of that form with $m$ at most 14, giving six values 7, 14, 28, 35, 56, and 70. Together there are twenty-one values, found without dividing one hundred fractions.

The key insight is midpoint doubling combined with a divisibility step count. Since $M=(29,47)$ is the midpoint, the missing endpoint satisfies $(5+Bx)\mathrm{/}2=29$ and $(11+By)\mathrm{/}2=47$, so the other endpoint is $(53,83)$. The run is 48 and the rise is 72. Lattice points divide the segment into equal integer steps, so the step vector must divide both 48 and 72. The greatest common divisor is 24, giving minimal steps of $(2,3)$. Starting at $t=0$ through $t=24$ yields $24+1=25$ integer points. Listing every point would be long, but the divisor makes it one division.

The insight is to turn each speed into a lap time with a common unit and take the least common multiple by prime factorization. Since 22 ft/s equals 15 mi/h, 33 ft/s equals 22.5 mi/h. For 3 miles the lap times are 3 divided by 22.5 hours equals 8 minutes, 3 divided by 30 hours equals 6 minutes, and 3 divided by 18 hours equals 10 minutes. The joint return is the least common multiple of 8, 6, and 10, whose prime factors give $2^3$ times 3 times 5 equals 120 minutes. Listing multiples to 120 would take far longer than factoring.