ISEE Upper · Competition-style problems · Part 1 of 8

Midpoint coordinates are averages, so the midpoint is integral exactly when both coordinate sums are even, which happens exactly when the two points share the same parity class $(x$ mod $2,y$ mod $2)$. There are four classes: even-even, even-odd, odd-even, and odd-odd. With seven points, some class must repeat. To guarantee as few pairs as possible spread the points as evenly as possible as 2,2,2,1. Pairs inside a class are $1+1+1+0=3$, and any less even spread creates more pairs, for example 4,1,1,1 gives 6. Hence at least three pairs must share a class and have integral midpoints.

Compute the net change over one full two-move cycle: $x$ changes by $-8+3=-5$ and $y$ changes by $-30+31=+1$. After $2k$ moves the marker is at $(100-5k,10+k)$, and after $2k+1$ moves it is at $(92-5k,-20+k)$. Quadrant II needs $x<0$ and $y>0$. For even moves this needs $100-5k<0$, so $k$ is at least 21, giving move 42 with $(-5,31)$. For odd moves this needs both $92-5k<0$ and $-20+k>0$, so $k$ is at least 21, giving move 43 with $(-13,1)$. Move 39 gives $(-3,-1)$ and move 41 gives $(-8,0)$, so 42 is first. Simulating all steps is long, while the net formulas give it by division.

The insight combines an equal-area invariance with a ratio. Triangles $ABD$ and $ABC$ share the same base $AB$ and the same height, which is the distance between the parallels, so they have equal area. Removing their common part $ABO$ leaves triangles $ADO$ and $BCO$ with equal area, call it $x$. Along the diagonals, areas with the same altitude are in the ratio of their bases, so $9\mathrm{/}x=x\mathrm{/}25$. Thus $x^2=225$ and $x=15$. The trapezoid is $9+25+15+15=64$. A student who only adds the two given areas never finds the two side triangles.

The insight is to avoid finding the edges individually and instead use the square of the edge sum together with the three dimensional version of the Pythagorean relationship for the diagonal. The sum of one length plus one width plus one height is 40 divided by 4, or 10, and the sum of the three pairwise products is 64 divided by 2, or 32. Since the square of the edge sum equals the sum of the squares plus twice the pairwise sum, the squared diagonal equals 100 minus 64, which is 36, so the diagonal is 6. Trying to factor the edges from volume style guessing leads to a long failing search, while the identity makes the arithmetic immediate.

The insight combines supplementary angles with a half-area re-framing. In a parallelogram consecutive angles are supplementary, so angle $A$ plus angle $B$ is 180 degrees. Their halves sum to 90 degrees, so angle $AEB$ is 90 degrees because the angles of triangle $AEB$ sum to 180 degrees. Hence triangle $AEB$ is right with legs 6 and 8, so its area is $(6)(8)\mathrm{/}2=24$ and its hypotenuse is 10. Triangle $AEB$ has base $AB$ and height equal to the distance between the parallels $AB$ and $CD$ since $E$ is on $CD$, so it occupies exactly half the parallelogram. Doubling gives 48. A student who misses the angle sum tries coordinate placement and gets stuck.

The insight is to cut and unfold the two faces the path crosses so the surface route becomes a straight segment, then choose the smallest of the three possible unfoldings with the Pythagorean theorem. Pairing 3 with 5 gives a right triangle with legs 8 and 6 and hypotenuse 10, pairing 3 with 6 gives legs 9 and 5 with hypotenuse the square root of 106, and pairing 5 with 6 gives legs 11 and 3 with hypotenuse the square root of 130. The smallest is 10, while the interior diagonal the square root of 70 is shorter but illegally passes through the box and the edge walk of 14 is much longer.

The insight is to split the total into positive and negative parts. Let $P$ be the sum of the positive entries and $N$ be the sum of the absolute values of the negative entries, ignoring zero which contributes nothing. Then $P-N=20$ and $P+N=24$, so $P=22$ and $N=2$. Since the negatives are distinct integers whose absolute values sum to 2, the only possibility is a single $-2$, because two distinct negatives need at least $1+2=3$. Thus the other five numbers are nonnegative, distinct from $-2$, and sum to 22. To make the largest as large as possible, make the other four as small as possible, namely 0,1,2,3 with sum 6, giving $22-6=16$, achieved by $-2,0,1,2,3,16$.

The insight combines a radius-to-chord right triangle with parity. Drop the perpendicular from the center to the chord. It bisects the chord, so with half-chord $m$, distance $d$, and radius 10, we have $d^2+m^2=100$. If the full chord $c$ is odd then $m$ ends in $\mathrm{.}5$ and $m^2$ ends in $\mathrm{.}25$, so $d^2=100-m^2$ cannot be an integer square, hence $d$ cannot be an integer. So $c$ is even and $(d,m,10)$ is an integer right triangle. The squares below 100 give only $0+100=100$ and $36+64=100$, so $(d,m)$ is $(0,10)$, $(6,8)$, or $(8,6)$. These give chords 20, 16, and 12, which is 3 lengths. Checking all 20 lengths one by one is far longer.

The insight is that $\mathrm{\mid }n-4\mathrm{\mid }$ dividing $n^2+8$ forces it to divide a fixed remainder, and the solutions pair symmetrically around 4. Write $d=\mathrm{\mid }n-4\mathrm{\mid }>0$, so $n=4\pm d$ and $n$ is congruent to 4 modulo $d$. Then $n^2+8$ is congruent to $16+8=24$ modulo $d$, so $d$ divides 24. Conversely every positive divisor $d$ of 24 gives two valid solutions $4+d$ and $4-d$, all distinct. Since 24 has 8 positive divisors, there are 8 pairs each summing to 8, so the total is $8\cdot 8=64$ without adding sixteen numbers.

There are $9\cdot 9\cdot 8=648$ three-digit numbers with all different digits (the first digit cannot be 0). The insight is to count sets of digits instead of numbers: each set of three different digits can be written in increasing order in exactly one way. But a set containing 0 would have to start with 0, which is not allowed, so only sets drawn from 1 through 9 count. There are $\left(\genfrac{}{}{0px}{}{9}{3}\right)=84$ of them. The probability is ${\displaystyle \frac{84}{648}}={\displaystyle \frac{7}{54}}$, a little less than ${\displaystyle \frac{1}{6}}$ because of the excluded zero sets.

The insight is that the boundary between “closer to $(0,0)$” and “closer to $(6,4)$” is a straight line. Setting $x^2+y^2<(x-6{)}^2+(y-4{)}^2$, the squares cancel and leave $12x+8y<52$, or $3x+2y<13$. Inside the rectangle this line runs from $({\displaystyle \frac{13}{3}},0)$ on the bottom edge to $({\displaystyle \frac{1}{3}},6)$ on the top edge, so the favorable region is a trapezoid with parallel sides ${\displaystyle \frac{13}{3}}$ and ${\displaystyle \frac{1}{3}}$ and height 6. Its area is ${\displaystyle \frac{1}{2}}({\displaystyle \frac{13}{3}}+{\displaystyle \frac{1}{3}})\cdot 6=14$. The rectangle’s area is 48, so the probability is ${\displaystyle \frac{14}{48}}={\displaystyle \frac{7}{24}}$.

The key insight is to use the red plus green total as an invariant multiple plus divisibility elimination with the original total interval. Originally red is $2k$, blue is $3k$, green is $4k$, so the total is $9k$ and red plus green is $6k$. After adding 60 with no blue, red plus green is $6k+60$ and must be a multiple of 11 from the $4:7$ split. Since $9k$ lies strictly between 100 and 120, $k$ is 12 or 13. Only $k=12$ makes $6k+60=132$ a multiple of 11, so blue is 36.

The insight is alternating-order symmetry with careful counting of the partial hour. Let Rosa do $r$ of the fence per hour and Tom do $t$ per hour. Starting with Rosa for 9 hours gives 5 Rosa hours and 4 Tom hours, so $5r+4t=1$. Starting with Tom, 9 full hours give 5 Tom hours and 4 Rosa hours, then the next 30 minutes fall in a Rosa hour, so $4.5r+5t=1$. Multiply the first by 5 to get $25r+20t=5$ and the second by 4 to get $18r+20t=4$. Subtracting gives $7r=1$, so $r=1\mathrm{/}7$ and Rosa alone needs 7 hours, with Tom needing 14 hours as a check.

The insight is to start from all pairs and subtract what parallelism and concurrency collapse. Any two nonparallel lines meet once, so with no conditions 6 lines give $6\times 5\mathrm{/}2=15$ pairs. The single parallel pair removes one point, leaving 14. The three concurrent lines contribute 3 pairs but only one point, so they collapse 3 points to 1 and remove two more, leaving 12. Trying to draw all cases instead is long and error prone.

The insight is to solve the system once with $k$ as a parameter and then enforce both Quadrant II inequalities with attention to sign. Setting $2x+5=kx-3$ gives $(k-2)x=8$ so $x=8\mathrm{/}(k-2)$ when $k\mathrm{\ne }2$, and $y=16\mathrm{/}(k-2)+5$. Quadrant II needs $x<0$ so $k<2$, and $y>0$ gives $16\mathrm{/}(k-2)+5>0$. With $k<2$ multiply by the negative $k-2$ to get $16+5(k-2)<0$, so $k<-6\mathrm{/}5$, hence integers $k\le -2$. In $-6\le k\le 6$ this is $-6,-5,-4,-3,-2$ for 5 values.

The insight is to use addition of polynomials together with base-ten place value instead of trying to split $R$ into $P$ and $Q$, which has many possibilities and fails. Write $R(x)=a{x}^2+bx+c$ where $a$, $b$, $c$ are integers from 0 to 8 because each is a sum of two coefficients from 0 to 4. Then $R(10)=100a+10b+c=352$ with each coefficient below 10, so no carrying occurs and the decimal digits are the coefficients, giving $a=3$, $b=5$, $c=2$ and $R(x)=3x^2+5x+2$. Therefore $R(2)=12+10+2=24$, and the computation is one short line once the decoding is seen.

Use midpoint symmetry. If $(0,0)$ is the midpoint then the endpoints are opposites: write $A=(x,y)$ with integers and $B=(-x,-y)$. The length condition gives twice the distance from $A$ to the origin equals 10, so $x^2+y^2=25$. Integer solutions are $(5,0)$ and $(0,5)$ with signs and $(3,4)$ and $(4,3)$ with signs, twelve points in all: four axial and eight diagonal. Opposite points define the same segment, so divide by two. The axial points give two segments and the diagonal points give four segments, for six total. Searching over pairs of endpoints without symmetry would be far longer.

The insight is to count how many times each gap contributes, then push points to the ends. Order the chosen integers $w<x<y<z$. Adding the six distances gives $(x-w)+(y-w)+(z-w)+(y-x)+(z-x)+(z-y)=3(z-w)+(y-x)$, so the outer range counts three times and the inner gap once. Since all lie in an interval of length 10, $z-w\le 10$, and distinctness gives $x\ge w+1$ and $y\le z-1$, so $y-x\le (z-w)-2\le 8$. Hence the sum is at most $3\cdot 10+8=38$, achieved by $-5,-4,4,5$ with distances $1+9+10+8+9+1=38$.

The idea is that the three midpoints form the midpoint triangle, whose sides are half as long as the original sides, so its area is one-fourth of the original. Using the vectors $(4,2)$ and $(2,6)$ from $(1,2)$, the midpoint triangle has area ${\displaystyle \frac{1}{2}}\mathrm{\mid }4\cdot 6-2\cdot 2\mathrm{\mid }=10$, so the original triangle has area 40. You can confirm by rebuilding the vertices, each equal to the sum of two midpoints minus the third: $(3,-2)$, $(-1,6)$, $(7,10)$; the box method gives area 40.

The insight is to combine the commutativity of successive multipliers with the asymmetry of intermediate divisibility. Both orders multiply the start by $6\mathrm{/}5$ and $5\mathrm{/}4$, so both end at $3\mathrm{/}2$ times the start, but the middle year differs. The 20 percent then 25 percent order needs the start to be a multiple of 5 and the middle to be a multiple of 4, forcing the start to be a multiple of 10. The reverse order needs the start to be a multiple of 4 with the middle automatically a multiple of 5, forcing only a multiple of 4. A start working for both must be a common multiple of 10 and 4, hence a multiple of 20. Below 50 the possibilities are 20 and 40, so the largest is 40, giving yearly counts 40, 48, 60 in one order and 40, 50, 60 in the other.

The insight is to avoid computing 6561 minus 256 and factoring 6305 by trial division, and instead to apply difference of squares twice using the distributive property. First 3 to the eighth minus 2 to the eighth equals 3 to the fourth minus 2 to the fourth times 3 to the fourth plus 2 to the fourth, which is 65 times 97 because 81 minus 16 is 65 and 81 plus 16 is 97. Then 65 is 5 times 13, and 97 has no divisor except 1 and itself after checking 3, 5, and 7 since its square root is below 10. Hence the prime factors are 5, 13, and 97, the largest of which is 97, while trial division of 6305 from scratch needs many more tests.

Outside: the cube had $6\cdot 36=216$ square centimeters, and each face loses a 2 by 2 opening, so $216-24=192$ remain. Inside, the insight is that the tunnels share a central 2 by 2 by 2 space with no walls, so only the six arms leading in from the faces have walls. Each arm is 2 cm long with four walls of $2\times 2=4$, so 16 square centimeters per arm and 96 in all. The total is $192+96=288$ square centimeters.

The key idea is that the two centers and either intersection point form an equilateral triangle, since every side is a radius of length 6. So each center sees the shared region through an angle of $60^{\circ }+60^{\circ }=120^{\circ }$. The two $120^{\circ }$ sectors, each with area ${\displaystyle \frac{1}{3}}\cdot 36\pi =12\pi$, together cover the shared region and also cover the rhombus formed by the two centers and the two intersection points twice. That rhombus is two equilateral triangles of side 6, with area $2\cdot 9\sqrt{3}=18\sqrt{3}$. So the shared region has area $24\pi -18\sqrt{3}$.

The insight is double-counting memberships to bound extras combined with an extremal construction. Union is $32-3=29$ students with at least one sport. Total memberships are $18+16+14=48$, so extras beyond one per student are $48-29=19$. Each all-three student contributes 2 extras and each exactly-two student contributes 1, so with $t$ triples, $2t$ is at most 19, giving $t$ at most 9. Nine is attainable with 9 triple, 1 soccer-basketball double, plus 8 only-soccer, 6 only-basketball, and 5 only-tennis, which uses 29 union members and exactly 18, 16, and 14 memberships. Dividing extras by 3 or using only the smallest group misses the double-count.

The insight is to reframe the absolute sum by sign cases and then use the bound. Suppose the larger of $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }$. If $a$ and $b$ have the same sign then $\mathrm{\mid }a+b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ and $\mathrm{\mid }a-b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }$, summing to twice $\mathrm{\mid }a\mathrm{\mid }$. If they have opposite signs the two roles swap, still summing to twice $\mathrm{\mid }a\mathrm{\mid }$. The same holds with $a$ and $b$ swapped, so in all cases $\mathrm{\mid }a+b\mathrm{\mid }+\mathrm{\mid }a-b\mathrm{\mid }$ equals twice the larger of $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$. Here that gives twice the larger equals 12, so the larger equals 6. But $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$ are at most 5, so the larger is at most 5, impossible. Hence no ordered pairs satisfy the equation.

Write the intercept form $6\mathrm{/}a+4\mathrm{/}b=1$ and clear denominators to get $6b+4a=ab$. Rearranged, $ab-6b-4a+24=24$, so $(a-6)(b-4)=24$. Since the intercepts are positive and the point has positive coordinates, $6\mathrm{/}a<1$ and $4\mathrm{/}b<1$, so $a>6$ and $b>4$, making both factors positive divisors of 24. Hence $a$ can be $7,8,9,10,12,14,18,$ or 30 with $b$ equal to $28,16,12,10,8,7,6,$ or 5. Only 28 is a multiple of 14, so $a$ is 7. Trying values one by one without the factored form needs many divisions, while the factored list is short.

The insight is that two reflections through points combine into one slide. Reflecting $P$ through $A$ gives $2A-P$; reflecting that through $B$ gives $2B-(2A-P)=P+2(B-A)$. So each pair of reflections moves the point by $2(B-A)=2(3,2)=(6,4)$, no matter where it starts. Ten reflections are five pairs, so the point ends at $5(6,4)=(30,20)$. Its distance from the origin is $\sqrt{900+400}=\sqrt{1300}=10\sqrt{13}$.

The idea is that every point of the crease is equally far from the two corners, so the crease is the perpendicular bisector of the diagonal joining them. The diagonal is 10, and the crease crosses it at the center of the rectangle. Half the crease, half the diagonal and the long side form a right triangle similar to the right triangle with legs 6 and 8: half the crease is to 5 as 6 is to 8, so half the crease is 3.75 and the crease is 7.5. Check: the crease ends 6.25 in from the corners along the long sides, and $\sqrt{4.5^2+6^2}=7.5$.

The insight is the longest-side bound plus a short case list. Write sides $a\le b\le c$ with $a+b+c=15$. Triangle inequality $c<a+b$ gives $2c<15$, so $c\le 7$, and $c\ge 5$ or the sum cannot reach 15. For $c=7$, $a+b=8$ gives $(1,7,7)$, $(2,6,7)$, $(3,5,7)$ and $(4,4,7)$. For $c=6$, $a+b=9$ gives $(3,6,6)$ and $(4,5,6)$. For $c=5$, only $(5,5,5)$. That is $4+2+1=7$ triangles. Listing every partition of 15 and testing each without the bound is much longer.

The insight is to split the base into primes and trap the hard part between easy squares instead of multiplying exactly. Write $6^{12}=2^{12}\cdot 3^{12}$ with $2^{12}=4096$, so $4000<2^{12}<5000$, and $3^6=729$ gives $3^{12}=729^2$ with $700<729<800$, hence $700^2<3^{12}<800^2$, that is $490000<3^{12}<640000$. Multiplying the bounds gives $4000\cdot 490000<N<5000\cdot 640000$, so $1960000000<N<3200000000$. Both bounds lie between $10^9$ and $10^{10}$, and $10^9$ is the smallest ten-digit number while $10^{10}$ is the smallest eleven-digit number, so $N$ has 10 digits. Exact multiplication needs a six-digit product, while the bounds need only three-digit squares.

The insight is to pull out the minimal shared prime powers and then test the leftover bracket for hidden factors instead of adding huge numbers. Write $12^5=2^{10}\cdot 3^5$, $18^6=2^6\cdot 3^{12}$, and $24^5=2^{15}\cdot 3^5$, so the common part is $2^6\cdot 3^5$ and $N=2^6\cdot 3^5(2^4+3^7+2^9)=2^6\cdot 3^5(16+2187+512)$. The bracket sums to 2715, which is odd, so no extra factor 2 hides there, but its digit sum is 15, so it is a multiple of 3; indeed $2715=3\cdot 905$ with 905 coprime to 6. Thus $N=2^6\cdot 3^6\cdot 905$, so the balanced 6-power is limited by the 2-count, giving $k=6$. Direct addition needs eight-digit arithmetic, while the factored residue check is short.

The insight is to combine solving a linear system by addition with a discount-threshold case split. If both totals were discounted, the pre-discount totals would be 100 and 125, since $80\mathrm{/}0.8=100$ and $100\mathrm{/}0.8=125$. Then $3N+2P=100$ and $2N+3P=125$ add to $5(N+P)=225$, so $N+P=45$, and subtracting gives $N-P=-25$, so $N=10$ and $P=35$. This satisfies the threshold, with pre-totals 100 and 125 both at least 100. The alternative with the first purchase undiscounted gives $3N+2P=80$ and $2N+3P=125$, yielding $N=-2$, impossible for a positive price, while two undiscounted totals would make the second pre-total 100, which should have been discounted. Hence only the both-discounted case survives, so the notebook price is 10.

The insight is weighted extremal bounding with an explicit construction. Simple interest makes one dollar for one year earn 0.10 dollars, so with deposits $P1$ less than $P2$ less than $P3$ the interest condition is $P1$ plus $2P2$ plus $3P3$ equals 220. To have three deposits above 40 dollars the cheapest distinct choice is 41, 42, 43, giving 41 plus 84 plus 129 equals 254 which exceeds 220, so three is impossible. Two is possible with 1, 42, 45 since 1 plus 84 plus 135 equals 220 for 22 dollars interest, so the maximum is two.

The insight is to factor twice by difference of squares and then count factors of 2 with parity. Write n to the fourth minus 1 as n squared minus 1 times n squared plus 1, and n squared minus 1 as n minus 1 times n plus 1. For odd n, both n minus 1 and n plus 1 are consecutive even numbers, so one is a multiple of 4, giving at least three factors of 2, hence a multiple of 8. Also n squared plus 1 is even but exactly 2 modulo 4, contributing exactly one more factor of 2, so 16 always divides the product. The bound is sharp because n equals 3 gives 80, which is divisible by 16 but by neither 32 nor 48, and 8 always divides but is not greatest.

The insight is to use prime factors to test terminating and exponents to test exact length. Since $60=4$ times 3 times 5, the reduced denominator still contains 3 unless 3 divides $n$. Hence terminating forces $n=3m$ and the fraction reduces to $m\mathrm{/}20$. Write the reduced denominator as 2 to a power times 5 to a power. Exactly two places means the larger exponent is 2, so the reduced denominator is 4 or 20. With $m$ at most 10, denominator 20 occurs for $m=1,3,7,9$ and denominator 4 occurs for $m=5$. That gives $n=3,9,15,21,27$, five values. Listing all 30 decimals would be long, while factor plus length analysis is short.

The unpainted cubes form an inner box whose dimensions are each 2 less than the solid’s, so if the solid is $a$ by $b$ by $c$, then $(a-2)(b-2)(c-2)=24$. The insight is extremal: to make the whole solid as small as possible, the inner box should be as close to a cube as possible. Checking the factorizations of 24: $1\cdot 1\cdot 24$ gives $3\cdot 3\cdot 26=234$, $1\cdot 4\cdot 6$ gives $3\cdot 6\cdot 8=144$, $2\cdot 2\cdot 6$ gives $4\cdot 4\cdot 8=128$, and $2\cdot 3\cdot 4$ gives $4\cdot 5\cdot 6=120$, the least. The painted cubes number $120-24=96$.

The insight is double-counting memberships combined with solving a small linear system for exactly-two and exactly-three. Union is $30-5=25$, so exactly-two plus exactly-three equals $25-9=16$. Total memberships are $18+16+14=48$, which also equal 9 times 1 plus exactly-two times 2 plus exactly-three times 3, so twice exactly-two plus three times exactly-three equals 39. Substituting exactly-two as 16 minus exactly-three gives 32 plus exactly-three equals 39, so exactly-three is 7. Counting each triple once or confusing none and exactly-one with triples misses the two-way count.