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

The key is the membership-excess invariant plus a parity extremal bound finished by a construction. The three enrollments total $60+55+50=165$ while 100 students each contribute at least one enrollment, so the excess 65 must come from students counted extra times. A student in exactly two subjects contributes one extra enrollment and a student in all three contributes two extra, so $E2+2E3=65$. Hence $E2$ and 65 have the same parity, so $E2$ is odd and at least 1, giving $E3$ at most 32. This bound is attainable with $E2=1$ and $E3=32$, for example with singletons 27, 22, and 18 and one shared pair, which totals 100 and meets each subject count.

The key is a frequency cap from the single mode plus a range link, settled by a case split on the mode frequency. Since the mode 15 is below the median 20, the smallest value must be 15, so the largest is 35. Let the mode appear $k$ times. Only the three values at or below the median can be 15, so $k$ is 2 or 3. With $k=3$ the list is forced toward 15, 15, 15, 20 with at most two 35s, totaling 169. With $k=2$ we can use 15, 15, 19, 20 with distinct top values 33, 34, 35, totaling 171, and 15 still occurs more than any other value. No larger total is possible, so the greatest mean is $171\mathrm{/}7$, which is 24 and three sevenths.

The insight is to combine the triangular total for 1 to $n$ with a bounding estimate that forces $n$ near twice the mean, after which parity eliminates the middle case. The total of 1 to $n$ is $n(n+1)\mathrm{/}2$, and after erasing $x$ the remaining total is $14.5(n-1)$. Since $x$ is between 1 and $n$, the value $n(n+1)\mathrm{/}2-14.5(n-1)$ lies between 1 and $n$, which forces $n$ between 27 and 29 because the mean 14.5 is about half of $n$. Checking gives $n=27$ with total 378 and remaining 377, so $x=1$, $n=28$ with total 406 and remaining 391.5, so $x=14.5$ impossible for an integer, and $n=29$ with total 435 and remaining 406, so $x=29$. The possible erased values are 1 and 29, whose sum is 30.

The insight pairs outer terms to get fixed lower bounds and uses odd-even parity to bound the middle term away from zero. For any x, |x-10|+|x-50| is at least 40, with equality when x lies between 10 and 50, and |x-20|+|x-40| is at least 20, with equality between 20 and 40. Since x is odd and 30 is even, |x-30| is at least 1. Adding gives at least 61. For example x=31 gives 21+11+1+9+19=61. Since 61 exceeds 60 and the bound holds for every odd integer with only small additions after the pairing, Column A is always greater.

The insight is a modular re-frame combined with systematic counting. The average of three integers is an integer exactly when their sum is divisible by 3, so only residues modulo 3 matter. From 1 through 9 there are three numbers of each residue: 1,4,7 leave 1, 2,5,8 leave 2, and 3,6,9 leave 0. A triple sums to 0 mod 3 either with all three residues equal, giving 3 triples $(1,4,7)$, $(2,5,8)$, $(3,6,9)$, or with one of each residue, giving $3\times 3\times 3=27$ triples. That is 30 favorable triples out of $\left(\genfrac{}{}{0px}{}{9}{3}\right)=84$, so the probability is $30\mathrm{/}84=5\mathrm{/}14$.

The key is symmetry plus square growth. Since a quadratic takes equal values only symmetrically about its axis, $f(2)=f(8)$ forces the axis to be $x=(2+8)\mathrm{/}2=5$, where the given minimum 3 sits. Write $x=5+k$, so $f(5+k)=3+a{k}^2$ for some $a$. From $x=2$, $k=-3$, so $3+9a=12$, hence $a=1$ and $f(5+k)=3+k^2$. The sum over $k=-3,-2,-1,0,1,2,3$ is $7(3)+2(1+4+9)=21+28=49$, found without solving for $b$ and $c$ or evaluating seven separate quadratics.

Draw one diagonal, say $AC$. In triangle $ABC$ the segment joining the midpoints of two sides is parallel to $AC$ with linear ratio one half, so its small corner triangle has area one quarter by squaring the linear ratio. The same holds for the corner at $D$ in triangle $CDA$, so those two opposite corners together occupy one quarter of the whole area. Repeating with the other diagonal shows the remaining two corners occupy another quarter. The four corners total one half, leaving one half of 48, namely 24, for the midpoint quadrilateral.

The idea is repeating-decimal period with a decimal shift for percent. Dividing 5 by 7 gives 0.714285 repeating, so multiplying by 100 to make a percent gives 71.428571 repeating with the same fractional repetend 4, 2, 8, 5, 7, 1 of length 6. Position in the repetend is determined by the remainder modulo 6. Since 40 leaves remainder 4 upon division by 6, the wanted digit is the fourth entry of the cycle, which is 5. Writing out 40 places by hand invites alignment error, while remainder reasoning needs only the 6-digit cycle.

The insight is that a positive proper fraction cannot be an integer, combined with quadratic growth dominating linear growth. For $n>6$, $n^2+1-(5n+7)=n^2-5n-6=(n-6)(n+1)>0$, so $0<{\displaystyle \frac{5n+7}{n^2+1}}<1$, which is rational but cannot be an integer. Hence only $1\le n\le 6$ can possibly give integers. Checking gives $12\mathrm{/}2=6$, $17\mathrm{/}5$, $22\mathrm{/}10$, $27\mathrm{/}17$, $32\mathrm{/}26$, $37\mathrm{/}37=1$, so only $n=1$ and $n=6$ give integers. Since the denominator is never zero, every value is rational, so the non-integers are $50-2=48$.

The insight is that perpendicular integer slopes are divisors of 1, collapsing infinitely many possibilities to two cases, and the height bound then selects. Perpendicular means slopes $a$ and $c$ satisfy $ac=-1$. With integers this forces $(a,c)=(1,-1)$ or $(-1,1)$. Through $(1,2)$, $L$ is $y=x+1$ in the first case and $y=-x+3$ in the second. Through $(9,6)$, $M$ is $y=-x+15$ in the first case and $y=x-3$ in the second. Intersections are $x+1=-x+15$ giving $(7,8)$ and $-x+3=x-3$ giving $(3,0)$. Only $(7,8)$ has $y$ greater than 5, so the surviving $L$ is $y=x+1$ with $y$-intercept 1.

The insight is a frequency inequality from the mode combined with ordered counting and an algebraic bound. The mode being 2 forces the unknown frequency $k$ to exceed 5, since 5 is the next largest frequency. Then 9 families have fewer than 2 children and 5 have more than 2, so with $k$ exceeding 5 the middle position or middle pair must lie among the 2s for both odd and even totals, giving median 2. The total children equal $5+2k+9+8=22+2k$ over $14+k$ families, and $22+2k$ is less than twice $14+k$ because 22 is less than 28, so the mean is less than 2 and therefore less than the median.

The insight is the extremal push: moving one marble from a smaller color pile to a larger pile always increases the number of same-color pairs, so the maximum occurs at the most lopsided allowed split. If one pile has $m$ and another has $n$ with $m$ at least $n$, changing them to $m+1$ and $n-1$ adds $m$ new pairs but removes only $n-1$, for a net gain of $m-n+1$, which is positive. Repeating this forces the distribution to 11, 2, and 2, the only extreme using 15 marbles with at least 2 of each color. Then same-color pairs are $55+1+1=57$ since $11\ast 10\mathrm{/}2=55$, out of $15\ast 14\mathrm{/}2=105$ total pairs, so $57\mathrm{/}105=19\mathrm{/}35$, and any balancing step lowers this count.

Count strings by number of heads using gaps instead of listing. With no adjacent heads, heads must sit in gaps between tails. For length 5, zero heads gives 1 string, one head gives 5, two heads gives 6 ways to choose nonadjacent positions, and three heads gives only 10101. More than three is impossible. The total is $1+5+6+1=13$ favorable strings. Since each of the $2^5=32$ toss sequences is equally likely, the probability is 13 over 32.

The three angles add to 180 degrees. Every prime larger than 2 is odd. If none of the three angles were 2, all three would be odd, and odd plus odd plus odd is odd, which cannot equal the even total 180. Therefore at least one angle equals 2. There cannot be two angles equal to 2, since 2 plus 2 leaves 176, which is not prime, and three angles of 2 add to only 6. Hence exactly one angle is 2, realized for instance by 2, 89 and 89, so the smallest angle is forced to be 2.

The insight is projection invariance: all the small horizontal ledges tile the footprint, combined with summing an arithmetic sequence for the sides. From above, the exposed top plus all ledges exactly cover a 6 by 6 square, area 36; the bottom is another 36. From the front, each layer contributes a strip of its full width, so the silhouette area is $6+5+4+3+2+1=21$. The same holds for the back, left, and right, giving $4(21)=84$ of vertical area. The total is $36+36+84=156$. Counting every cube face and subtracting contacts is far longer and invites missing hidden faces.

The key is order preservation plus linearity of averaging, with slope from ranges. Because $x$-coordinates are distinct and slope is positive, sorting by $x$ sorts by $y$ the same way, so the middle $x$ and middle $y$ belong to the same point; thus $(10,22)$ is on the line. Averaging the five point equations shows the mean point is also on the line. The slope magnitude is range over range, $24\mathrm{/}8=3$, and positivity makes it positive 3. Hence moving $x$ from 10 to its mean 11 lifts $y$ by 3, so the mean of $y$ is $22+3=25$. Recomputing five individual $y$ values would be long, while these two points finish it.

The decisive observation is about prime exponents: write $n$ with prime exponents $e$, so $n^2$ has exponents $2e$, and a cube root is integral exactly when every exponent is a multiple of 3. Thus 3 divides $2e$ for each prime, and since 3 does not divide 2, the number 3 must divide $e$, so every exponent of $n$ is a multiple of 3 and $n$ itself is a perfect cube. Conversely every perfect cube works because if $n=t^3$ then $n^2=t^6=(t^2{)}^3$. Counting cubes with $1\le t^3\le 1000$ gives $t=1$ through 10, for 10 values. Checking all 1000 integers or confusing squares with cubes leads to a long or wrong count.

The insight is to form one ratio with the quotient rule and then use the linear equation to find opposite extremal cases. The ratio of the first quantity to the second is $8^{m-4}{9}^{n-7}$, so after canceling the comparison becomes $9^{n-7}$ against $8^{4-m}$. Both $m=1,n=13$ and $m=9,n=1$ satisfy $3m+2n=29$ because $3+26=29$ and $27+2=29$. In the first case the ratio is $9^6\mathrm{/}{8}^3=(81\mathrm{/}8{)}^3$, and $81\mathrm{/}8>1$ so the first quantity is larger. In the second case the ratio is $8^5\mathrm{/}{9}^6=(8\mathrm{/}9{)}^5(1\mathrm{/}9)$, and since $(8\mathrm{/}9{)}^5<1$ the ratio is below $1\mathrm{/}9<1$, so the second quantity is larger. Two admissible cases give opposite orders, so the relationship varies.

The interior angles of any 8-sided polygon add to $(8-2)$ times 180, which is 1080 degrees. Call the smallest angle $s$. Because the eight measures are distinct integers larger than $s$, the other seven are at least $s+1$ through $s+7$ in some order. Adding those lower bounds gives a total of at least $8s+28$. Since the true total is 1080, we have $8s+28$ at most 1080, so $8s$ is at most 1052 and $s$ is at most 131. Hence the smallest angle is always below 132, and the fixed number 132 is always larger.

The clever move is to count faces by direction instead of by cube. Every vertical stack contributes exactly one top face, so the top shows 16 squares and the bottom shows 16 squares. From the front the steps show $4+3+2+1=10$ squares, and the same count appears from the back, left, and right, giving 40 side squares and 72 in total. The volume is the sum of square layers $16+9+4+1=30$. Twice that volume plus ten is 70, which is two less than the surface count, so the surface-area number is larger. A student who counts cube by cube faces 30 separate tallies, while the projection finishes in seconds.

The only prime that matters for divisibility by 9 is 3. On one die the faces 3 and 6 each supply exactly one factor 3, while the other four faces supply none; no face supplies two. So the product has two factors 3 exactly when at least two dice show 3 or 6. Count ordered outcomes: three multiples gives $2^3=8$; exactly two multiples gives $3\cdot 2^2\cdot 4=48$. The total is 56 of 216 equally likely ordered triples, which reduces to 7 over 27.

Factor first: $f(n)=(n-2)(n-3)$. For $n$ at least 5 both factors are at least 2, so the product is composite. Also $n^2-5n=n(n-5)$ is always even because $n$ and $n-5$ have opposite parity, so $f(n)$ is even for every integer $n$. An even prime must equal 2. Solving $f(n)=2$ gives $n^2-5n+4=0$, or $(n-1)(n-4)=0$, so $n=1$ or $n=4$, both giving value 2. The zeros at $n=2,3$ are not prime. Hence 2 of 20 values work.

The key reframe is to sort the numbers by remainder upon division by 4 and then split the favorable pairs by type. From 1 to 21 there are 6 numbers leaving remainder 1 and 5 numbers in each of the other three remainder classes. A sum is a multiple of 4 exactly for remainder pairs $1+3$, $2+2$, and $0+0$. Counting unordered pairs without replacement gives $6\ast 5=30$ of the first type and $5\ast 4\mathrm{/}2=10$ of each of the other two types, for $30+10+10=50$ favorable pairs. The total is $21\ast 20\mathrm{/}2=210$, so the probability is $50\mathrm{/}210=5\mathrm{/}21$, a little below $1\mathrm{/}4$ because the remainder classes are uneven.

The insight is symmetry: drop perpendiculars from the short base to the long base. The overhang on each side is $(14-8)\mathrm{/}2=3$, so with feet on the long base the diagonal from the top vertex projects to $14-3=11$ on the long base. Hence the height $h$ satisfies $h^2+11^2=13^2$, so $h^2=169-121=48$ and $h=4\sqrt{3}$. The area is the average of the bases times the height, $(8+14)\mathrm{/}2\cdot 4\sqrt{3}=11\cdot 4\sqrt{3}=44\sqrt{3}$.

The insight combines the less-used lower form of the triangle inequalities with the perimeter to cap every side, then rounds with integrality. From a+c>b and b+c>a we get c>b-a and c>a-b, so c exceeds |a-b|. From a+b>c and a+b+c=20 we get 20-c>c, so c is below 10. Since all sides are integers, |a-b| is at most c-1 and c is at most 9, hence |a-b| is at most 8. For example 2,9,9 gives |2-9|=7. Since 8 is below 9 with only small integer comparisons after the two bounds, Column B is always greater.

The key is to factor each factor as a difference of squares and then cancel diagonally across the product. Write $1-1\mathrm{/}{k}^2$ as $(k^2-1)\mathrm{/}{k}^2=(k-1)(k+1)\mathrm{/}{k}^2$. For $k=2$ to 9 the numerator collects 1 through 8 times 3 through 10 while the denominator collects 2 through 9 twice. Canceling 2 through 9 from both top and bottom leaves 1 times 10 over 2 times 9, which is $10\mathrm{/}18=5\mathrm{/}9$. Expanding everything over one common denominator or multiplying eight decimals would be extremely long, but factoring reduces the whole comparison to four one-digit numbers.

The key is parity collapse plus double counting. With three clubs an odd membership count can only be one or three, so every student is either in exactly one club or in all three. Count memberships: $41+43+45=129$. If $t$ is the number in all three and $s$ is the number in exactly one, then $s+t=99$ and $s+3t=129$. Subtracting eliminates $s$ and leaves $2t=30$, so $t=15$. This is attainable with only-club counts 26, 28, and 30, which sum to 84 and meet each club total. The arithmetic is short once the two region types are seen.

The key is double counting the overlap plus divisibility pinned by a range. Let $t$ be the number in both clubs. Then chess has $4t\mathrm{/}3$ members because $t$ is three fourths of it, and debate has $6t\mathrm{/}5$ members because $t$ is five sixths of it. For these to be whole numbers $t$ must be a multiple of 3 and of 5, so $t=15k$. By inclusion exclusion the union is $4t\mathrm{/}3+6t\mathrm{/}5-t=23t\mathrm{/}15=23k$, and with 7 in neither the grade has $23k+7$ students. Being more than 50 and fewer than 70 forces $k=2$, since 30, 53, and 76 are the nearby totals. Thus $t=30$, giving 40 chess and 36 debate members with union 46.

The sheet can become either side of the cylinder. Unrolling says circumference times height is the sheet, so if 6 is the circumference the radius is $3\mathrm{/}\pi$ and the volume is $90\mathrm{/}\pi$, while if 10 is the circumference the radius is $5\mathrm{/}\pi$ and the volume is $150\mathrm{/}\pi$. Against 40, the first needs 90 versus $40\pi$, or 9 versus $4\pi$, which holds as below because $\pi$ exceeds 2.25. The second needs 150 versus $40\pi$, or 15 versus $4\pi$, which holds as above because $\pi$ near 3.14 is below 3.75. One admissible roll sits below the threshold and the other sits above, so the comparison cannot be settled. Assuming one rolling direction is the shortcut that traps most solvers.

The insight is to unfold the two faces that meet at the shared base vertex into a flat rhombus and then use coordinates with the Pythagorean Theorem. The faces containing $M$ and $N$ share a lateral edge from the base vertex to the apex, so any route on the triangles from $M$ to $N$ must cross that edge. Lay those two equilateral triangles flat about that edge to get a rhombus with side 8 and diagonal 8. Place the shared base vertex at $(0,0)$ and the apex image at $(8,0)$; the other vertices are at $(4,4\sqrt{3})$ and $(4,-4\sqrt{3})$ since each altitude is $\sqrt{8^2-4^2}=4\sqrt{3}$. Then $M=(2,2\sqrt{3})$ and $N=(2,-2\sqrt{3})$, so the straight segment has length $4\sqrt{3}$. Any route crossing the fold corresponds to a bent path in the net, hence longer.

The trick is to split the unknown functions through their sum and difference and then parametrize the difference by its roots. Let $S(x)=P(x)+Q(x)=2x^2+4x+6$ and $D(x)=P(x)-Q(x)$. Then $P=(S+D)\mathrm{/}2$. Since $D(1)=0$ and $D(2)=0$ and $D$ has degree at most two, $D(x)=a(x-1)(x-2)$ for some real $a$. At $x=3$, $S(3)=36$ and $D(3)=2a$, so $P(3)=18+a$. This is undetermined because $a$ can vary. For example $P(x)=3x^2-4x+7$ with $Q(x)=-x^2+8x-1$ satisfies all conditions and gives $P(3)=22$, which exceeds 18, while swapping the two polynomials satisfies all conditions and gives $P(3)=14$, which is below 18. Hence different admissible pairs give different orders.

The invariant is the second difference of a quadratic, which does not depend on where it is taken. Since $P$ and $Q$ are quadratics with positive leading coefficients, their sum has the form $S(x)=a{x}^2+bx+c$ with $a>0$. Expanding gives $S(n+1)-2S(n)+S(n-1)=a((n+1{)}^2-2n^2+(n-1{)}^2)=a(2)=2a$ because the terms in $b$ and $c$ cancel completely. Hence the first quantity equals $2a$, which is positive since $a>0$, while the second quantity is 0. Trying to recover $S$ from unknown $P$ and $Q$ is impossible, but the algebra shows the result is always a fixed positive number.

Drop perpendiculars from the ends of the short diagonal side to the base of length 6, calling the horizontal offset $x$ and the height $h$. Then the leg of length 8 gives $x^2+h^2$ equals 64 by the right-triangle relation. The two diagonals span $6+x$ and $6-x$ horizontally with the same height, so $p^2$ equals $(6+x{)}^2+h^2$ and $q^2$ equals $(6-x{)}^2+h^2$, with signs handling acute and obtuse cases. Adding cancels the cross terms, leaving 2 times 36 plus $2x^2+2h^2$, which is 72 plus 2 times 64, equal to 200 for every shape.

The insight is to create the missing factor $2-1=1$ and telescope with difference of squares. Multiplying the first quantity by $2-1$, which equals 1, gives $(2-1)(2+1)=2^2-1$. Multiplying by $(2^2+1)$ gives $2^4-1$, then by $(2^4+1)$ gives $2^8-1$, then by $(2^8+1)$ gives $2^{16}-1$, and finally by $(2^{16}+1)$ gives $2^{32}-1$, using $(a-1)(a+1)=a^2-1$ and $(2^k{)}^2=2^{2k}$ at each step. Since the extra factor is 1, the first quantity equals $2^{32}-1$, which is one less than the second quantity, so the second quantity is larger. Direct multiplication would need a nine digit product.

The insight is the gap construction: place the tails first and choose gaps for the heads, after which the power count is immediate. Exactly 3 heads means 5 tails, and 5 tails in a row create 6 gaps including both ends. Choosing 3 of those gaps for single heads guarantees no adjacency, giving $6\ast 5\ast 4\mathrm{/}6=20$ favorable sequences. The total number of sequences is $2\ast 2\ast 2\ast 2\ast 2\ast 2\ast 2\ast 2=256$ since each toss has 2 outcomes, so the probability is $20\mathrm{/}256=5\mathrm{/}64$, while listing all triples of positions would be long.