ISEE Upper Quant challenge

The idea is fixed-sum product maximization by symmetry. Expanding gives (a + 1)(b + 1) equals ab plus a plus b plus 1, which is ab plus 61 because the sum is 60, so maximizing the expression is the same as maximizing ab. Writing the pair symmetrically about the mean 30 as 30 plus t and 30 minus t gives ab equals 900 minus t squared, largest when the absolute value of t is smallest. Different even integers summing to 60 force t to be a nonzero even integer, so the smallest allowed absolute value is 2, attained at 28 and 32, giving 29 times 33 equals 957. Guessing nearby integers without enforcing parity misses the restriction.

The insight is to convert percents to reduced fractions and pin the total with a least common multiple plus the interval. Since $37.5\mathrm{\%}=375\mathrm{/}1000=3\mathrm{/}8$, the total $T$ is a multiple of 8, and $28\mathrm{\%}=28\mathrm{/}100=7\mathrm{/}25$, so $T$ is a multiple of 25. Hence $T$ is a multiple of 200. The only multiple strictly between 300 and 450 is 400. Then band is $3\mathrm{/}8$ of 400, which is 150. Testing every total from 301 to 449 for whole counts is long, while the fraction plus interval reasoning is short.

Let the radius be $r$ and height $h$ with both positive integers. Canceling $\pi$ from the volume gives $r^{2}h=36$, so $r^2$ must divide 36. Hence $r$ is 1, 2, 3, or 6 with heights 36, 9, 4, and 1. The curved area is $2\pi rh$, which simplifies with $h=36\mathrm{/}{r}^2$ to $72\pi \mathrm{/}r$, giving $72\pi$, $36\pi$, $24\pi$, and $12\pi$. Against $24\pi$ these are above, above, exactly at, and below. For instance radius 2 gives above while radius 6 gives below, so admissible cylinders fall on different sides and the comparison cannot be settled. Stopping after one divisor is the trap.

The insight is that two pairs with the same midpoint $x$-value must give the same total on any straight line, because the line value preserves averages. Write the line as $y=mx+b$, so $r=3m+b$, $s=4m+b$, $t=10m+b$, and $u=11m+b$. Then $r+u=14m+2b$ and $s+t=14m+2b$, since $3+11=14$ and $4+10=14$. Equivalently, both pairs are centered at $x=7$, so each sum is twice the line value there. The totals match for every slope and intercept, even though the four values themselves are unknown.

The insight is to draw both diagonals of the quadrilateral. In the triangle cut off by one diagonal, the segment joining two side midpoints is parallel to the diagonal and half its length, so the corner triangle is similar with linear ratio $1\mathrm{/}2$ and area ratio $1\mathrm{/}4$ of that large triangle. The same holds at the opposite corner, so those two opposite corner triangles together are $1\mathrm{/}4$ of the whole quadrilateral. Repeating with the other diagonal shows the other two corners are also $1\mathrm{/}4$ of the whole. Hence the four corners are $1\mathrm{/}2$ of the area and the midpoint quadrilateral is the other half, $48\mathrm{/}2=24$.

The insight is size weighting combined with comparing totals over a common count. The first club has $10+2=12$ members with total 10 times 80 plus 2 times 60, which is $800+120=920$, so its overall average is 920 over 12. The second club has $2+10=12$ members with total 2 times 90 plus 10 times 70, which is $180+700=880$, so its overall average is 880 over 12. Since 920 exceeds 880 with the same denominator, the first average is greater even though the second club leads in each grade separately, so averaging the two grade averages gives the wrong order.

The insight is to order the sides and bound the largest. Write $a\le b\le c$ to avoid double-counting congruent triangles. Since $a+b>c$ and $a+b+c=12$, replacing $a+b$ by $12-c$ gives $12-c>c$, so $c<6$. Also $c\ge 4$ since it is largest. For $c=5$, $a+b=7$ with $a\le b\le 5$ gives $(2,5,5)$ and $(3,4,5)$, both satisfying the inequality. For $c=4$, $a+b=8$ forces $(4,4,4)$. No other $c$ works, so there are 3 triangles.

The insight is to multiply the geometric sum by 3 so most terms cancel. Let $S$ be the first quantity $3^{-1}+\cdots +3^{-10}=1\mathrm{/}3+\cdots +1\mathrm{/}{3}^{10}$. Then $3S=1+3^{-1}+\cdots +3^{-9}=1+S-3^{-10}$, because shifting the exponents adds the leading 1 and drops the last term. Hence $2S=1-3^{-10}$. Since $3^{-10}=1\mathrm{/}59049>0$, $2S$ is 1 minus a tiny positive amount, so $2S<1$ and $S<1\mathrm{/}2$. Thus the first quantity is below the second quantity $1\mathrm{/}2$. Adding ten fractions over 59049 directly would be infeasible without a calculator.

The insight is to convert repeating percents to fractions and turn the comparison into integer bounds. Since $16.\bar{6}\mathrm{\%}=1\mathrm{/}6=7\mathrm{/}42$ and $27.\overline{27}\mathrm{\%}=3\mathrm{/}11$, the condition is $1\mathrm{/}6<k\mathrm{/}42<3\mathrm{/}11$, so $7<k<126\mathrm{/}11$. Here $126\mathrm{/}11=11.45\dots$. With $1\le k\le 10$ this leaves $k=8,9,10$, which is 3 values. Comparing decimals digit by digit is long and fails at repeating tails, while cross-multiplication gives two short inequalities with strict endpoints doing real work.

The move is to divide by $x$ to expose reciprocal symmetry and then square it. Since $x=0$ does not satisfy $x^2-6x+1=0$, $x\mathrm{\ne }0$, so dividing gives $x-6+1\mathrm{/}x=0$, hence $x+1\mathrm{/}x=6$. Squaring gives $x^2+2+1\mathrm{/}{x}^2=36$, so $x^2+1\mathrm{/}{x}^2=34$. Trying to factor $x^2-6x+1$ over integers fails and the quadratic formula is out of bounds, so solving for $x$ directly is a dead end. The reciprocal reframe plus squaring gives 34 versus 6 with almost no arithmetic, so Column A is greater.

The insight is to combine the geometric definition with prime-exponent factoring instead of guessing an integer ratio. Let the ratio be $r$, so $a_6=a_1{r}^5$, giving $r^5=729\mathrm{/}96=243\mathrm{/}32$. Since $243=3^5$ and $32=2^5$, this is $(3\mathrm{/}2{)}^5$, and positivity forces $r=3\mathrm{/}2$. Then work forward three steps: 96 times $3\mathrm{/}2$ is 144, times $3\mathrm{/}2$ again is 216, and once more is 324, so the fourth term is 324. Students who insist on an integer ratio get stuck or miscount positions.

The key is a counting re-frame of range plus averaging. A pair of scores can be the minimum and maximum of a triple exactly when the third score is chosen from between them. Hence each unordered pair contributes its difference multiplied by the number of scores strictly between its endpoints. The nonzero contributions are 7 once, 38 twice, 39 three times, 32 once, 33 twice, and 32 once, summing to 330. Dividing by the 10 triples gives mean 33. Listing all ten triples and their ranges is much longer, while weighting pairs by interior choices collapses it.

The insight is a geometric re-cut showing lattice area is always a half-integer, combined with rational closure. Enclose the triangle in the smallest axis-aligned rectangle with integer side lengths; its area is an integer. Removing up to three right triangles with integer legs, each of area $(leg1\times leg2)\mathrm{/}2$, leaves the triangle area as an integer minus half-integers, hence a multiple of 0.5 and therefore rational. So $5.25=21\mathrm{/}4$ can never occur. The other values do occur, for example $(0,0)$, $(12,0)$, $(0,1)$ gives area 6, $(0,0)$, $(13,0)$, $(0,1)$ gives 6.5, and $(0,0)$, $(14,0)$, $(0,1)$ gives 7 by base times height over two.

The key insight is that similarity does not say which side matches which, so all three correspondences must be tested and then filtered by the integer requirement. Divide 9,12,15 by 3 to get shape 3,4,5. If 20 matches 9, the sides are $20,80\mathrm{/}3,100\mathrm{/}3$ with perimeter 80, rejected for non-integers. If 20 matches 12, the sides are 15,20,25 with perimeter 60, all integers. If 20 matches 15, the sides are 12,16,20 with perimeter 48, all integers. The integer options give perimeters 60 and 48, so the greatest is 60.

The insight is that the fixed total turns maximizing the largest square into minimizing three distinct smaller squares, and square sparsity makes the top candidates fail quickly before one construction succeeds. The total is 30 times 4, which is 120. If the largest were 100, the other three would total 20, but the smallest three distinct squares 1, 4, 9 already total 14 and the next option 1, 4, 16 totals 21, so 20 is impossible. If the largest were 81, the other three would total 39, yet with 36 the remainder 3 cannot be two distinct squares, with 25 the remainder 14 cannot be two distinct squares, and smaller top choices leave remainders that also fail, so 81 is impossible. With largest 64, the other three total 56, achieved by 4, 16, 36, giving the set 4, 16, 36, 64 with mean 30.

Write a divisor as $2^a{3}^b$ with $0\le a\le 4$ and $0\le b\le 4$. There are $5\cdot 5=25$ such pairs, so 25 divisors. A divisor is a square exactly when both exponents are even. Even choices for $a$ are 0,2,4 and likewise for $b$, giving $3\cdot 3=9$ squares. Hence 9 of 25 divisors are squares. Listing all divisors is long, but the exponent test makes it a short product count.

Walking once around any convex pentagon turns through 360 degrees, so the five exterior angles add to 360 and their mean is 360 divided by 5, which is 72. The median need not equal the mean. For example the exterior measures 60, 65, 71, 72 and 92 add to 360, use only angles below 180, and have median 71 below the mean 72. By contrast 50, 60, 80, 85 and 85 also add to 360, use only angles below 180, and have median 80 above the mean 72. Both lists can occur as the exterior angles of some convex pentagon, so sometimes the mean is larger and sometimes the median is larger, and no fixed order holds.

The insight is the shared side gives an isosceles triangle. The square contributes 90 degrees at $A$ and the equilateral contributes 60 degrees, leaving $90-60=30$ degrees for angle $DAE$. Since $AD$ equals $AB$ and $AB$ equals $AE$, $AD$ equals $AE$, so triangle $ADE$ is isosceles with vertex 30 degrees and base angles $(180-30)\mathrm{/}2=75$ degrees. Thus angle $ADE$ is 75 degrees, leaving $90-75=15$ degrees for the lower angle, and by symmetry angle $DCE$ is also 15 degrees, with 150 degrees at the top of that lower triangle.

The key is remainder periodicity plus proportional counting. Divisibility by 3 depends only on $n$ mod 3, so test the three remainders. If $n=3k$, then $f(n)=9k^2+3k+1$ leaves remainder 1. If $n=3k+1$, then $f(n)=9k^2+9k+3$, a multiple of 3. If $n=3k+2$, then $f(n)=9k^2+15k+7$ leaves remainder 1. Hence exactly the integers with $n$ one more than a multiple of 3 work. From 1 to 99 there are 33 of them, and 100 also works, for 34 total, without evaluating one hundred quadratics.

Insight is to align the two ratios to a three-part ratio and then use primality to pin the scale. Simplify $14:21$ to $2:3$, which is $10:15$, and simplify $35:28$ to $5:4$, which is $15:12$, so with a common middle term the triple is $a:b:c=10:15:12$. Hence for some positive integer $k$ the values are $10k$, $15k$ and $12k$, and the sum is $37k$. Since $37k$ is prime and $k$ is a positive integer, $k$ cannot exceed 1, because any $k>1$ makes $37k$ divisible by both 37 and $k$, so $k=1$ and the sum equals 37, matching the second quantity exactly with short arithmetic after the alignment.

The insight is that integer scores force the totals to be integers, which pins the count, after which the missing value is a short balance. Write $82.4=412\mathrm{/}5$ and $85.5=171\mathrm{/}2$. If there are $n$ scores, the original total $82.4n$ is an integer only when $n$ is a multiple of 5, and the remaining total $85.5(n-1)$ is an integer only when $n-1$ is even, so $n$ is odd. With $n$ fewer than 12, the only odd multiple of 5 is $n=5$. Then the original total is 82.4 times 5, which is 412, and the remaining total is 85.5 times 4, which is 342. The dropped score is the difference $412-342=70$, and 80, 84, 88, 90 with 70 indeed give those means.

The insight combines clearing denominators with Simon-type factoring and a divisor restriction to force uniqueness. Since x and y satisfy 1/x+1/y=1, neither is zero and multiplying by xy gives x+y=xy, so xy-x-y=0. Adding 1 gives (x-1)(y-1)=1. Because x-1 and y-1 are integers whose product is 1, each is 1 or each is -1. The second case would give x=y=0, which cannot satisfy the original equation, so x-1=y-1=1 and x=y=2. Hence x+y=4, matching the right quantity with only short integer arithmetic after the factorization, so the two quantities are equal.

The clever move is to square both sides to get rid of the roots and then see the leftover as a perfect square. Both quantities are positive, so squaring keeps the order. Squaring gives $(\sqrt{a}+\sqrt{b}{)}^2=a+b+2\sqrt{ab}$ and $(\sqrt{2a+2b}{)}^2=2a+2b$. Subtracting gives $(2a+2b)-(a+b+2\sqrt{ab})=a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b}{)}^2$. Since $a$ and $b$ are distinct positives, $\sqrt{a}\mathrm{\ne }\sqrt{b}$, so this square is strictly positive. Hence the square of the second quantity exceeds the square of the first, and because both are positive, Column B is greater.

The key re-frame is to rationalize each difference since $\sqrt{a}-\sqrt{b}=(a-b)\mathrm{/}(\sqrt{a}+\sqrt{b})$. Hence $\sqrt{20}-\sqrt{18}=2\mathrm{/}(\sqrt{20}+\sqrt{18})$ and $\sqrt{5}-\sqrt{4}=1\mathrm{/}(\sqrt{5}+2)$ and $\sqrt{15}-\sqrt{13}=2\mathrm{/}(\sqrt{15}+\sqrt{13})$ and $\sqrt{10}-\sqrt{8}=2\mathrm{/}(\sqrt{10}+\sqrt{8})$. Now factor $\sqrt{2}$ from the last denominator: $\sqrt{10}+\sqrt{8}=\sqrt{2}\sqrt{5}+\sqrt{2}\cdot 2=\sqrt{2}(\sqrt{5}+2)$, so $\sqrt{10}-\sqrt{8}$ equals $\sqrt{2}$ times $\sqrt{5}-\sqrt{4}$ and is therefore larger. Moreover $\sqrt{15}+\sqrt{13}$ exceeds $\sqrt{10}+\sqrt{8}$ termwise and $\sqrt{20}+\sqrt{18}$ exceeds $\sqrt{10}+\sqrt{8}$ termwise, so with the same numerator 2 the number $\sqrt{10}-\sqrt{8}$ has the smallest denominator among the gap-two numbers. Thus it exceeds the other two gap-two numbers as well, making it the overall greatest. Decimal approximation without this re-frame is long and unreliable without a calculator.

Reduce powers of 2 modulo 5. They cycle 2,4,3,1 and repeat with period 4, since multiplying by 2 permutes residues. Then $2^n+1$ is 0 modulo 5 exactly when $2^n$ is 4 modulo 5, which happens exactly when $n$ leaves remainder 2 upon division by 4. Among 1 through 30 those values are 2,6,10,14,18,22,26,30, a total of 8. With 30 equally likely choices the probability is 8 over 30, which reduces to 4 over 15.

The insight is to collapse both sides to the same exponent and then use parity. By the power rule the first quantity is $(-2{)}^{n(n+1)}$ and since $4=2^2$ the second is $2^{2n(n+1)\mathrm{/}2}=2^{n(n+1)}$, where $n(n+1)\mathrm{/}2$ is an integer because the product of two consecutive integers is even. Let $k=n(n+1)$; then $k$ is even, so write $k=2t$. The first quantity is $(-2{)}^{2t}=((-2{)}^2{)}^t=4^t$ and the second is $2^{2t}=4^t$, hence they are the same number for every integer $n$. Trying only whether $n$ itself is even misses that the exponent $k$ is always even.

The insight is that terminating depends only on the reduced denominator combined with counting multiples. Write $m\mathrm{/}120$ in lowest terms with $g=\gcd (m,120)$ and denominator $120\mathrm{/}g$. Since $120=2^3\cdot 3\cdot 5$, this denominator terminates exactly when it has no factor 3, so $g$ must contain 3 and hence 3 divides $m$. Conversely every multiple of 3 cancels the factor 3 and leaves a divisor of 40, which terminates. Counting multiples of 3 from 1 to 50 gives 16. Converting all 50 fractions is long, while the prime-factor invariant makes it one division.

The insight is that the diagonal must satisfy two triangle inequalities at once. Let the diagonal be $d$. In the triangle with sides $4,12,d$ the triangle inequality gives $8<d<16$, since $12-4=8$ and $12+4=16$. In the triangle with sides $5,13,d$ it gives $8<d<18$, since $13-5=8$ and $13+5=18$. The diagonal must satisfy both, so $8<d<16$. With $d$ an integer, $d=9,10,11,12,13,14,15$. Their sum is an arithmetic sequence with 7 terms, $(9+15)\cdot 7\mathrm{/}2=24\cdot 7\mathrm{/}2=84$.

The insight is to turn acuteness into a two-sided bound. Let the smaller doubled angle be $x$, so the angles are $x$, $2x$, and $180-3x$. Acuteness gives $x<90$, $2x<90$, and $180-3x<90$, plus positivity, so $30<x<45$. With integer degrees $x=31,\dots ,44$. The largest angle is the maximum of $2x$ and $180-3x$: at $x=31$ it is 87, at $x=44$ it is 88, and 89 would require $2x=89$ or $180-3x=89$, neither solvable in integers. The triple 44,88,48 is acute and works, so the greatest possible largest angle is 88.

The trick is to never find the edges. The squared space diagonal comes from two right triangles as $l^2+w^2+h^2$. Squaring the given sum gives $(l+w+h{)}^2=l^2+w^2+h^2+2(lw+lh+wh)=64$. The surface condition says $2(lw+lh+wh)=36$, so the cross term is 36. Subtracting leaves $64-36=28$ for the sum of squares, which is exactly the squared diagonal. Hence the diagonal-squared number matches the threshold 28. Trying to solve for three edges from two equations, including hunting for integers where none work, is the long failing route, while the symmetric expansion finishes in one subtraction.

The insight is the integer-or-irrational dichotomy for square roots combined with squeezing between consecutive squares. Since $n^2+15$ is an integer, its square root is rational exactly when it is an integer, which happens exactly when $n^2+15$ is a perfect square. For $n>7$, $n^2<n^2+15<n^2+2n+1=(n+1{)}^2$ because $15<2n+1$, so the radicand lies strictly between consecutive squares and cannot be a square, hence the root is irrational. That leaves $1\le n\le 7$ to check: the values are 16,19,24,31,40,51,64, of which only 16 and 64 are squares, giving rational roots 4 and 8 for $n=1$ and $n=7$. So 2 values give rational roots and the remaining $100-2=98$ give irrational roots.

The insight is that the closest point on the continuous line is the foot of the perpendicular from the origin, and integer points can only depart from it in fixed steps. The given line has slope $-1\mathrm{/}2$, so a perpendicular through the origin has slope 2, equation $y=2x$. Solving $x+2(2x)=101$ gives $5x=101$, so the foot is $x=20.2$, $y=40.4$. For $(x,y)$ with integers on $x+2y=101$, $x$ must be odd, so the two lattice points straddling the foot are $(19,41)$ and $(21,40)$. Squared distance from the origin is $x^2+y^2$, and distance along the line grows monotonically away from the foot, so no farther lattice point can beat both neighbors. Computing $19^2+41^2=361+1681=2042$ and $21^2+40^2=441+1600=2041$ shows the second is smaller by 1, so the minimizer has $x=21$.

The insight is to factor the quadratic inside the absolute value and split by sign to expose a hidden threshold. Write n^2-10n=n(n-10). Since n exceeds 5, n is positive, so the sign is decided by n-10. For n=6, n^2-10n=-24, so the left quantity is 24 and the right quantity is 10n-n^2=24, giving equality. For n=12, n^2-10n=24, so the left quantity is 24 while the right quantity is -24, so the left quantity is greater. Because one admissible integer gives equality and another gives a strict lead for the left quantity, the relationship cannot be determined from the given information alone, and the work after factoring is only substituting two small integers.

The insight is the symmetry that the sum pins the average, combined with a parity plus extremal bound on the difference. Let the first term be $a$ and the nonzero difference be $d$, both integers. The sum is $6a+15d=66$, so $2a+5d=22$. Hence $5d$ is even, so $d$ is even, and distinctness gives $d$ is not zero, so the absolute value of $d$ is at least 2. The average is 11, and the largest term is $11+2.5$ times the absolute value of $d$, because the extremes are 2.5 steps from the center. With the absolute value of $d$ at least 2 the smallest possible largest term is $11+5=16$, achieved by 6,8,10,12,14,16 and also reversed. Since even the smallest possible largest term already exceeds 15, every admissible sequence has its largest term above 15.

Write length $l$ and width $w$. Perimeter words give $2(l+w)=16$ so $l+w=8$, and area gives $lw=10$. By Pythagoras the square of the diagonal is $l^2+w^2$. Do not solve for $l$ and $w$ since that needs messy roots. Instead use $(l+w{)}^2=l^2+2lw+w^2$, so $l^2+w^2=(l+w{)}^2-2lw$. Then $(l+w{)}^2-2lw=64-20=44$. The computed diagonal square matches 44 exactly, even though the sides are not whole numbers.

The insight is to convert the volume condition into a square-divisor condition via prime factorization. From $V=(1\mathrm{/}3)s^{2}h=900$, we need $s^{2}h=2700=2^2\cdot 3^3\cdot 5^2$. Since $s^2$ is a perfect square, its prime exponents must be even and bounded by those of 2700: exponent of 2 is 0 or 2, of 3 is 0 or 2, of 5 is 0 or 2, giving $2\cdot 2\cdot 2=8$ choices. They occur at $s=1,2,3,5,6,10,15,30$ with $h=2700,675,300,108,75,27,12,3$. Trying every $s$ up to 51 is far longer, and treating $s^2$ as an arbitrary divisor overcounts.

Sort the digits. A strictly decreasing three-digit number is determined by its set of three distinct digits, because the digits can be placed in decreasing order in only one way, and the largest digit is at least 2 so the hundreds digit is never zero. Thus decreasing numbers correspond bijectively to 3-element subsets of 10 digits. There are 10 · 9 · 8 over 6, which is 120, such subsets. With 900 three-digit numbers total, the probability is 120 over 900, which reduces to 2 over 15.