ISEE Upper · Competition-style problems · Part 5 of 8

Insight: reframe probability bounds as a compound inequality and use exponent growth with reversal. Let $r=32-2^k$ and $p=r\mathrm{/}30$, so $0<p<1$ gives $0<32-2^k<30$. From $0<32-2^k$ get $2^k<32$, so $k<5$ and $k\le 4$ since $2^4=16$ and $2^5=32$. From $32-2^k<30$ get $-2^k<-2$, so $2^k>2$ after multiplying by $-1$ and flipping, giving $k>1$ and $k\ge 2$ since $2^1=2$ and $2^2=4$. With $k\ge 0$ this leaves $k=2$, 3, and 4 with red counts 28, 24, and 16, all strictly between 0 and 30. Trying each $k$ without the bound structure is longer.

The key insight is a bottleneck bound combined with divisibility for integer sides. Let the scale be $k$, so sides are $6k$ by $9k$. Fitting gives $6k$ at most 60 and $9k$ at most 80, so $k$ is at most 10 and at most $80\mathrm{/}9$, leaving $80\mathrm{/}9$ as the binding limit. For $6k$ and $9k$ to both be integers, $k$ in lowest terms can have denominator only 1 or 3 since those divide both 6 and 9. The largest such $k$ not exceeding $80\mathrm{/}9$ is $26\mathrm{/}3$, giving 52 by 78 with area 4056 square inches.

The insight is an enclosing-box re-cut that avoids slant heights, followed by quadratic rescaling. The triangle fits in the 6 by 6 box from 0 to 6 in each direction, whose area is 36 square centimeters. The three outside right triangles have areas 6 times 2 over 2, which is 6, 4 times 4 over 2, which is 8, and 2 times 6 over 2, which is 6, totaling 20. So the drawing area is 36 minus 20, which is 16. Trying base times height on slant sides would need square roots, while the box uses only whole numbers. Linear scale 4 gives area factor 16, so the true area is 16 times 16, which is 256 square meters.

The insight is excess double-counting plus an upper cap with construction. Each triple member is counted three times in $14+15+13=42$ but once in the union 25, contributing 2 to the excess, while each exactly-two member contributes 1, so exactly-two plus twice triple equals 17. Hence triple is at most 8. This bound is attainable, for example with eight in all three, one in drama and robotics only, six only chess, six only drama, and four only robotics, which meets 14, 15, 13 and 25.

The insight is to turn range into distance plus pigeonhole pressure from the mode. The interval must contain 10 and 12, so range 0 or 1 is impossible by distance. If the range were 2, the whole list would lie from 10 to 12 with minimum 10. Since the fourth of seven numbers is 10, the first four would all be 10, giving at least four 10s but at most three 12s, so the mode would be 10, a contradiction. Thus the range is at least 3. It is attainable with 9, 9, 10, 10, 12, 12, 12, which has median 10, mode 12, and range 3.

The insight is to combine the translated difference with the algebra of reciprocals so the product appears without finding the numbers. Let the smaller be $s$ and the larger be $l$, so $l-s=6$ and $1\mathrm{/}s-1\mathrm{/}l=3\mathrm{/}16$. Since $1\mathrm{/}s-1\mathrm{/}l=(l-s)\mathrm{/}sl$, the left side is $6\mathrm{/}sl$ and the equation becomes $6\mathrm{/}sl=3\mathrm{/}16$. Multiplying gives $96=3sl$, so $sl=32$. Trying to find $s$ and $l$ themselves leads to $s^2+6s-32=0$, whose discriminant 164 is not a perfect square and cannot be factored simply, so the direct product route is the only short path.

The key insight is to locate the middle by cumulative counts and to enforce the even-count median rule that both middle numbers must cooperate. The total is $16+k$, with 12 scores at or below 20 and 4 scores at or above 40. For $k=9$ the total is 25 with median the 13th score: the first 12 are 10s and 20s, so the 13th is a 30. For $k=8$ the total is 24 with median the average of the 12th and 13th: the 12th is 20 and the 13th is 30, averaging 25, not 30. Any smaller $k$ puts the middle entirely at or below 20, so 9 is least.

The insight is to split by where $x$ falls in the order because the median formula changes. Sort the known values as 3, 5, 12, 16. If $x$ is at most 3, the median is 5, so $(36+x)\mathrm{/}5=5$ gives $x=-11$. If $x$ is between 5 and 12, the median is $x$, so $(36+x)\mathrm{/}5=x$ gives $x=9$. If $x$ is at least 16, the median is 12, so $(36+x)\mathrm{/}5=12$ gives $x=24$. The other intervals give no valid $x$. The three consistent lists have ranges 27, 13, and 21. Only the last matches the given range, so the missing number must be 24.

The insight is to balance deviations from the mean and then filter by order and maximality. Let counts of 70, 80, 90 be $a$, $b$, $c$ with $a+b+c=20$ and $70a+80b+90c=1640$. Then $c=a+4$ and $b=16-2a$, giving possibilities from $a=1$ to 7. Median 80 needs the tenth and eleventh values to average 80, and a 70 plus 90 median would need at least ten 70s, impossible here, so the middle two must be 80s. Mode 90 needs $c$ strictly largest. Checking leaves only $a=5$, $b=6$, $c=9$, with tenth and eleventh both 80 and nine nineties on top. Thus the number of 80s is 6.

The idea is to place the right angle at the origin, so the hypotenuse is the line ${\displaystyle \frac{x}{15}}+{\displaystyle \frac{y}{10}}=1$, and the square’s far corner is $(s,s)$. Putting it on the line gives $s({\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{10}})=1$, so ${\displaystyle \frac{s}{6}}=1$ and $s=6$. (Equivalently, the small triangle above the square is similar to the whole one.) The triangle’s area is ${\displaystyle \frac{1}{2}}\cdot 10\cdot 15=75$, and the square’s area is 36, so the part outside the square has area $75-36=39$.

The insight is parity of the middle pair plus a maximum-total bound. For six integers with median 11.5, the third and fourth sum to 23, so the third is at most 11. If the largest were at most 13, then the first two are at most 11 each and the last two at most 13 each, so the total is at most $11+11+23+13+13=71$, but mean 12 needs 72, impossible. Hence the largest is at least 14. The list 11, 11, 11, 12, 13, 14 has total 72, median 11.5, and largest 14.

The insight is an extremal minimal-total bound: to fit many large values under a fixed mean, make every value as small as the rules allow and see when even that minimum is too big. The total must be 10 times 12, which is 120. Suppose $k$ values exceed 15. The cheapest distinct list with $k$ such values uses $1,2$ and so on for the small values and $16,17$ and so on for the large values. For $k$ equal to 6, that cheapest list is $1,2,3,4,16,17,18,19,20,21$, whose total is 121, already above 120, so 6 or more large values are impossible. For $k$ equal to 5, the cheapest total is 105, and raising the largest value to 35 gives $1,2,3,4,5,16,17,18,19,35$ with total 120, so 5 can occur.

Group by the sum $s=x+y$. For a fixed $s$, the positive pairs are $(1,s-1)$ through $(s-1,1)$, a total of $s-1$ points. Without parity the total for $s=2$ through 11 is $1+2+\mathrm{.}\mathrm{.}\mathrm{.}+10=55$. Both even or both odd means the sum is even, since odd plus odd is even and even plus even is even, while mixed parity gives an odd sum. So keep only even $s=2,4,6,8,10$ with counts 1,3,5,7,9. Their sum is 25. Every even $s$ automatically gives same parity because $x$ and $s-x$ share parity when $s$ is even, so no further filtering is needed.

The insight is to clear denominators and complete a rectangle into factored form, then count unordered factor pairs. Multiply by $6xy$ to get $6y+6x=xy$, so $xy-6x-6y=0$. Adding 36 gives $xy-6x-6y+36=36$, which factors as $(x-6)(y-6)=36$. With $x\le y$, the factor $x-6$ is at most $y-6$, so each solution corresponds to an unordered factor pair of 36. Since $36=1\times 36=2\times 18=3\times 12=4\times 9=6\times 6$, there are 5 pairs, giving $(7,42)$, $(8,24)$, $(9,18)$, $(10,15)$, and $(12,12)$.

The key is that only the change in the total matters, with percents as fractions, plus the range as a sum-difference system. The original total is $5(34)=170$ and the new total is $5(36)=180$, a rise of 10. Let the smallest be $s$ and the greatest be $g$. The 60 percent rise adds $(3\mathrm{/}5)s$ and the 20 percent cut subtracts $(1\mathrm{/}5)g$, so $(3\mathrm{/}5)s-(1\mathrm{/}5)g=10$, or $3s-g=50$. The range gives $g=s+10$. Substituting yields $3s-(s+10)=50$, so $2s=60$ and $s=30$. Hence $g=40$. The middle three do not change and never enter the change equation.

The insight is to expand both determinants and then use divisibility together with the digit bound instead of guessing pairs. Expanding gives $5a-6$ for the first matrix and $24-3b$ for the second, so equality means $5a+3b=30$, or $5a=3(10-b)$. Thus 5 divides $10-b$. With $b$ from 1 to 9, the quantity $10-b$ ranges from 1 to 9, whose only multiple of 5 is 5, so $b=5$ and $5a=15$, hence $a=3$. The sum is 8, and both determinants equal 9. Trying all 81 digit pairs would take far too long without a calculator.

The insight is that the five pair totals form an odd ring, so an alternating sum isolates one printer because every other rate appears once with each sign. Subtracting and adding in order gives 22 minus 26 plus 30 minus 34 plus 28 equals 20, and each of printers 2 through 5 cancels while printer 1 appears twice, so twice its rate is 20 and its rate is 10 pages per minute. Solving the five equations by substitution is long, while the parity cancellation is immediate and the arithmetic uses only small whole numbers.

The insight is to combine a counting re-frame for new pairs with percent re-basing. With $n$ students originally there are $n(n-1)\mathrm{/}2$ pairs. The newcomer pairs with each of the $n$ old students, so exactly $n$ new handshakes appear. The percent increase is therefore $n$ divided by $n(n-1)\mathrm{/}2$, which simplifies to $2\mathrm{/}(n-1)$. Setting $2\mathrm{/}(n-1)=1\mathrm{/}5$ gives $n-1=10$, so $n=11$. Indeed 55 pairs become 66, an increase of 11, which is 20 percent of 55.

The insight is to rewrite every quotient as a multiple of the same power $6^n$ before adding. Write $12^{n+1}=12^n\cdot 12$ with $12^n=2^{2n}{3}^n$, so dividing by $2^n$ leaves $2^n{3}^n\cdot 12=12\cdot 6^n$. Similarly $18^{n+1}=2^n{3}^{2n}\cdot 18$, and dividing by $3^{n+2}=3^n\cdot 9$ leaves $2^n{3}^n\cdot 2=2\cdot 6^n$. Thus $E=(12+2)6^n=14\cdot 6^n$. Trying to compute the powers directly fails because $n$ is unspecified, while converting to a common base makes the like terms visible.

The insight is a weighted-average equation plus an integer-count divisibility step. Let $A$ be apples and $B$ oranges. Then $(3\mathrm{/}8)A+(5\mathrm{/}12)B=(2\mathrm{/}5)(A+B)$, so $(3\mathrm{/}8-2\mathrm{/}5)A=(2\mathrm{/}5-5\mathrm{/}12)B$. Since $3\mathrm{/}8-2\mathrm{/}5=15\mathrm{/}40-16\mathrm{/}40=-1\mathrm{/}40$ and $2\mathrm{/}5-5\mathrm{/}12=24\mathrm{/}60-25\mathrm{/}60=-1\mathrm{/}60$, this gives $A\mathrm{/}40=B\mathrm{/}60$, hence $A\mathrm{/}B=40\mathrm{/}60=2\mathrm{/}3$. Write $A=2k$ and $B=3k$. The green counts $(3\mathrm{/}8)A$ and $(5\mathrm{/}12)B$ must be integers, so $A$ is a multiple of 8 and $B$ is a multiple of 12. Thus $2k$ is a multiple of 8 and $3k$ is a multiple of 12, forcing $k$ to be a multiple of 4. The smallest such $k$ is 4, giving $A=8$, $B=12$, and total 20, with green counts 3 and 5 matching 8/20.

The insight is to turn the story into a partition equation and then use divisibility inside a strict bound. Write $p=2f$ for the number who pass and $a$ for the number absent, so $2f+f+a=32$ and hence $3f=32-a$. Fewer than 5 means $0\le a\le 4$, so $28\le 3f\le 32$. But $3f$ must be a multiple of 3, and the only multiple of 3 from 28 to 32 is 30, so $f=10$ with $a=2$. The counts 20 passes, 10 fails, and 2 absents total 32 and meet every condition, so the only possible value is 10.

The insight is to square the sum instead of approximating each root, then squeeze between consecutive perfect squares. Let $s=\sqrt{12}+\sqrt{13}$, so $s^2=25+2\sqrt{156}$. Since $144<156$, we have $\sqrt{156}>12$ and thus $s^2>25+24=49$, so $s>7$. Also $2\sqrt{156}<39$ because $4\cdot 156=624<1521=39^2$, so $s^2<25+39=64$ and $s<8$. Hence $7<s<8$, and the greatest integer strictly less than $s$ is 7. Decimal estimation without a calculator is long and unreliable, while one squaring plus two integer comparisons finishes in seconds.

The insight is to factor the difference of squares to expose the slope and intercept, then use integer divisors with the range to pin the pair. The slope is $(b^2-a^2)\mathrm{/}(b-a)=a+b$ after cancelling $(b-a)$, and the line equation gives $y-a^2=(a+b)(x-a)$ so the $y$-intercept is $-ab$. Thus $ab=21$ with $1\le a<b\le 9$, forcing $a=3$ and $b=7$ since 1 × 21 leaves the range, and the slope is $3+7=10$.

The insight is to collapse the system to a single expression and then use parity to rule out primes. Let cars be $c$ and motorcycles be $m$. Then $c+m=n$ and $c-m=t$, so $c=(n+t)\mathrm{/}2$ and $m=(n-t)\mathrm{/}2$, which are whole numbers because $n+t$ is even. Total wheels equal $4c+2m=2(n+t)+(n-t)=3n+t$. Since $3n+t=2n+(n+t)$ is the sum of two even numbers when $n+t$ is even, the total is even. Because $0<t<n$ forces $n$ at least 3 and the total at least 10, the total is an even number greater than 2 and therefore composite, never prime. Hence no pair works.

The insight is to eliminate the small weight by doubling the first total and subtracting the second, then use divisibility and positivity to bound the rest. Let small, medium, and large weigh $s$, $m$, and $l$. Then $2s+3m+4l=67$ and $4s+m+6l=73$. Doubling the first gives $4s+6m+8l=134$. Subtracting the second leaves $5m+2l=61$. So $5m=61-2l$, with $m$ and $l$ positive whole numbers. Thus $61-2l$ must be a positive multiple of 5, and using $2s+3m+4l=67$ gives $s=(76-7l)\mathrm{/}5$, so $s>0$ forces $l$ at most 10. Hence only $l=3$ gives $m=11$ and $l=8$ gives $m=9$. The greatest possible medium weight is 11 pounds.

The insight is position symmetry among draws. Imagine all 15 cubes labeled, then shuffled and laid out; each cube is equally likely to land in the sixth position, so the color of the sixth cube has the same distribution as the color of the first cube. There are 7 red cubes out of 15 total cubes, so the probability the cube in that position is red is ${\displaystyle \frac{7}{15}}$. A tree conditioned on the first five draws would need many branches, but symmetry makes the earlier draws irrelevant.

The insight is to recognize both quadratics as perfect squares and then use the interval to resolve absolute values instead of squaring blindly. Since $4x^2-12x+9=(2x-3{)}^2$ and $4x^2+4x+1=(2x+1{)}^2$, the sum is $\mathrm{\mid }2x-3\mathrm{\mid }+\mathrm{\mid }2x+1\mathrm{\mid }$. For $0<x<1$, we have $2x-3<0$ and $2x+1>0$, so this equals $(3-2x)+(2x+1)=4$, a constant. Expanding or squaring the sum directly would be long and messy, while factoring makes it immediate.

The insight is that slope forces equal fractional steps, so lattice points correspond to common divisors of the run and rise. The run is $63-3=60$ and the rise is $52-7=45$, so slope is $45\mathrm{/}60=3\mathrm{/}4$. Moving from one lattice point to the next must increase $x$ by a multiple of 4 and $y$ by the same multiple of 3, otherwise $y$ would not stay integral. Hence the number of steps equals a common divisor of 60 and 45, and the finest stepping uses their greatest common divisor, which is 15. Those 15 steps connect 16 lattice points, from $(3,7)$ by repeated $(+4,+3)$ to $(63,52)$.

The insight is symmetry: equal angles placed in vertically opposite corners force OE and OF to form a single straight line. Vertical angles give angle AOD equal to angle BOC, call it $x$. Then the remainders are angle EOD $=x-50$ and angle FOC $=x-50$, so those remainders are equal. Angle AOC is supplementary to angle AOD, so it measures $180-x$. Then angle AOF, which goes from OA through OC to OF, measures $(180-x)+(x-50)=130$. Adding angle AOE gives $50+130=180$, so points E, O, and F are collinear and angle EOF measures $180^{\circ }$. Trying to find $x$ first stalls because the crossing angle was never given.

The insight is to never solve for the legs but to get their product from the sum and the Pythagorean relation, then add areas. Let the legs be $a$ and $b$ with $a+b=33$ from $58-25$ and $a^2+b^2=625$ from $25^2$. Squaring the sum gives $(a+b{)}^2=a^2+2ab+b^2$, so $1089=625+2ab$ and $ab=232$. The triangle has area $ab\mathrm{/}2=116$ and the two squares have total area $a^2+b^2=625$, so the shaded total is $625+116=741$. Trying to solve for $a$ and $b$ leads to an irrational quadratic with discriminant 161, a long failing route, while the product shortcut finishes quickly.

The insight is to compute average speed as total distance over total time rather than averaging speeds, then use monotonicity and integer feasibility. For second-half speed $v$, the overall speed is 2 divided by $1\mathrm{/}40+1\mathrm{/}v$, which equals $80v\mathrm{/}(40+v)$ and increases with $v$. At the limit $v=96$ this equals $960\mathrm{/}17$, about 56.47, so no integer above 56 is possible. The value 56 occurs since $80v\mathrm{/}(40+v)=56$ gives $v=280\mathrm{/}3$, about 93.33, within the limit, while the arithmetic mean misses the time weighting entirely.

The insight is ratio reduction to the smallest whole-number bag combined with unordered counting without replacement. The observed $60:50:40$ reduces by 10 to $6:5:4$, summing to 15, which is the only multiple below 20 since the next multiple 30 is too large, so the bag must be 6 red, 5 blue, and 4 green. Then both blue without replacement is choosing 2 of the 5 blues out of choosing 2 of 15: 5 times 4 over 2 divided by 15 times 14 over 2 equals $10\mathrm{/}105=2\mathrm{/}21$. Using the large experimental totals directly would force huge combinations and miss the integer re-frame.

The insight is to assume the denested form and match whole-number and radical parts, turning radicals into an integer factor problem. Squaring gives $m+n+2\sqrt{mn}=11+2\sqrt{24}$, so $m+n=11$ and $mn=24$. Thus $m$ and $n$ are two positive integers with product 24 and sum 11; checking factor pairs of 24 leaves only 3 and 8. With $m<n$, $m=3$ and $n=8$, and $(n-m{)}^2=(m+n{)}^2-4mn=121-96=25$, so $n-m=5$. Approximating nested decimals is slow and inexact, while matching parts reduces everything to factoring 24.

The insight is the center of the segment. The inequality means ${\displaystyle \frac{5-k}{2}}\le x\le {\displaystyle \frac{5+k}{2}}$, a segment centered at 2.5, halfway between 2 and 3. So the integers it contains come in symmetric pairs: $\{2,3\}$, then $\{1,4\}$, then $\{0,5\}$. Exactly 6 integers means 0 through 5 are in and $-1$ is out: ${\displaystyle \frac{5-k}{2}}\le 0$ gives $k\ge 5$, and ${\displaystyle \frac{5-k}{2}}>-1$ gives $k<7$. So $k=5$ or $k=6$, and the sum is 11.

The insight is to express both intercepts through the slope and turn integrality into a divisor condition. Write the slope as $m$ with $m$ a nonzero integer. Through $(6,12)$ the equation is $y-12=m(x-6)$. Setting $x=0$ gives $y=12-6m$, always an integer. Setting $y=0$ gives $x=6-12\mathrm{/}m$, which is an integer exactly when $m$ divides 12. For both intercepts to be positive, $m$ must be negative, because a positive slope through $(6,12)$ drives the $y$-intercept down. The negative divisors of 12 are $-1$, $-2$, $-3$, $-4$, $-6$, and $-12$, a total of 6 lines.

The insight combines a diagonal re-cut with parity. Draw diagonal $AC$ to split the quadrilateral into triangles $ABC$ and $ADC$. Triangle $ABC$ with sides 5, 6, and 8 satisfies $5+6>8$, so the figure exists. In triangle $ADC$ the third side must satisfy the strict triangle inequalities $8-7<DA<8+7$, so $1<DA<15$. With integer $DA$ this gives 2 through 14, which is 13 values. The perimeter is $5+6+7+DA=18+DA$, and oddness forces $DA$ to be odd. The odd integers from 2 to 14 are 3, 5, 7, 9, 11, and 13, which is 6 values. A student who never draws the diagonal uses only the weak quadrilateral bound and overcounts badly.

The insight is a two-hour block plus remainder case work for who finishes. The six-hour pipe does $1\mathrm{/}6$ tank per hour and the four-hour pipe does $1\mathrm{/}4$, so one cycle of each gives $5\mathrm{/}12$ in 2 hours. Two cycles give $10\mathrm{/}12$ in 4 hours, leaving $2\mathrm{/}12=1\mathrm{/}6$, exactly one hour of the starting pipe, so 5 hours total. Treating the pipes as simultaneous gives 2.4, averaging gives 4.8, and rounding to full cycles gives 6.