ISEE Upper · Competition-style problems · Part 3 of 8

The insight is to equate experimental frequencies to theoretical ratios and solve the resulting linear system. The first experiment gives red fraction $50\mathrm{/}80=5\mathrm{/}8$, so reds $r$ and total $t$ satisfy $r=5t\mathrm{/}8$. After adding 12 blue, total is $t+12$ and red fraction $25\mathrm{/}60=5\mathrm{/}12$, so $r=5(t+12)\mathrm{/}12$. Setting equal gives $5t\mathrm{/}8=5t\mathrm{/}12+5$, and multiplying by 24 gives $15t=10t+120$, so $5t=120$ and $t=24$. Guessing totals without linking the two experiments through $r$ leaves two unknowns with one equation.

The insight is to convert the repetend with a power-of-ten shift combined with divisor analysis of 999. Let $ABC$ be the integer value of the block. Then $1000N-N=ABC$, so $N=ABC\mathrm{/}999$. Since $999=27\times 37$, write the reduced denominator as $999\mathrm{/}\gcd (ABC,999)$. For this to equal 37 we need $\gcd (ABC,999)=27$, so 27 divides $ABC$. Write $ABC=27k$. Then $\gcd (27k,27\times 37)=27\times \gcd (k,37)$, so we need $\gcd (k,37)=1$, hence 37 does not divide $k$. With $100\le 27k\le 999$ we get $4\le k\le 37$. The value $k=37$ gives $ABC=999$ and $N=0.\overline{999}=1$ with denominator 1, so it is excluded. The remaining $k=4$ through 36 give 33 values such as $108\mathrm{/}999=4\mathrm{/}37$, all with reduced denominator 37.

Insight: reverse the inequality when dividing by a negative and then use an extremal counting argument. From $-4x+k>5$ get $-4x>5-k$, so $x<(k-5)\mathrm{/}4$ after dividing by $-4$ and flipping the direction. Together with $x\ge -3$, the admissible integers start at $-3$ and run up to but not including $(k-5)\mathrm{/}4$. To contain exactly three multiples of 3, they must be the first three encountered from the bottom, namely $-3$, 0, and 3, so the upper bound must satisfy $3<(k-5)\mathrm{/}4\le 6$. Hence $12<k-5\le 24$, so $17<k\le 29$, giving integers 18 through 29. Among multiples of 11 only 22 lies there, and it gives $x<4.25$ with multiples $-3$, 0, and 3.

The insight is to substitute the repeated exponential and use the definition of the zero exponent. Since $3^{2k}=(3^k{)}^2$, setting $y=3^k$ turns the base into $y^2-13y+36=(y-4)(y-9)$, so the whole expression is undefined exactly when the base is zero, that is when $3^k=4$ or $3^k=9$. Powers $3^0,3^1,3^2,\dots$ run $1,3,9,27,\dots$ and skip 4, so $3^k=4$ has no integer solution, while $3^k=9$ gives $k=2$ only, which lies in the allowed range. Hence exactly one value makes the base zero. Trying values blindly still needs the factorization to be sure no other $k$ works.

The insight is that 25 is the arithmetic mean of 20 and 30, so with average defined by total distance over total time the two driving times must be equal, not the distances. Let the sections be $d1$ and $d2$. Then $(d1+d2)\mathrm{/}(d1\mathrm{/}20+d2\mathrm{/}30)=25$, which gives $d1\mathrm{/}d2=2\mathrm{/}3$. Hence the total $D=d1+d2$ is 5 times an integer. Since $D\mathrm{/}25$ is the total time, $D\mathrm{/}25$ lies between 4 and $25\mathrm{/}6$, so $D$ lies between 100 and 104.16. The only multiple of 5 there is 100. A student who thinks average 25 means equal distances is pushed to several possibilities and cannot finish.

The insight is an extremal bound plus an explicit construction. Any two distinct positive integers summing to $90^{\circ }$ cannot both be at most $45^{\circ }$, since $45+45=90$ would repeat a value, so the larger of the complementary pair is at least $46^{\circ }$ and therefore the largest of the five is at least $46^{\circ }$. This bound is attainable with distinct integers summing to $180^{\circ }$: $44+46=90$ uses the pair and $28+30+32=90$ uses three more distinct values, giving the set 28, 30, 32, 44, 46 whose total is $180^{\circ }$ and whose largest is $46^{\circ }$. Hence the minimum is achieved.

The insight is to combine the mean total with the strict triangle inequality to bound the longest side. The perimeter is 30, so with longest $c$ and smallest $a$ equals $c$ minus 9, the middle is 39 minus $2c$. Ordering alone would allow up to 16, but a triangle needs smallest plus middle greater than longest, so 30 minus $c$ greater than $c$, giving $c$ less than 15. With integers the greatest is 14, realized by 5, 11, 14 with mean 10, range 9, and 5 plus 11 greater than 14.

The idea is to find the sum of the diagonals without finding each one. The diagonals of a rhombus are perpendicular bisectors of each other, so with diagonals $p$ and $q$: ${\left({\displaystyle \frac{p}{2}}\right)}^2+{\left({\displaystyle \frac{q}{2}}\right)}^2=10^2$, which gives $p^2+q^2=400$. The area is ${\displaystyle \frac{pq}{2}}=96$, so $pq=192$. Then $(p+q{)}^2=p^2+q^2+2pq=400+384=784$, and $p+q=28$. (The diagonals are 12 and 16.)

The insight is that lowest-terms terminating decimals are controlled by the denominator combined with a short interval bound. A fraction in lowest terms terminates exactly when its denominator has no prime factors other than 2 and 5, so with exactly 3 places $n=2^a{5}^b$ with $\max (a,b)=3$, giving $n$ in $8,40,125,200,250,500,1000$. The interval $(0.4,0.41)$ has length 0.01. For $n=8$ we need an integer $m$ with $3.2<m<3.28$, impossible. For $n=40$ we need $16<m<16.4$, impossible since $m=16$ gives the excluded endpoint 0.40. For $n=125$ we need $50<m<51.25$, so $m=51$ works, and $51\mathrm{/}125=0.408$ is in lowest terms with three places. Thus the smallest working denominator is 125.

The insight is a counting re-frame by how many acute angles are chosen plus a mean equation and a feasibility bound. Let the acute be $a$ and the obtuse $180-a$, and let $k$ of the four chosen be acute, so $k$ is 0 to 4 with at most four of each available. The mean condition is $(ka+(4-k)(180-a))\mathrm{/}4=110$, so $(2k-4)a=440-720+180k$. Checking $k$ gives $k=0$ with $a=70$ using four obtuses of 110, $k=1$ with $a=50$ using one 50 and three 130s, $k=2$ impossible, and $k=3,4$ obtuse for $a$. Only $k=1$ uses both types, so the acute is $50^{\circ }$. Enumerating all 70 subsets is long.

The center is the same distance from all three points. Because $(0,0)$ and $(8,0)$ are on a horizontal line, the center must lie on their perpendicular bisector, the line $x=4$. Call the center $(4,k)$ and set its squared distance to $(0,0)$ equal to its squared distance to $(2,6)$: $16+k^2=4+(k-6{)}^2=4+k^2-12k+36$. The $k^2$ terms cancel, leaving $12k=24$, so $k=2$. The radius is the distance from $(4,2)$ to $(0,0)$, which is $\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$.

The insight is to pair complementary fractions to one and count the survivors by inclusion-exclusion. For denominator 50, ${\displaystyle \frac{p}{50}}+{\displaystyle \frac{50-p}{50}}=1$, so survivors pair to 1 and the sum equals half the count. Count $1\le p\le 49$ coprime to 50, meaning not divisible by 2 or 5. There are 24 evens and 9 multiples of 5, with 4 numbers divisible by both, so the count is $49-24-9+4=20$. Twenty numbers form ten complementary pairs, each summing to 1, so the total is 10. Listing and adding twenty numerators directly is far longer.

The insight is that the circular base must fit inside the rectangular cross section perpendicular to the axis, so its diameter is limited by the smaller side of that rectangle, and then the volume tradeoff across the three orientations must be compared because radius is squared. Along 10 the base is 6 by 8 so radius 3 and volume $90\pi$, along 8 the base is 6 by 10 so radius 3 and volume $72\pi$, and along 6 the base is 8 by 10 so radius 4 and volume $96\pi$. The shortest axis wins at $96\pi$ since squaring the larger radius outweighs the shorter height, while assuming the tallest cylinder is largest leads to the wrong orientation.

The key is to connect the sum and the difference through their squares and use $x>1$ to pick the positive root. Squaring gives $(x+{\displaystyle \frac{1}{x}}{)}^2=x^2+2+{\displaystyle \frac{1}{x^2}}=16$ and $(x-{\displaystyle \frac{1}{x}}{)}^2=x^2-2+{\displaystyle \frac{1}{x^2}}$, so the squares differ by 4 and $(x-{\displaystyle \frac{1}{x}}{)}^2=16-4=12$. Since $x>1$ gives $x>{\displaystyle \frac{1}{x}}$, the difference is positive, so $x-{\displaystyle \frac{1}{x}}=\sqrt{12}=2\sqrt{3}$. Solving $x^2-4x+1=0$ and forming the difference with radicals would be much longer.

The insight is to bound the short side by half the half-perimeter and show the stadium area strictly increases with that side, so the maximum is the square. Write the short side as $w$ with $w$ at most 9 since the sides sum to 18, and the long side is $18-w$. The shaded area is $w(18-w)$ plus one full circle of radius $w\mathrm{/}2$, namely $w(18-w)+\pi w^2\mathrm{/}4$. Write $w=9-d$ with $d$ at least 0; the rectangle part is $81-d^2$ and the circle part is $\pi (9-d{)}^2\mathrm{/}4$, and the drop from the square is $d^2+\pi (18d-d^2)\mathrm{/}4$ which is positive for $d$ positive, so $w=9$ is best. Then the area is $81+81\pi \mathrm{/}4$. Trying all widths is long, while the shift finishes quickly.

The insight is to recover the hidden red count from the without-replacement product by using consecutive-integer factoring, then use complementary counting for blue. Total pairs are 13 times 12 divided by 2, which is 78. Favorable red pairs are 78 times $6\mathrm{/}13$, which is 36. Since 36 equals 9 times 8 divided by 2, there are 9 red marbles and 13 minus 9, which is 4, blue marbles. Blue pairs are 4 times 3 divided by 2, which is 6. So the blue probability is 6 divided by 78, which is $1\mathrm{/}13$.

The insight is to shift each fraction by 1 to manufacture a common factor and then use the imbalance of the reciprocals. Subtract 1 from each term: ${\displaystyle \frac{x+7}{5}}-1={\displaystyle \frac{x+2}{5}}$ and similarly ${\displaystyle \frac{x+2}{6}}$, ${\displaystyle \frac{x+2}{4}}$, ${\displaystyle \frac{x+2}{7}}$. The equation becomes $(x+2)(1\mathrm{/}5+1\mathrm{/}6)=(x+2)(1\mathrm{/}4+1\mathrm{/}7)$. Since $1\mathrm{/}5+1\mathrm{/}6=11\mathrm{/}30$ and $1\mathrm{/}4+1\mathrm{/}7=11\mathrm{/}28$ are different, the parenthesized sums cannot cancel, so $x+2$ must be 0 and $x=-2$. Clearing denominators 4, 5, 6, 7 directly needs denominator 420 and large products, a long error-prone route without a calculator.

The insight is that the change depends on position because outer faces lost and inner walls gained differ. A corner cube loses 3 outer faces and gains 3 inner walls for net 0; an edge cube loses 2 and gains 4 for net $+2$; a face cube loses 1 and gains 5 for net $+4$, and non-touching removals add. The original cube has area $6\cdot 49=294$, so the increase is $310-294=16$. Let $c$, $e$, $f$ be the numbers of corner, edge, and face cubes. Then $c+e+f=9$, $e=2f$, and $2e+4f=16$. Hence $8f=16$, so $f=2$, $e=4$, and $c=9-6=3$. Counting every exposed square separately is much longer.

The insight is to rotate the side vector to get the adjacent side and then add vectors to get the diagonal. Side $AB$ is $(2,1)$ with slope $1\mathrm{/}2$. A 90-degree counterclockwise turn sends $(p,q)$ to $(-q,p)$, so the next side is $(-1,2)$ with slope $-2$. Adding gives diagonal $AC=(2-1,1+2)=(1,3)$, so its slope is $3\mathrm{/}1=3$. Solving with distance equations instead would be a long system with squares.

The insight is to factor the trinomial and then use the definition of prime to force a factor of 1 in absolute value, with a sign split. First $n^2+14n+40=(n+4)(n+10)$. If the product is prime it is positive, so either both factors are positive or both are negative. If both are positive, the smaller factor $n+4$ must equal 1, giving $n=-3$ and product 7, which is prime. If both exceed 1, the product is composite. If both are negative, the larger factor $n+10$ must equal $-1$, giving $n=-11$ and product 7, which is prime. A negative product cannot be prime, and factors below $-1$ in absolute value give composite absolute products. Hence exactly two integers work, while blind trial over all integers would never finish.

The insight is to bound the variables before checking divisibility, turning an infinite search into six cases. With $x\le y$, adding gives $2\mathrm{/}x\ge 1\mathrm{/}6$, so $x\le 12$, and $1\mathrm{/}x<1\mathrm{/}6$ gives $x>6$, hence $x=7,8,9,10,11,12$. From $1\mathrm{/}y=1\mathrm{/}6-1\mathrm{/}x=(x-6)\mathrm{/}6x$ we get $y=6x\mathrm{/}(x-6)$, which yields 42,24,18,15,12 for $x=7,8,9,10,12$ and a non-integer $66\mathrm{/}5$ for $x=11$. Thus there are five unordered solutions, and ordering both coordinates doubles all but $(12,12)$, giving $5\cdot 2-1=9$ ordered pairs. Guessing pairs without the bound has no stopping point, while the bound plus one division per case finishes quickly.

The insight is a parity elimination followed by a modulo-3 pigeonhole that bounds $n$ to six candidates. If $n$ is odd, then $n-1$ and $n+1$ are both even, so their absolute values are even integers; to be prime both would have to equal 2, impossible since they differ by 2. Hence $n$ is even and the three values are odd. Moreover $n-1$, $n+1$, $n+3$ cover all residues modulo 3 (they are $r,r+2,r+1$), so one is a multiple of 3; a prime multiple of 3 must be 3. Thus one absolute value equals 3, giving $n=4,-2,2,-4,0,-6$. Checking gives $\mathrm{\mid }3\mathrm{\mid },\mathrm{\mid }5\mathrm{\mid },\mathrm{\mid }7\mathrm{\mid }$ prime only for $n=4$ (3,5,7) and $n=-6$ (7,5,3), so there are 2 integers.

The insight is square divisibility with the angle sum. An even square is a multiple of 4 and an odd square leaves remainder 1 on division by 4. Since 180 is a multiple of 4, three squares summing to 180 cannot leave remainder 1 or 2, so all three must be even. Write them as squares of $2a$, $2b$ and $2c$; then $4(a^2+b^2+c^2)=180$, so $a^2+b^2+c^2=45$. Among squares below 45, 36 leads nowhere while 25 forces $25+16+4=45$. Hence the angles are 100, 64 and 16, with largest 100. Searching all square triples without the divisibility cut is much longer.

The key insight is that strict inequalities between integers become at-least-one gaps after clearing denominators, and a clever combination eliminates $p$. From $3\mathrm{/}7<p\mathrm{/}q<4\mathrm{/}9$ with positive denominators, $7p-3q$ and $4q-9p$ are positive integers, hence at least 1. Eliminate $p$ by forming $9(7p-3q)+7(4q-9p)=q$, so $q$ is at least $9+7=16$. The bound is attainable since $p=7$ and $q=16$ satisfy $49>48$ and $64>63$, meaning $3\mathrm{/}7<7\mathrm{/}16<4\mathrm{/}9$. Therefore the smallest possible denominator is 16.

The insight is to bound the smaller denominator and then enforce integrality. Subtracting gives $1\mathrm{/}n+1\mathrm{/}m=1\mathrm{/}6$. Since $n$ is smaller, $1\mathrm{/}n$ exceeds half of $1\mathrm{/}6$, so $2\mathrm{/}n$ exceeds $1\mathrm{/}6$ and $n$ is less than 12, while $1\mathrm{/}n$ is less than $1\mathrm{/}6$ so $n$ exceeds 6. Hence $n=7,8,9,10,11$. Writing $1\mathrm{/}m=1\mathrm{/}6-1\mathrm{/}n=(n-6)\mathrm{/}6n$ gives $m=6n\mathrm{/}(n-6)$. This is integral for $n=7,8,9,10$ giving $m=42,24,18,15$, while $n=11$ gives a noninteger. With $n$ smaller than $m$, the possible larger denominators are 42,24,18,15, whose smallest is 15.

The insight is to isolate the familiar near-power of ten $2^{10}=1024$ and then squeeze between powers of ten rather than computing $2^{34}$. Write $2^{34}=2^4\times (2^{10}{)}^3=16\times 1024^3$. Since $1000<1024$, the value exceeds $16\times 1000^3=1.6\times 10^{10}$, so it is at least $10^{10}$. Since $1024<1100$ and $1100^3=1.331\times 10^9$, the value is below $16\times 1.331\times 10^9=2.1296\times 10^{10}$, which is below $10^{11}$. Trapped between $10^{10}$ and $10^{11}$, the number has the form $a\times 10^{10}$ with $1\le a<10$, so $n=10$. Direct multiplication is infeasible, while one regrouping plus two integer bounds settles the exponent.

Think of distance as a multiple of each circumference and handle the offset as a congruence. Let the distance be $D$. The 6-inch wheel needs $D\mathrm{/}(6\pi )$ to be a whole number, while the 8-inch wheel starts half a turn away, so $D\mathrm{/}(8\pi )$ must be a half-integer. Hence $D\mathrm{/}\pi$ is a multiple of 6 and 4 more than a multiple of 8: $6n=8m+4$, or $3n=4m+2$. The smallest positive solution is $n=2$ with $m=1$, giving $D=12\pi$. Anyone ignoring the half turn takes the ordinary least common multiple $24\pi$ and rolls too far.

The insight is to subtract tunnels by inclusion-exclusion and notice all three pairwise intersections coincide in the same central cube. Each tunnel is 9 by 3 by 3 equals 81, so three give 243. Any two tunnels meet in the central 3 by 3 by 3 equals 27, but all three pairs are the same 27, and the triple intersection is that 27 again, so the union is $243-81+27=189$. Hence the remainder is $729-189=540$. Subtracting without correction gives 486 after a long miscount, while the coincident-overlap shortcut finishes quickly.

The insight is a case split for intersecting lines combined with a divisibility filter on the parameter. Opposite angles are equal while adjacent angles are supplementary, so either $5x+50=9x+46$ or $5x+50+9x+46=180$. The first gives $4x=4$ and $x=1$, which is not a multiple of 3. The second gives $14x+96=180$, so $14x=84$ and $x=6$, which is a multiple of 3. Then the two angles are $5(6)+50=80^{\circ }$ and $9(6)+46=100^{\circ }$, so the smaller is $80^{\circ }$. A solver who assumes one position without checking the multiple condition fails.

The insight is to multiply by 10 to expose a remainder near 1, then split on the sign. Ten times ${\displaystyle \frac{10^k-1}{10^{k+1}-1}}$ equals $(10^{k+1}-10)\mathrm{/}(10^{k+1}-1)=1-9\mathrm{/}(10^{k+1}-1)$, which increases when $k$ increases because less is subtracted. Ten times ${\displaystyle \frac{10^k+1}{10^{k+1}+1}}$ equals $(10^{k+1}+10)\mathrm{/}(10^{k+1}+1)=1+9\mathrm{/}(10^{k+1}+1)$, which decreases when $k$ increases because less is added. Every plus-fraction exceeds 1/10 and every minus-fraction is below 1/10, so the order from smallest is small minus, large minus, large plus, small plus. Hence the plus-fraction with the smaller exponent is greatest, and cross-multiplying 11-digit numbers is never needed.

Let the base area be $B$ and the height $h$. The volume gives $Bh=120$. The base perimeter is 20, so the four side faces have total area $20h$, and the surface area gives $2B+20h=148$, or $B=74-10h$. Testing values in $Bh=120$: $h=5$ gives $B=24$ and $24\cdot 5=120$, and $h={\displaystyle \frac{12}{5}}$ gives $B=50$ and $50\cdot {\displaystyle \frac{12}{5}}=120$ too, while $h=4$ and $h=6$ fail. But a base with length plus width 10 has area at most $5\cdot 5=25$, so $B=50$ is impossible. With $B=24$ the base is 4 by 6, the box is $4\times 6\times 5$, and the height is 5.