ISEE Upper · Competition-style problems · Part 2 of 8

First convert the ratio to arcs: $2+3+4=9$ parts of $360^{\circ }$, so the arcs are $80^{\circ }$, $120^{\circ }$, and $160^{\circ }$, and arc $CA$, the one not containing $B$, is $160^{\circ }$. The key idea is the quadrilateral $OAPC$, where $O$ is the center: a tangent is perpendicular to the radius, so the angles at $A$ and $C$ are both $90^{\circ }$, and the angle at $O$ is the central angle $160^{\circ }$. The four angles total $360^{\circ }$, so angle $APC=360^{\circ }-90^{\circ }-90^{\circ }-160^{\circ }=20^{\circ }$.

The insight is to collapse the two congruences into one progression and then enforce the negative bound. Write $n=6k+2$ to get remainder 2 upon division by 6. Needing remainder 5 upon division by 9 gives $6k+2$ congruent to 5 modulo 9, so $6k$ is 3 more than a multiple of 9, which forces $k$ to leave remainder 2 upon division by 3. Hence $k=3t+2$ and $n=6(3t+2)+2=18t+14$. Needing $n$ less than $-100$ gives $18t+14$ less than $-100$, so $18t$ less than $-114$ and $t$ at most $-7$. The value $t=-6$ gives $-94$, which is greater than $-100$, while $t=-7$ gives $-112$, the greatest integer below $-100$ in the progression.

The insight is that a fixed linear combination kills $n$ and traps every common divisor, turning the question into parity and multiples of 3. Compute $7A-5B=7(5n+3)-5(7n+3)=21-15=6$, so any common divisor of $A$ and $B$ must divide 6, hence is 1, 2, 3, or 6. Now $A=5n+3$ and $B=7n+3$ have the same parity as $n+1$, so both are even exactly when $n$ is odd. Modulo 3, $A$ is $2n$ and $B$ is $n$, so both are multiples of 3 exactly when $n$ is a multiple of 3. Thus the greatest common divisor exceeds 1 exactly when $n$ is odd or a multiple of 3. Up to 30 there are 15 odds and 10 multiples of 3, with 5 odd multiples of 3 counted twice, giving $15+10-5=20$.

The insight is that division by a parenthesized quotient inverts it, so a divisor placed in the denominator of the denominator becomes a multiplier, while the first divisor after 81 can never leave the denominator. Write everything as powers of 3: 81 is 3 to the fourth, 27 is 3 to the third, 9 is 3 squared, and each 3 is 3 to the first. A parenthesization gives 3 to the fourth minus third plus or minus second plus or minus first plus or minus first. To maximize, make every later exponent positive, giving 3 to the fifth. The construction 81 divided by the quantity 27 divided by 9 divided by 3 divided by 3 has inner value one third, and 81 divided by one third is 243, and no larger exponent sum is possible because 27 stays underneath.

The insight combines divisor enumeration with the Pythagorean condition. Let the sides be integers $a$ and $b$ with $ab=120$, so $(a,b)$ is a factor pair of 120: $(1,120)$, $(2,60)$, $(3,40)$, $(4,30)$, $(5,24)$, $(6,20)$, $(8,15)$, $(10,12)$. For an integer diagonal, $a^2+b^2$ must be a perfect square. Checking gives $1+14400$, $4+3600$, $9+1600$, $16+900$, $25+576=601$, $36+400=436$, $64+225=289=17^2$, and $100+144=244$. Only 8 by 15 works, with diagonal 17. The perimeter is $2(8+15)=46$. A student who tries to solve $ab=120$ with $a^2+b^2=c^2$ directly faces a hard Diophantine system.

The insight is to convert the mixed repetend with a shift combined with divisibility by 9 and then merge duplicates with the carry $0.\bar{9}=1$. Let $x=0.a\bar{b}$. Then $10x=a\mathrm{.}\bar{b}$ and $100x=10a+b\mathrm{.}\bar{b}$, so $90x=9a+b$ and $x=(9a+b)\mathrm{/}90$. As a percent this is $100x=10(9a+b)\mathrm{/}9=10a+10b\mathrm{/}9$, which is an integer exactly when 9 divides $b$, so $b=0$ or $b=9$, giving 20 pairs. But values repeat because $0.a\bar{9}=0.(a+1)$ for $a=0$ through 8 and $0.9\bar{9}=1$. Hence $b=0$ gives $0\mathrm{\%}$, $10\mathrm{\%}$, through $90\mathrm{\%}$, while $b=9$ gives $10\mathrm{\%}$ through $100\mathrm{\%}$. The union is $0\mathrm{\%}$, $10\mathrm{\%}$, through $100\mathrm{\%}$, which is 11 distinct whole-number percents.

The insight is to invert to tanks per hour and clear denominators to a factorable quadratic, then keep only the admissible positive root. Let the faster take $n$ hours, so the slower takes $n+2$. Then $1\mathrm{/}n+1\mathrm{/}(n+2)=1\mathrm{/}2.4=5\mathrm{/}12$. Clearing gives $12(2n+2)=5n(n+2)$, or $5n^2-14n-24=0$, which factors as $(5n+6)(n-4)=0$. The positive root is $n=4$, so the slower takes 6 hours. A student who adds or averages times never forms the reciprocal equation and is stuck with decimals.

The insight is to combine square area scaling with a strict unit-price inequality. The areas are in the ratio $(10\mathrm{/}20{)}^2=1\mathrm{/}4$, since area scales as the square of the diameter and pi cancels. Per square inch the smaller costs $10\mathrm{/}25=0.40$ (up to pi) and the larger costs $50\mathrm{/}100=0.50$ before discount. After a fractional discount $r$, the larger unit cost is $0.50(1-r)$. Requiring $0.50(1-r)<0.40$ gives $1-r<0.80$, so $r>0.20$. The least whole-number percent exceeding 20 percent is 21 percent, while 20 percent only ties and comparing totals instead of unit costs leads far higher.

The idea is to measure both rays from side $AB$. The altitude from $A$ meets $BC$ at a right angle, so it makes an angle of $90^{\circ }-75^{\circ }=15^{\circ }$ with $AB$. The bisector makes an angle of ${\displaystyle \frac{A}{2}}$ with $AB$. The angle between them is the difference: $\mid {\displaystyle \frac{A}{2}}-15^{\circ }\mid =15^{\circ }$, so ${\displaystyle \frac{A}{2}}=30^{\circ }$ or ${\displaystyle \frac{A}{2}}=0^{\circ }$. The second is impossible, so $A=60^{\circ }$. Check: $C=45^{\circ }$, and the bisector at $30^{\circ }$ and the altitude at $15^{\circ }$ from $AB$ differ by $15^{\circ }$.

The insight is to turn the mean into a total and couple the minimum and maximum through the range and distinctness. The total is 100 and with sorted values the largest equals the smallest plus 18. To make the median as large as possible, keep the neighbors as tight as possible, namely the two above are median plus 1 and median plus 2 in the cheapest arrangement, with smallest median minus 16 and next median minus 15, giving minimal total 5 times median minus 28. Requiring this minimum to not exceed 100 gives median at most 25, and 10, 11, 25, 26, 28 totals 100 with range 18 and median 25.

The key is to pair outer and inner factors and substitute to reveal a perfect square under the root, then use $m>0$ to take the positive root. Pairing gives $m(m+6)=m^2+6m$ and $(m+2)(m+4)=m^2+6m+8$. With $t=m^2+6m$, the inside becomes $t(t+8)+16=t^2+8t+16=(t+4{)}^2$. Thus the square root equals $\mathrm{\mid }t+4\mathrm{\mid }$, and $m>0$ makes $t+4$ positive, so the simplified form is $t+4=m^2+6m+4$. Expanding the quartic directly would be very long, while pairing makes it short.

First list the lattice points on the circle of radius 5: they come from $3^2+4^2=5^2$ and $5^2+0^2$, giving the 12 points $(\pm 5,0)$, $(0,\pm 5)$, $(\pm 3,\pm 4)$, $(\pm 4,\pm 3)$. Then test each against $(6,0)$ with the distance formula. $(5,0)$ gives 1 and $(-5,0)$ gives 11. $(3,\pm 4)$ give $\sqrt{9+16}=5$. The rest fail: $(0,\pm 5)$ give $\sqrt{61}$, $(-3,\pm 4)$ give $\sqrt{97}$, $(4,\pm 3)$ give $\sqrt{13}$, and $(-4,\pm 3)$ give $\sqrt{109}$. So exactly 4 points work.

The insight is Gauss pairing to write the sum in closed form and then a coprime prime-power split. Pair 1 with $n$, 2 with $n-1$, and so on to get $S(n)=n(n+1)\mathrm{/}2$. Since $n$ and $n+1$ are consecutive, they share no prime factor, so the prime powers 2 and 25 dividing 50 cannot split across them arbitrarily: because $S(n)$ is a multiple of 25 exactly when $n(n+1)$ is a multiple of 50, each of 2 and 25 must divide wholly one factor, with the only mixed split being 2 on one side and 25 on the other. Checking $n<50$ gives $n+1=50$ so $n=49$, $n+1=25$ with $n$ even so $n=24$, and $n=25$ with $n+1$ even, and $n=50$ would be a fourth solution but is excluded by $n<50$. Hence there are 3 values, namely 24, 25, and 49.

The insight is to bound the middle edge by its cube root and then construct the triple by testing consecutive integers rather than factoring blindly. Since 13 cubed is 2197, the product 2184 is just 13 less, which suggests middle value 13, and 12 times 13 times 14 indeed equals 2184. The surface area is then twice the sum of 156 plus 168 plus 182, which is twice 506, or 1012. A solver who multiplies out a general cubic or lists many divisors is pushed into long arithmetic, while the cube root estimate finds the triple at once.

The insight is to avoid testing one hundred numbers for divisibility and adding about fifty survivors, and instead to combine Gauss pairing with inclusion and exclusion using the distributive property. The total 1 through 100 is 100 times 101 over 2, which is 5050. Multiples of 3 sum to 3 times 1 through 33, which is 1683, multiples of 5 sum to 5 times 1 through 20, which is 1050, and multiples of both sum twice, namely multiples of 15 totaling 315. Hence the desired sum is 5050 minus 1683 minus 1050 plus 315, which is 2632, while forgetting to add back the double-subtracted multiples leaves a smaller total.

The insight is to separate nonrepeating from repeating digits by denominator factors and then count inside the period. Since $140=4$ times 5 times 7, the factors 2 and 5 contribute at most two nonrepeating places. Long division gives $11\mathrm{/}140=0.0785714285714$, so after 0.07 the block 857142 of length 6 repeats from the third place onward. The 100th place is 97 steps past the start of the period at place 3, and 97 leaves remainder 1 upon division by 6, so it is the second digit of 857142, which is 5. Direct division to 100 places would be very long, while factoring plus modular counting is short.

The idea is to spot the pair of similar triangles. Since $DM$ is parallel to $AB$, triangles $DPM$ and $BPA$ are similar, and $DM$ is half of $AB$, so $MP:PA=1:2$. Triangle $ADM$ has base $DM$, half of $DC$, and the same height as the parallelogram, so its area is ${\displaystyle \frac{1}{4}}\cdot 72=18$. Triangle $DPM$ shares vertex $D$ with it, and its base $PM$ is ${\displaystyle \frac{1}{3}}$ of $AM$, so its area is ${\displaystyle \frac{1}{3}}\cdot 18=6$.

The insight is to use parity to see when the strip endpoints are integers and then count by cases. For each integer $x$ the allowed $y$ form an interval of length 2. If $x$ is even then $3x\mathrm{/}2$ is an integer, so the interval contains three integers. If $x$ is odd then $3x\mathrm{/}2$ is a half integer, so the interval contains two integers. From 0 to 19 there are ten even and ten odd values, giving $10\times 3+10\times 2=50$ points. Listing all twenty intervals instead is long.

The insight is to combine evaluation of the sum $R=P+Q$ with bounding from nonnegative integer coefficients, rather than trying to recover $P$ and $Q$ separately, which is impossible. Write $R(x)=a{x}^2+bx+c$ with integers $a$, $b$, $c$ at least 0. Then $a+b+c=R(1)=6$ and $9a+3b+c=R(3)=48$, so subtracting gives $8a+2b=42$ and $4a+b=21$. Hence $b=21-4a$ with $b$ at least 0 gives $a$ at most 5, while $a+b$ at most 6 gives $21-3a$ at most 6 and $a$ at least 5. Thus $a=5$, $b=1$, $c=0$, so $R(x)=5x^2+x$ and $R(2)=20+2=22$. The bound and the construction together pin the answer.

The key move is to locate the median by cumulative counting and then push every score to its interval top under the order constraint. Positions 1 to 3 lie in 1 to 10, 4 to 8 in 11 to 20, 9 to 17 in 21 to 30, 18 to 22 in 31 to 40, and 23 to 25 in 41 to 50, so the 13th score is the fifth in the 21 to 30 block and equals 22. To maximize the total, set the first three to 10, the next five to 20, the first five of the middle block to 22 including the median, the last four of that block to 30, the next five to 40, and the last three to 50. The total is 30 plus 100 plus 110 plus 120 plus 200 plus 150 equals 710, and 710 divided by 25 equals 28.4.

The insight is to complete the square so the radical becomes a difference of squares, then use factor pairs. Write $n^2+8n=(n+4{)}^2-16$, so if $\sqrt{n^2+8n}=k$ is an integer then $(n+4{)}^2-k^2=16$, hence $(n+4-k)(n+4+k)=16$. Both factors are positive integers with the same parity since their sum $2(n+4)$ is even. The positive factor pairs of 16 are $(1,16)$, $(2,8)$, and $(4,4)$, but only $(2,8)$ and $(4,4)$ have matching parity. They give $n+4=5$ and $n+4=4$, so $n=1$ and $n=0$; only $n=1$ is positive, so there is exactly one such integer. Trying values without the factorization leaves a solver with an unbounded search.

The water volume is $8\cdot 12\cdot 2.5=240$ cubic inches. The insight is to decide whether the cube ends up fully under water. If it were, the depth would be $(240+64)\mathrm{/}96=3{\displaystyle \frac{1}{6}}$ inches, which is less than the cube’s height of 4, so the cube sticks out of the water. Then the water fills only the floor space around the cube, $96-16=80$ square inches, up to some depth $h$ below 4: $80h=240$, so $h=3$ inches. Since $3<4$, this is consistent.

The insight is to combine the fractions first and then watch one prime factor that powers of 10 can never supply. Since $2^{-2}=1\mathrm{/}4$ and $3^{-1}=1\mathrm{/}3$, their sum is $1\mathrm{/}4+1\mathrm{/}3=7\mathrm{/}12$, so the expression equals $7\cdot 10^n\mathrm{/}12$. For this to be an integer, $12=2^2\cdot 3$ must divide $7\cdot 10^n$, but $\gcd (7,12)=1$ and $10^n=2^n{5}^n$ contains only primes 2 and 5, never the needed factor 3. Hence no positive integer $n$ works, however large. Clearing only the 4 with $n\ge 2$ leads to a long decimal check, while the prime-factor argument settles all twenty cases at once.

The key combines a weighted-average balance with common-multiple minimality under the bound. Let the fraction of girls in January be $r$. Then $0.25r+0.10(1-r)=0.16$, so $0.15r=0.06$ and $r=2\mathrm{/}5$. Hence January girls are $2T\mathrm{/}5$ and boys $3T\mathrm{/}5$ for total $T$, so $T$ is a multiple of 5. February girls are $5T\mathrm{/}10=T\mathrm{/}2$ and boys $33T\mathrm{/}50$, so $T$ must make $T\mathrm{/}2$ and $33T\mathrm{/}50$ whole, forcing $T$ to be a multiple of 50. With $T$ fewer than 60, the only possibility is 50, which indeed gives 20 girls to 25 and 30 boys to 33. A student who checks only one integer condition is pushed into testing multiples one by one.

The insight is to group squares by their diagonal sum $s=r+c$ and count each diagonal with a triangular number instead of checking 400 squares. Here $s$ runs from 2 to 40 and for $s$ up to 21 there are $s-1$ squares while for $s$ above 21 there are $41-s$ squares. The primes in range are 2,3,5,7,11,13,17,19,23,29,31,37. Adding $1+2+4+6+10+12+16+18=69$ for the lower diagonals and $18+12+10+4=44$ for the upper gives 113. Checking squares one by one is a long failing route, while the diagonal reframe finishes with a short sum.

The key is to add the two nonlinear equations to complete the square for the sum and then use the sum-of-cubes identity without solving the system by substitution. Adding gives $a^2+2ab+b^2=36$, so $(a+b{)}^2=36$, and $a$ and $b$ positive forces $a+b=6$. Then $a(a+b)=12$ gives $a=2$ and $b(a+b)=24$ gives $b=4$, so $ab=8$. Since $a^3+b^3=(a+b{)}^3-3ab(a+b)$, the value is $216-144=72$. Substitution would lead to a quartic and very long arithmetic, while adding keeps every step short.

The insight is that the unknown route length cancels, so average speed must be total distance over total time with extra weight on the slower headwind leg rather than a simple mean of speeds. Ground speeds are 120 miles per hour one way and 80 miles per hour the other way. For a convenient 240 mile leg, the times are 2 hours and 3 hours, so the round trip covers 480 miles in 5 hours for 96 miles per hour. The same value follows from two divided by the sum of one over 120 and one over 80. Averaging the two speeds ignores unequal times and gives the trap value.

The insight is that “equally far from two points” is a straight line, because the squared terms cancel. Write $x^2+y^2=(x-6{)}^2+(y-4{)}^2$. Expanding, $x^2$ and $y^2$ disappear, leaving $0=-12x+36-8y+16$, so $12x+8y=52$, which simplifies to $3x+2y=13$. Now use parity: $2y=13-3x$ must be even, so $3x$ is odd and $x$ is odd. The odd values from 0 to 10 are 1,3,5,7,9, giving $y=5,2,-1,-4,-7$. That is 5 points.

The insight is to simplify the geometric percent to a divisor test combined with polygon angle analysis. Each interior angle is $I_n=180(n-2)\mathrm{/}n$ degrees. As a fraction of 360 this is $(n-2)\mathrm{/}(2n)$, and as a percent it is $100(n-2)\mathrm{/}(2n)=50(n-2)\mathrm{/}n=50-100\mathrm{/}n$. Since 50 is an integer, the percent is an integer exactly when $100\mathrm{/}n$ is an integer, so $n$ must divide 100. The positive divisors of 100 are $1,2,4,5,10,20,25,50,100$. Those satisfying $3\le n\le 50$ are $4,5,10,20,25,50$, for example $n=4$ gives $25\mathrm{\%}$ and $n=25$ gives $46\mathrm{\%}$. Hence there are 6 such values.

The insight is to align the two repetends to their least common period and handle the infinite carry through nines. The periods have lengths 3 and 2, so the sum repeats with period dividing 6. Write $A=876\mathrm{/}999=876876\mathrm{/}999999$ by $1000A-A=876$, and $B=59\mathrm{/}99=595959\mathrm{/}999999$ by $100B-B=59$. Adding gives $(876876+595959)\mathrm{/}999999=1472835\mathrm{/}999999=1+472836\mathrm{/}999999$, so $A+B=1.472836472836\dots$ with repeating block 472836. A student who adds only the first six digits gets 472835 and misses the carry arriving from the infinite tail. Counting positions in 472836 repeated, the 102nd place satisfies $102=17\times 6$, so it is the 6th digit of the block, which is 6.

The insight is an alternate-interior re-cut plus isosceles symmetry with a two-side case split filtered by acuteness. If $C$ is to the left of $B$, the vertex angle $ABC$ equals the $40^{\circ }$ interior angle by alternate interior angles, so the base angles are $(180-40)\mathrm{/}2=70^{\circ }$ and the triangle has angles 40, 70, 70, which is acute. If $C$ is to the right, the vertex is supplementary, $180-40=140^{\circ }$, so the base angles are $(180-140)\mathrm{/}2=20^{\circ }$ and the triangle has angles 140, 20, 20, which is obtuse. Since the triangle is acute, only the first case survives and angle $ACB$, a base angle, is $70^{\circ }$. Assuming one side without checking acuteness fails.

The insight is to complete the square in the binomial product and then use a difference of squares with divisor pairs. Expanding gives $(n+2)(n+12)=n^2+14n+24=(n+7{)}^2-25$. If this equals $k^2$ for integer $k$ at least 0, then $(n+7{)}^2-k^2=25$ and $(n+7-k)(n+7+k)=25$. Both factors are positive odd integers with the same parity since their sum $2n+14$ is even. The positive factor pairs of 25 are 1 times 25 giving $n+7=13$ and $n=6$ with $k=12$, and 5 times 5 giving $n+7=5$ and $n=-2$. Only $n=6$ is positive, with 8 times $18=144=12^2$. Testing $n=1,2$ and so on one by one would never finish.

The insight is to extract perfect squares and then use the bound $n<0$ to replace absolute value correctly. Since $\sqrt{18n^2}=\mathrm{\mid }n\mathrm{\mid }\sqrt{18}=\mathrm{\mid }n\mathrm{\mid }\cdot 3\sqrt{2}$ and $\sqrt{8n^2}=\mathrm{\mid }n\mathrm{\mid }\cdot 2\sqrt{2}$, and $\mathrm{\mid }n\mathrm{\mid }=-n$ for negative $n$, the first term is $2(-n)3\sqrt{2}=-6n\sqrt{2}$ and the second is $3(-n)2\sqrt{2}=-6n\sqrt{2}$, so $E=-12n\sqrt{2}$, which is positive as a sum of principal roots must be. Adding radicands directly or dropping roots misses the perfect-square structure and gives unlike or rootless results.