SHSAT Math challenge

The insight is to reduce the unit rate to lowest terms and reframe counting as multiples of the denominator. The line is $y=(30\mathrm{/}84)x=(5\mathrm{/}7)x$ after dividing by 12. For $y$ to be a whole number, $5x\mathrm{/}7$ must be whole, and since 5 and 7 share no factor, 7 must divide $x$. Positive multiples of 7 less than 84 are 7,14,21,28,35,42,49,56,63,70,77, which is 11 values, each giving a whole $y$ such as 5,10,15 and so on. Including $x=84$ would add one more, and including 0 would add another, while multiples of 5 swap rise and run.

The key insight is to parametrize the ratio with a common multiplier and use without replacement counting where the multiplier partly cancels. Write red as $2k$, blue as $3k$, green as $5k$, so the total is $10k$. The chance both are red is $(2k\mathrm{/}10k)$ times $(2k-1)\mathrm{/}(10k-1)$, which is $(1\mathrm{/}5)$ times $(2k-1)\mathrm{/}(10k-1)$. Setting this equal to $1\mathrm{/}29$ gives $29(2k-1)=5(10k-1)$, so $58k-29=50k-5$ and $k=3$. Thus blue is 9.

The insight is to factor each pair instead of computing one hundred squares. Each difference is $(2k-(2k-1))(2k+2k-1)=1\cdot (4k-1)$ by the difference of squares, so with $k=1$ to 50 the sum is $4(1+\cdots +50)-50$. Since $1+\cdots +50=50\cdot 51\mathrm{/}2=1275$, this is $4\cdot 1275-50=5100-50=5050$. Computing all squares directly without a calculator is extremely long, while factoring reduces each pair to a linear term and leaves only one short triangular sum.

The insight is conditional inversion to express each marginal through the overlap combined with complement and inclusion-exclusion. Let both be $x$. Since half of soccer is both, soccer is $2x$; since $40\mathrm{\%}$ of basketball is both, basketball is $x$ divided by $2\mathrm{/}5$, which is $5x\mathrm{/}2$. Union is total minus neither, $70-7=63$, and union also equals soccer plus basketball minus both, so $2x+5x\mathrm{/}2-x=7x\mathrm{/}2=63$, giving $x=18$. Soccer is 36 and basketball is 45, which check as 18 is half of 36 and $40\mathrm{\%}$ of 45. Omitting the subtracted overlap or using total 70 as union are the natural slips.

The insight is symmetry: add the two supplementary equations instead of solving for each variable. Same-side interior angles cut by parallel lines are supplementary, so $7x+10+3y+20=180$ and $3x+30+7y+10=180$, giving $7x+3y=150$ and $3x+7y=140$. Adding eliminates the asymmetry at once: $10x+10y=290$, so $x+y=29$. Solving each variable separately forces fractions $x=15.75$ and $y=13.25$ with long decimal work, while the sum is an integer found in one division. The helper addition is the whole shortcut.

The insight is a periodic block with careful remainder handling and no trailing rest. One full cycle is $45+12=57$ minutes with 45 minutes riding. Two cycles use 114 minutes and give 90 minutes riding. In 120 minutes 6 minutes remain, all riding, for 96 minutes riding total, which is 1.6 hours. At 15 miles per hour while moving, that is 1.6 times 15 equals 24 miles. Ignoring rests gives 30, dropping the partial ride gives 22.5, and adding a third rest gives 21.

The insight is to use parity of the total together with an explicit construction. The total is $7\times 12=84$, which is even, and the parity of a sum equals the parity of the count of odd terms, so the number of odds must be even; seven odds would sum to an odd number and can never make 84. Hence at most six integers can be odd. Six is attainable with $1,3,5,7,9,11,48$, which are seven distinct positive integers with six odds summing to $36+48=84$ for mean 12, so the greatest possible odd count is six.

The key combines the total-sum invariant with the fact that the median must be one of the missing numbers. The five total $5(10)=50$ and the three known total $4+12+16=32$, so the other two sum to 18. In order the five contain 4, 12, and 16, and the third value must be 10. Since none of 4, 12, or 16 equals 10, one missing number must be 10. The other is then $18-10=8$. The ordered list is 4, 8, 10, 12, 16, with mean $50\mathrm{/}5=10$ and third value 10. Splitting 18 equally would ignore the ordering condition.

The insight is to recover the single-draw red ratio from the with-replacement square and then use the without-replacement adjustment to pin the scale with a linear equation. With replacement gives red ratio squared equals $1\mathrm{/}4$, so the red ratio is $1\mathrm{/}2$, which means red equals total divided by 2. Write total as 2 times $m$ and red as $m$ for a whole number $m$. Without replacement gives $m$ times $m$ minus 1 divided by 2 times $m$ times 2 times $m$ minus 1 equals $1\mathrm{/}5$. Canceling $m$ leaves $m$ minus 1 divided by 2 times 2 times $m$ minus 1 equals $1\mathrm{/}5$, so 5 times $m$ minus 5 equals 4 times $m$ minus 2, so $m$ equals 3 and total 2 times 3 equals 6.

The insight is to expand the rotation with place value instead of listing all three-digit numbers with digit sum 12. Write $M-N=(100b+10c+a)-(100a+10b+c)=90b+9c-99a=9(10b+c-11a)$. Setting this equal to 270 gives $10b+c-11a=30$, so $10b+c=30+11a$. Since $10b+c$ is a two-digit number from 10 to 99 and $a$ is 1 to 9, only 41,52,63,74,85,96 are possible, with digit sums 6,9,12,15,18,21. Only 63 gives total 12, so $a=3,b=6,c=3$ and $N=363$. Listing all numbers with sum 12 would mean checking dozens of rotations.

The insight is to factor the difference of squares and filter by parity. Let the larger be $m$ and the smaller $n$, so $m^2-n^2=(m-n)(m+n)=48$. Since their sum is $2m$, an even number, the two factors $m-n$ and $m+n$ have the same parity. Their product 48 is even, so both must be even. Positive factor pairs of 48 with both even are 2 by 24, 4 by 12, and 6 by 8, giving $m=(2+24)\mathrm{/}2=13$, $m=(4+12)\mathrm{/}2=8$, and $m=(6+8)\mathrm{/}2=7$. Pairs like 1 by 48 mix odd and even and give half-integers, so parity eliminates them. The greatest possible larger integer is 13.

The insight is to convert repeating decimals to fractions before adding and then use parity for the digit position. Since 0.666 repeating equals $6\mathrm{/}9=2\mathrm{/}3$ and 0.272727 repeating equals $27\mathrm{/}99=3\mathrm{/}11$, the sum is $22\mathrm{/}33+9\mathrm{/}33=31\mathrm{/}33$, which is 0.939393 repeating with period 93 of length 2. Odd places after the point are 9 and even places are 3. The 50th place is even, so it is 3. Adding 50 digits with carries would be long and error prone, while fraction conversion plus parity is short.

The insight is to benchmark every fraction against $4\mathrm{/}5$ and 1 and then add the bounds. Each numerator is smaller than its denominator, so each fraction is below 1 and the sum of five fractions is below 5. Cross-multiplying shows each exceeds $4\mathrm{/}5$: $5\cdot 5>4\cdot 6$, $7\cdot 5>4\cdot 8$, $9\cdot 5>4\cdot 11$, $11\cdot 5>4\cdot 13$, and $13\cdot 5>4\cdot 16$. Hence the sum exceeds $5\cdot {\displaystyle \frac{4}{5}}=4$. Trapped strictly between 4 and 5, the greatest integer below it is 4. Finding the common denominator 6864 and adding five four-digit numerators is far longer.

The idea is periodic counting with inclusion-exclusion and density inversion corrected for remainders. Multiples of two occur every other integer and multiples of ten every tenth integer, so desired numbers are evens minus multiples of ten, counted with floors as floor of N over two minus floor of N over ten. When the range length is a multiple of ten this equals two-fifths of the range, but the stem makes the range a multiple of six, so the simple proportion predicts a non-multiple and cannot be trusted directly. Solving the floor equation for the target 63 forces checking remainders modulo thirty, which singles out one multiple of six where the two floor counts differ by exactly 63, and the count is verified by direct division. Every given, including the multiple-of-six condition, is needed to resolve the remainder ambiguity.

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 double-bound the longest side by the average from below and by half the perimeter from the triangle inequality, with ordering to avoid duplicates. Write sides with $a$ at most $b$ at most $c$ and $a$ plus $b$ plus $c$ equals 12 with $a$ plus $b$ greater than $c$. Then 3 times $c$ is at least 12, so $c$ is at least 4, and 2 times $c$ is less than 12, so $c$ is less than 6. Thus $c$ is 4 or 5. For $c$ equals 4, only 4, 4, 4 works. For $c$ equals 5, $a$ plus $b$ equals 7 with $a$ at most $b$ at most 5 gives 2, 5, 5 and 3, 4, 5. So there are 3 triangles.

The insight is to re-cut the surface instead of using the cube alone. The full 4-inch cube has $6\times 4\times 4=96$ square inches. Drilling removes two 2 by 2 openings, subtracting $2\times 4=8$ to leave 88. It adds four interior walls, each 2 by 4, adding $4\times 8=32$. The painted area is $88+32=120$ square inches. Since one ounce covers 8 square inches, the number of ounces is $120\mathrm{/}8=15$. Forgetting the interior or the openings gives the smaller quotients.

The insight is difference-of-squares pairing: each neighboring pair has the form $a^2-b^2=(a-b)(a+b)$, and since the numbers are consecutive $a-b=1$, each pair collapses to the sum $a+b$. So $20^2-19^2=39$, $18^2-17^2=35$, and so on down to $2^2-1^2=3$. Adding those ten sums gives $39+35+31+27+23+19+15+11+7+3$, which is the same as $1+2+\cdots +20=20\times 21÷2=210$. Without this collapse a student must square twenty numbers and alternate signs, which is long and error prone without a calculator.

The idea is absolute value as distance on the number line with outer-pair constancy plus median minimization. For any number strictly between the outer anchors the two outer distances sum to the distance between those anchors, here 80, because one expression drops its absolute value positively and the other negatively and the variable cancels. The total is therefore 80 plus the middle distance, which is minimized to zero only at the middle anchor 60, which lies inside the allowed interval, giving 80 as attainable minimum. Testing the center of the allowed interval fails because the anchors are asymmetric, and testing an endpoint leaves a positive middle distance plus the full outer span. Every number in the stem is used to locate the constant sum and the minimizing point.

The insight is to group complete cycles of units digits instead of computing powers. The units digits of powers of 3 cycle 3,9,7,1 with length 4, since each step multiplies by 3 modulo 10. Each block of four consecutive powers therefore contributes $3+9+7+1=20$, which ends in 0. With $100=25\cdot 4$ there are 25 complete blocks, so the total ends in 0 as well. Computing one hundred powers directly is impossible without a calculator, while the periodic grouping leaves only one short addition.

The insight is to translate the verbal rule into triples and sum the triple values by pairing instead of adding sixty numbers. Group as $(1+2-3)+(4+5-6)$ and so on. The group starting at $3k+1$ equals $3k$, giving values 0,3,6 through 57 for $k=0$ to 19. Factoring 3 leaves 3 times $0+1+\mathrm{.}\mathrm{.}\mathrm{.}+19$. Pairing $1+19$, $2+18$, and so on gives nine pairs of 20 plus 10, so 190. Hence the total is 3 times $190=570$. Adding term by term is the long failing route; grouping plus pairing is short.

Split the square by its diagonal and use containment plus inclusion-exclusion. The diagonal divides the square into two triangles. The triangle with vertex $A$ has all vertices within 10 of $A$, and since a disk is convex the whole triangle lies in the quarter centered at $A$; likewise the opposite triangle lies in the quarter centered at $C$. Thus the two quarters cover the square, so their union is 100. Each quarter is $100\pi \mathrm{/}4=25\pi$, summing to $50\pi$. The overlap counted twice is sum minus union: $50\pi -100$. Trying to integrate the lens directly is hopeless, while covering plus correcting the double count is immediate.

Visualize the tunnel as removing caps and exposing walls. The cube alone has $6\times 36=216$. Cutting the hole removes one 4 square from each of two faces, subtracting 8 to leave 208. The interior contributes four 2 by 6 rectangles, adding $4\times 12=48$. The total is $208+48=256$. Forgetting the interior gives only the subtraction, while forgetting the subtraction overcounts by the missing caps; both corrections are needed and together they finish the computation in seconds.

First convert by the linear scale, then count whole tiles along each side separately rather than dividing areas. The drawing of 5 cm by 7 cm at 1 cm to 3 feet gives a true hall 15 feet by 21 feet. Along the 15-foot side, $15\mathrm{/}2=7.5$ needs 8 tiles across, and along the 21-foot side, $21\mathrm{/}2=10.5$ needs 11 tiles down, because a leftover strip still needs a full tile and scraps are not reused. The grid therefore needs 8 by 11, which is 88 tiles, and that arrangement achieves it. Dividing total areas gives only $315\mathrm{/}4=78.75$, which misses that partial rows and columns cannot be combined.

The insight is that the median of five depends on position, so the median bound becomes an upper bound on $x$ while the mean becomes a lower bound. The sum of the known four is $10+20+30+40=100$, so mean at least 24 means $(100+x)\mathrm{/}5>=24$, giving $100+x>=120$, so $x>=20$. For the median, when $x$ is at most 20 the median is 20, when $x$ is between 20 and 30 the median is $x$, and when $x$ is at least 30 the median is 30. Requiring median at most 22 forces $x<=22$, because larger $x$ pushes the middle value above 22. Together $20<=x<=22$ with integer $x$ gives 20, 21, and 22. Ignoring the median leaves an unbounded ray, dropping 20 misses equality in at least, and adding 23 misses that the median moves with $x$.

The insight is to combine the two ratios with a common Beth term and then notice the unknown absolute speed cancels in the chase. Write Ann:Beth as $8:12$ and Beth:Cara as $12:15$, so the three speeds are $8k$, $12k$, $15k$. Cara gains on Ann at $7k$ per unit time, so catch-up time is $28\mathrm{/}(7k)=4\mathrm{/}k$. In that time Beth runs $12k$ times $4\mathrm{/}k=48$ meters. A student who tries to find $k$ thinks the problem has insufficient data, and a student who chains ratios as $2:3:5$ gets lost in a long failing route.

Insight: combine the triangle inequality as a bound with reversal when dividing by a negative. Positivity gives $15-2x>0$, so $x<7.5$. No longer than 7 gives $15-2x\le 7$, so $-2x\le -8$ and $x\ge 4$ after dividing by $-2$ and flipping. The triangle conditions give $7+12>15-2x$, so $x>-2$, and $7+(15-2x)>12$, so $22-2x>12$ and $x<5$ after flipping, while $12+(15-2x)>7$ is weaker. Hence $x\ge 4$ and $x<5$, so with integer $x$ only $x=4$ works, giving sides 7, 12, and 7 with $7+7>12$. Checking cases separately without the reversal ordering is longer and invites sign mistakes.

The insight is pigeonhole forcing of overlap. If the apple group and banana group overlapped as little as possible, their union would be as large as possible, but it cannot exceed the 30 students in the class. So the overlap is at least $18+20-30=8$ students. Conditioning on the 18 apple lovers, the smallest possible share who also like bananas is therefore ${\displaystyle \frac{8}{18}}={\displaystyle \frac{4}{9}}$.

The insight is an overlap invariant that eliminates both the middle angle and the parameter plus a proportional relationship. Let $AOB=a$, $BOC=b$ and $COD=c$, so $a+b=x+100$ and $b+c=x+60$ with $a+b+c=180$ and $a=2c$. Subtracting the two window equations gives $a-c=40^{\circ }$ with $x$ cancelling, which is the hidden invariant. With $a=2c$, $2c-c=40$ gives $c=40^{\circ }$ and $a=80^{\circ }$. Then $b=180-a-c=60^{\circ }$, and indeed $x=40$ satisfies both windows. A direct attempt to solve for four unknowns stalls, while the subtraction finishes quickly.

The insight is to translate the order into a linear Diophantine equation and kill it with divisibility plus a size bound. Let $a$ be the number of 8-packs and $b$ the number of 15-packs, so $8a+15b=192$ with $a$ at least 1 and $b$ at least 1. Since 192 and $8a$ are multiples of 8, $15b$ must be a multiple of 8, and because 15 shares no factor 2 with 8, $b$ itself must be a multiple of 8. But $b$ is less than $192\mathrm{/}15$, so below 12.8, leaving only $b=8$. Then $192-120=72$ and $a=72\mathrm{/}8=9$, so the total number of packs is $9+8=17$. The at-least-one-each condition removes the pure 24-pack all-8s solution, and testing every $b$ without the divisibility step is the long failing route.

The insight is exponent parity for squares. Every positive divisor has the form $2^a\cdot 3^b$ with $0\le a\le 5$ and $0\le b\le 3$, giving $(5+1)(3+1)=24$ equally likely exponent pairs. A divisor is a square exactly when both exponents are even. The even choices are $a=0,2,4$ with 3 options and $b=0,2$ with 2 options, so $3\cdot 2=6$ divisors are squares. Thus the probability is ${\displaystyle \frac{6}{24}}={\displaystyle \frac{1}{4}}$.

The insight is that the least common multiple takes the maximum prime exponent, which turns the condition into independent choices. Write $12=2^2\times 3$ and $60=2^2\times 3\times 5$. If $N$ used any prime besides 2, 3, or 5, that prime would appear in the least common multiple, so $N=2^a\times 3^b\times 5^c$. Then $\max (2,a)=2$ gives $a=0,1,2$; $\max (1,b)=1$ gives $b=0,1$; and $\max (0,c)=1$ gives $c=1$. The choices are independent, giving $3\times 2\times 1=6$ values, namely $5,10,20,15,30,60$.

The insight is to bound $m$ from both sides with opposite extremal companion sets and then build examples attaining each end. With longest value $m$, the five smaller distinct positive integers are at most $m-5$ through $m-1$, so the greatest total any list with longest $m$ can reach is $6m-15$. Since the actual total 30 must be reachable, $6m-15\ge 30$, so $m\ge 7.5$ and hence $m\ge 8$ for integers. The cheapest companions are 1,2,3,4,5, so $15+m\le 30$ gives $m\le 15$. Both ends occur: 1,2,3,4,5,15 totals 30 with largest 15, and shifting one unit from the largest into the gaps gives lists such as 1,2,3,4,6,14 down to 1,3,5,6,7,8, so every integer from 8 through 15 occurs.

The insight is that a linear percent enlargement squares for area, combined with map rescaling that cancels the denominator. Increasing each length by 25 percent multiplies lengths by 1.25, so areas multiply by 1.25 squared, which is 1.5625, or $25\mathrm{/}16$. Hence the original drawing area is 100 times $16\mathrm{/}25$, which is 64 square centimeters. The map scale 1 to 5 gives area factor 25, so the true area is 64 times 25, which is 1600 square meters. Dividing by 1.25 or subtracting 25 percent misses the squaring, and ignoring the copier misses the first step entirely.

The insight is to combine a weighted-average ratio with lowest-terms divisibility. Let the January notebooks be $N$ and pens be $P$. Then $1.25N+0.95P=1.10(N+P)$, so $0.15N=0.15P$ and $N=P$. Write the multipliers as fractions in lowest terms: $5\mathrm{/}4$, $19\mathrm{/}20$, and $11\mathrm{/}10$. Hence 4 divides $N$, 20 divides $P$, and 10 divides the January total $T=N+P=2N$. Since $N=P$, $N$ must be a multiple of 20, so $T$ is a multiple of 40. With $0<T<50$ and positive counts, $T=40$, giving 20 and 20 in January and 25, 19, and 44 in February.

The key insight is to recover the hidden triple-overlap percent by three-set inclusion-exclusion and then invert that percent to find the whole. Let the triple percent be $T$. The percent liking at least one sport is $100-10=90$. By inclusion-exclusion this also equals $60+50+40-25-20-20+T=85+T$. Hence $85+T=90$ so $T=5$. Since $5\mathrm{\%}$ of the class is 15, the class total is $15\mathrm{/}0.05=300$. Indeed $5\mathrm{\%}$ of 300 is 15 and all given percents give whole-student counts.

The insight is to use the repeating-block sum together with division with remainder. Dividing 3 by 7 gives $0.\overline{428571}$ with cycle 4,2,8,5,7,1 summing to 27. Trying one hundred long-division steps by hand would run out of time, but cycles finish it quickly. Write $100=16\cdot 6+4$, so there are sixteen full cycles plus the first four digits of the next cycle. Sixteen cycles contribute $16\cdot 27=432$. The next four digits are $4+2+8+5=19$. Adding gives $432+19=451$, so the requested digit sum is 451.

The insight is percent-to-fraction conversion forcing the total by divisibility combined with inclusion-exclusion. Since $62.5\mathrm{\%}$ is $5\mathrm{/}8$, $60\mathrm{\%}$ is $3\mathrm{/}5$, and $7.5\mathrm{\%}$ is $3\mathrm{/}40$, the total must be a whole-number multiple of 8, 5, and 40, so a multiple of 40. Below 50 the only such total is 40, giving 25 art, 24 music, and 3 neither. Union is $40-3=37$, so both equals $25+24-37=12$. Forgetting to add neither or scaling to the next multiple 80 gives the familiar wrong counts, and the arithmetic is short once the denominator 40 is seen.

The insight is to pair the $k$th terms across the two brackets instead of summing each bracket separately. Both progressions use steps of 4 and each has 13 terms, since $(49-1)\mathrm{/}4+1=13$ and $(51-3)\mathrm{/}4+1=13$, so every vertical pair has the same difference $(4x+c)-(x+d)=3x-2$, because $1-3=5-7=\cdots =49-51=-2$. Hence $E$ is 13 copies of $3x-2$, which is $39x-26$. Summing each arithmetic series first with Gauss pairing is much longer and invites off-by-one counts of terms.

The insight is multiplicative complement scaling rather than adding removals. Let juice fraction after one pour be $(V-6)\mathrm{/}V$. After two identical pours the juice fraction multiplies, so $((V-6)\mathrm{/}V{)}^2=0.25$. Since $V$ exceeds 6, the fraction is positive, so $(V-6)\mathrm{/}V=0.5$. Then $V-6=0.5V$, so $0.5V=6$ and $V=12$. Checking, first leaves 6 liters of juice in 12 liters and second leaves 3 liters in 12 liters, which is 25 percent, with only halving and doubling.

The insight is loop closure plus relative rate with rates subtracted, not times. Rates are $1\mathrm{/}6$ and $1\mathrm{/}10$ lap per minute, so relative gain is $1\mathrm{/}6-1\mathrm{/}10=1\mathrm{/}15$ lap per minute. One full extra lap needs 15 minutes, when the faster runner has 2.5 laps and the slower has 1.5. Listing positions to 15 is long, while subtracting times gives 4, averaging gives 8, and least common multiple 30 meets at the start.

The key insight is that a 12 hour face repeats after gaining 12 hours, so convert that modular cycle into days using the rate. Twelve hours is 43200 seconds. At 10 seconds per hour this needs 4320 hours. Since one day is 24 hours, this is 180 days. After that gain the hands have advanced exactly one full dial, so the clock reads correctly again for the first time.

The insight is to bound the reduced step of three collinear grid points instead of listing lines. Three grid points on a line with at least three points contain two consecutive points differing by a reduced run $q$ and rise $p$ with no common divisor. From the first to the third is twice that step, so twice $\mathrm{\mid }q\mathrm{\mid }$ is at most 3 and twice $\mathrm{\mid }p\mathrm{\mid }$ is at most 3, forcing $\mathrm{\mid }q\mathrm{\mid }$ at most 1 and $\mathrm{\mid }p\mathrm{\mid }$ at most 1. Apart from vertical and horizontal, only slopes 1 and $-1$ survive. Counting gives 4 horizontal, 4 vertical, 3 with slope 1, and 3 with slope $-1$, for 14 total. Slopes such as $1\mathrm{/}2$ would need width 4, outside the grid.

The insight is to factor out the greatest common divisor into coprime cofactors and then compare sums extremally. Write the numbers as $12m$ and $12n$ with $m$ and $n$ relatively prime. Then $12mn=720$, so $mn=60$. The factor pairs of 60 are $1\times 60$, $2\times 30$, $3\times 20$, $4\times 15$, $5\times 12$, and $6\times 10$, of which the relatively prime ones are $(1,60)$, $(3,20)$, $(4,15)$, and $(5,12)$. Excluding $(1,60)$ because it makes one number a multiple of the other leaves sums 23, 19, and 17 times 12. The largest is $12\times 23=276$, from 36 and 240.

The insight is to turn place value into a tight digit equation and then count with bounds instead of testing 900 numbers. Write the number as $100h+10t+u$ with $h$ from 1 to 9 and $t$ and $u$ from 0 to 9. The condition gives $100h+10t+u=19h+19t+19u$, so $81h=9t+18u$, or $9h=t+2u$. Thus $t=9h-2u$ must lie between 0 and 9. For $h=1$, $u=0$ through 4 work, giving 5 numbers. For $h=2$, $u=5$ through 9 work, giving 5 more. For $h=3$, only $u=9$ with $t=9$ works, giving 1 more. For $h$ at least 4, $9h-2u$ exceeds 9 even at $u=9$. The total is $5+5+1=11$.

Let the radii be $r_1$ and $r_2$ with $r_1>r_2$. The insight is to avoid solving a quadratic by using the symmetric identity for the difference. From circumference, $2\pi (r_1+r_2)=60\pi$, so $r_1+r_2=30$. From area, $\pi (r_1^2+r_2^2)=458\pi$, so $r_1^2+r_2^2=458$. Then $(r_1-r_2{)}^2=2(r_1^2+r_2^2)-(r_1+r_2{)}^2=916-900=16$, so $r_1-r_2=4$. Solving $r_1+r_2=30$ and $r_1-r_2=4$ gives $r_1=17$. A student who substitutes $r_2=30-r_1$ must factor $r_1^2-30r_1+221=0$, which needs factoring $221=13\times 17$.

The insight is to expand the increased volume and read off the increase as pairwise sum plus edge sum plus one without finding the edges. Let edges be $a$, $b$, $c$ with $a+b+c=18$ and $2(ab+bc+ca)=208$ so $ab+bc+ca=104$. Then $(a+1)(b+1)(c+1)-abc=(ab+bc+ca)+(a+b+c)+1=104+18+1=123$. Trying to find $a$, $b$, $c$ first requires factoring to 8,6,4, a long failing route, while the expansion finishes in seconds with short arithmetic.

The key is that a median in the gap between histogram bins forces the cumulative count to equal exactly half the data. Half of 22 is 11, so the 11th and 12th scores must straddle the gap with the 11th at most 10 and the 12th at least 11 to average 10.5. The first bin holds 7 scores, so the second bin must supply scores 8 through 11, which is 4 scores. Hence $x=4$, leaving $y=11$. Then scores 1 through 7 are 1 to 5, 8 through 11 are 6 to 10, and 12 through 22 are 11 to 15, whose middle two average 10.5. Counting to the half avoids any listing.

The insight is a baseline-excess reframe plus divisibility counting. Simple interest gives $5\mathrm{\%}$ on the full 1000 dollars plus the extra $3\mathrm{\%}$ on the part in the $8\mathrm{\%}$ account, so total equals 50 dollars plus 0.03 times that part. Requiring 59 to 71 dollars forces that part between 300 and 700 dollars, and whole-dollar interest forces 0.03 times that part to be whole, so that part must be a multiple of 100 dollars since 3 and 100 are coprime. The multiples 300, 400, 500, 600, 700 give totals 59, 62, 65, 68, 71 dollars, so five splits work and the short count 12 divided by 3 plus one confirms it.

The insight is prime-exponent parity: squaring doubles exponents, so $r^2$ being a multiple of 12 forces $r$ to supply the missing primes. Write $12=2^2\times 3$. For $r^2$ to be divisible by $2^2$, $r$ must be even, and for $r^2$ to be divisible by 3, $r$ must be divisible by 3 since the exponent in a square is even. Hence $r$ must be a multiple of 6. Counting multiples of 6 with $1\le r\le 100$ gives $\lfloor 100\mathrm{/}6\rfloor =16$. Listing all 100 squares and testing divisibility by 12 is long.