SHSAT · Competition-style problems · Part 2 of 2

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.

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.

The insight is to split each term into a difference of reciprocals and subtract the excluded terms from the full telescoped total. Note ${\displaystyle \frac{1}{k(k+1)}}={\displaystyle \frac{1}{k}}-{\displaystyle \frac{1}{k+1}}$, so adding without exclusions telescopes: ${\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{9}}={\displaystyle \frac{8}{9}}$ for $1\le k\le 8$. The excluded values are $k=3$ giving ${\displaystyle \frac{1}{12}}$ and $k=6$ giving ${\displaystyle \frac{1}{42}}$, whose sum is ${\displaystyle \frac{7}{84}}+{\displaystyle \frac{2}{84}}={\displaystyle \frac{9}{84}}={\displaystyle \frac{3}{28}}$. Subtracting gives ${\displaystyle \frac{8}{9}}-{\displaystyle \frac{3}{28}}={\displaystyle \frac{224}{252}}-{\displaystyle \frac{27}{252}}={\displaystyle \frac{197}{252}}$. Adding six fractions with denominator 2520 directly is far longer, so the split plus sieve is the short route.

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 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 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 key combines turning the mean into a total with a Pythagorean difference of squares. The five total $5(10)=50$ and the three known total $8+12+14=34$, so the other two, call them $p$ and $q$, satisfy $p+q=16$. Together with 8 they form a right triangle. If 8 were the hypotenuse then $p^2+q^2=64$ would combine with $(p+q{)}^2=256$ to force $pq=96$, which has no integer pair summing to 16. So 8 is a leg and $q^2-p^2=64$, which factors as $(q-p)(q+p)=64$. Since $q+p=16$, we get $q-p=4$. Solving $p+q=16$ and $q-p=4$ gives $p=6$ and $q=10$. The full list is 6, 8, 10, 12, 14, whose smallest is 6.

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 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 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 insight is extremal ordering that any successful sixth score must beat the current minimum, so the dropped score is known and the denominator stays five. The sum of the first five is $82+85+88+90+92=437$ with minimum 82, so without the minimum the kept four sum to $437-82=355$. If $s$ were at or below 82, the best five would be the original five averaging $437\mathrm{/}5=87.4$ below 90, so success forces $s$ above 82 and the best five sum to $355+s$ over 5. Requiring at least 90 gives $(355+s)\mathrm{/}5>=90$, which needs $s>=95$ by short algebra. Averaging all six misses the drop, dividing the six-sum by five misses the subtraction, and flipping misses at least.

The insight is to reframe the absolute sum by sign cases and then use the bound. Suppose the larger of $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }$. If $a$ and $b$ have the same sign then $\mathrm{\mid }a+b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ and $\mathrm{\mid }a-b\mathrm{\mid }$ is $\mathrm{\mid }a\mathrm{\mid }-\mathrm{\mid }b\mathrm{\mid }$, summing to twice $\mathrm{\mid }a\mathrm{\mid }$. If they have opposite signs the two roles swap, still summing to twice $\mathrm{\mid }a\mathrm{\mid }$. The same holds with $a$ and $b$ swapped, so in all cases $\mathrm{\mid }a+b\mathrm{\mid }+\mathrm{\mid }a-b\mathrm{\mid }$ equals twice the larger of $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$. Here that gives twice the larger equals 12, so the larger equals 6. But $\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }b\mathrm{\mid }$ are at most 5, so the larger is at most 5, impossible. Hence no ordered pairs satisfy the equation.

The insight is to write the original total as a triangular number and trap $n$ by factoring out $n-1$. The original sum is $n(n+1)\mathrm{/}2$ and the remaining sum is $61(n-1)\mathrm{/}3$, so the erased integer is $k=n(n+1)\mathrm{/}2-61(n-1)\mathrm{/}3=(n-1)(3n-116)\mathrm{/}6$. Since $1\le k\le n$, we get $(n-1)(3n-116)\ge 6$ and $(n-1)(3n-122)\le 0$, which factor cleanly to force $39\le n\le 40$. The remaining sum must be an integer, so $n-1$ is a multiple of 3, leaving $n=40$. Then the original sum is $40\times 41\mathrm{/}2=820$ and the remaining sum is $61\times 39\mathrm{/}3=61\times 13=793$, so the erased integer is $820-793=27$.

The insight is to re-cut valid triples by gap compression and then count through a bijection with a smaller range. There are $\left(\genfrac{}{}{0px}{}{8}{3}\right)=56$ equally likely triples. Write a valid triple as $a<b<c$ with gaps of at least 2. Define $a^{\mathrm{\prime }}=a$, $b^{\mathrm{\prime }}=b-1$, and $c^{\mathrm{\prime }}=c-2$. Then $1\le a^{\mathrm{\prime }}<b^{\mathrm{\prime }}<c^{\mathrm{\prime }}\le 6$, so every valid triple compresses to a distinct triple from 1 through 6. Conversely adding 0, 1, 2 expands any triple from 1 through 6 to a valid triple with no consecutive numbers, so the correspondence is exact. Hence there are $\left(\genfrac{}{}{0px}{}{6}{3}\right)=20$ favorable triples, and the probability is ${\displaystyle \frac{20}{56}}={\displaystyle \frac{5}{14}}$ after dividing by 4.

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 key is to count factors of 3 and then make an extremal split on multiples of 9. The multiples of 9 up to 20 are 9 and 18, each already containing $3^2$, so any pair containing one of them has a product divisible by 9; a collection using one can therefore have size at most 1. A large collection must avoid both. The remaining multiples of 3 are 3, 6, 12, and 15, each with a single factor of 3, but the product of any two of them contains $3^2$ and is divisible by 9, so at most one of them can be kept. The 14 integers not divisible by 3 are safe together and safe with one such lone multiple. Thus the maximum is $14+1=15$, achieved for example by all non-multiples of 3 together with 3.

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 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 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 alternating-order symmetry with careful counting of the partial hour. Let Rosa do $r$ of the fence per hour and Tom do $t$ per hour. Starting with Rosa for 9 hours gives 5 Rosa hours and 4 Tom hours, so $5r+4t=1$. Starting with Tom, 9 full hours give 5 Tom hours and 4 Rosa hours, then the next 30 minutes fall in a Rosa hour, so $4.5r+5t=1$. Multiply the first by 5 to get $25r+20t=5$ and the second by 4 to get $18r+20t=4$. Subtracting gives $7r=1$, so $r=1\mathrm{/}7$ and Rosa alone needs 7 hours, with Tom needing 14 hours as a check.

The insight is to combine solving a linear system by addition with a discount-threshold case split. If both totals were discounted, the pre-discount totals would be 100 and 125, since $80\mathrm{/}0.8=100$ and $100\mathrm{/}0.8=125$. Then $3N+2P=100$ and $2N+3P=125$ add to $5(N+P)=225$, so $N+P=45$, and subtracting gives $N-P=-25$, so $N=10$ and $P=35$. This satisfies the threshold, with pre-totals 100 and 125 both at least 100. The alternative with the first purchase undiscounted gives $3N+2P=80$ and $2N+3P=125$, yielding $N=-2$, impossible for a positive price, while two undiscounted totals would make the second pre-total 100, which should have been discounted. Hence only the both-discounted case survives, so the notebook price is 10.

The insight is to conserve water volume on the reduced base and first check the partial-submersion bound. Water volume is 8 by 6 by 2 equals 96. If the cube sticks out, water covers $48-16=32$ square inches, so depth would be $96\mathrm{/}32=3$, which is at most 4, so the assumption holds and 3 is correct. Using the full base for the total gives $(96+64)\mathrm{/}48=10\mathrm{/}3$, adding the rise on the reduced base to the old depth gives $2+2=4$, and dividing the total by the reduced base gives $160\mathrm{/}32=5$. The bound selects the correct case in seconds while ignoring the footprint fails.

The key is to recover the three face areas and then use that their product is the volume squared. Let the face areas be $X$, $Y$, $Z$. One block has $2(X+Y+Z)=94$, so $X+Y+Z=47$. Gluing two blocks hides two copies of the glued face, so the combined area is $2\cdot 94-2\cdot \text{overlap}$. Hence the overlaps are $(188-164)\mathrm{/}2=12$ and $(188-148)\mathrm{/}2=20$. The third face is $47-12-20=15$. Since $(lw)(lh)(wh)=(lwh{)}^2$, the volume satisfies $V^2=12\cdot 20\cdot 15=3600$, so $V=60$. Trying to solve for length, width, and height separately is much longer.

The insight is that a digit sum is even exactly when the number of odd digits is even, so the last digit parity is forced by the first two and the leading-zero restriction changes the remaining counts. Count prefixes with hundreds nonzero and tens different. Hundreds odd and tens even gives 5 times 5 equals 25 prefixes needing an odd last digit with 4 odds left, giving 100. Hundreds odd and tens odd gives 5 times 4 equals 20 prefixes needing an even last digit with 5 evens left, giving 100. Hundreds even nonzero and tens odd gives 4 times 5 equals 20 prefixes needing an odd last digit with 4 odds left, giving 80. Hundreds even nonzero and tens even gives 4 times 4 equals 16 prefixes needing an even last digit with 3 evens left, giving 48. The total is 100 plus 100 plus 80 plus 48, which is 328.

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 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 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 key insight is to use the red plus green total as an invariant multiple plus divisibility elimination with the original total interval. Originally red is $2k$, blue is $3k$, green is $4k$, so the total is $9k$ and red plus green is $6k$. After adding 60 with no blue, red plus green is $6k+60$ and must be a multiple of 11 from the $4:7$ split. Since $9k$ lies strictly between 100 and 120, $k$ is 12 or 13. Only $k=12$ makes $6k+60=132$ a multiple of 11, so blue is 36.

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 insight is to rewrite chained division as one fraction and then minimize its denominator. The quotient $a÷b$ equals ${\displaystyle \frac{a}{b}}$, and parentheses decide whether each later divisor lands in the numerator or denominator, while the second number 3 must stay in the denominator. To maximize, keep only 3 below and flip 4 and 5 above, giving ${\displaystyle \frac{60\cdot 4\cdot 5}{3}}={\displaystyle \frac{1200}{3}}=400$. The left-to-right evaluation gives 1, flipping only $3÷4$ gives 16, and flipping only $4÷5$ gives 25, all smaller. Trying all five parenthesizations with fractions is far longer than the one extremal fraction.