SHSAT · Competition-style problems · Part 1 of 2

The insight is to use prime factors to test terminating and exponents to test exact length. Since $60=4$ times 3 times 5, the reduced denominator still contains 3 unless 3 divides $n$. Hence terminating forces $n=3m$ and the fraction reduces to $m\mathrm{/}20$. Write the reduced denominator as 2 to a power times 5 to a power. Exactly two places means the larger exponent is 2, so the reduced denominator is 4 or 20. With $m$ at most 10, denominator 20 occurs for $m=1,3,7,9$ and denominator 4 occurs for $m=5$. That gives $n=3,9,15,21,27$, five values. Listing all 30 decimals would be long, while factor plus length analysis is short.

The insight is to fill positions sequentially with conditional counts and then split on whether the two forbidden predecessors coincide. There are $10^4=10000$ equally likely PINs. The first position has 10 choices and the second has 9 choices different from the first. The third has 9 choices different from the second. If the third equals the first, which happens in 1 way, the last position has 9 choices different from that digit, giving $10\times 9\times 1\times 9=810$ codes. If the third differs from both the first and second, which happens in 8 ways, the last must avoid two different digits, giving 8 choices, for $10\times 9\times 8\times 8=5760$ codes. The total is $810+5760=6570$, so the probability is ${\displaystyle \frac{6570}{10000}}={\displaystyle \frac{657}{1000}}$ after dividing by 10.

The insight is to recut the perimeter with the triangle inequality for the top and with averaging for the bottom, then exhibit each value. Call the sides $x\le y\le L$. Since $x+y>L$, the perimeter $x+y+L$ exceeds $2L$, so $2L<20$ from the perimeter cap, giving $L<10$ and hence $L\le 9$ for integers. Since $x$ and $y$ are each at most $L$, the perimeter is at most $3L$, and the lower cap gives $18\le 3L$, so $L\ge 6$. Each end occurs: 6,6,6 gives 6 with perimeter 18, 6,7,7 gives 7, 6,6,8 gives 8, and 5,6,9 gives 9, each a genuine triangle with perimeter 18 to 20, so every integer 6 through 9 occurs.

The idea is to write each map side as $400\mathrm{/}n$ and $800\mathrm{/}n$ and then force the orientation with a size bound, plus exhibit the packing that meets it. For $n$ less than 50, each map has long side $800\mathrm{/}n$ greater than 16 cm, so no long side fits across the 16 cm width and every map must place its long side along the 25 cm height. Two such maps side by side need width 2 by $400\mathrm{/}n$ greater than 16 cm, and stacked need height 2 by $800\mathrm{/}n$ greater than 25 cm, so two maps cannot fit. For $n$ equal to 50, each map is 8 cm by 16 cm, and two maps with 16 cm sides along the page width stacked to height 16 cm fit inside 16 cm by 25 cm. Hence 50 is the smallest feasible value.

The insight is to turn means into totals and use inclusion-exclusion for the overlap. The art total is $20\times 84=1680$ and the music total is $15\times 88=1320$, so the two lists sum to 3000. If $o$ students take both, the number of distinct students is $35-o$ and their total is $85\times (35-o)=2975-85o$. Hence the overlap total is $3000-(2975-85o)=25+85o$ and its mean is $85+25\mathrm{/}o$. Since that mean is a whole number, $o$ divides 25, so with $o\le 15$ we get $o=1$ or $o=5$. The value $o=1$ would give mean 110, impossible for scores capped at 100, so $o=5$ and the mean is $85+5=90$.

The insight is to combine sign analysis with a size bound. Since the product is negative, there are one or three negatives, but three negatives cannot sum to zero, so there are exactly two positives $a$ and $b$ and one negative $-(a+b)$. Then $ab(a+b)=210$. If $a$ is the smaller positive then $2a^3$ is at most 210, so $a$ cubed is at most 105 and $a$ is 1, 2, 3, or 4. Checking gives $1+14-15=0$ with product $-210$ so $(1,14,-15)$ works and $3+7-10=0$ with product $-210$ so $(3,7,-10)$ works, while $a=2$ and $a=4$ give no integer $b$. The largest values are 14 and 7, so the greatest possible largest is 14.

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 insight is that trailing zeros count factors of $10=2\times 5$, combined with counting multiples to see that factors of 5 control the total. Each zero needs one 2 and one 5, but even numbers supply far more 2s than multiples of 5 supply 5s, so the number of zeros equals the number of factors of 5 in $P(n)$. Counting multiples of 5 gives one each for $5,10,15,20$, but $25=5^2$ contributes two. Hence $P(20)$ through $P(24)$ have four zeros, while $P(25)$ through $P(29)$ already have six zeros, and $P(30)$ has seven. The count jumps from four to six, so five never occurs.

Let the intercepts be $(p,0)$ and $(0,q)$. The point $(6,4)$ lies on the segment between them, so the slope from $(p,0)$ to $(6,4)$ equals the slope from $(p,0)$ to $(0,q)$: ${\displaystyle \frac{4}{6-p}}={\displaystyle \frac{q}{-p}}$, which gives $q={\displaystyle \frac{4p}{p-6}}=4+{\displaystyle \frac{24}{p-6}}$. For $q$ to be a whole number, $p-6$ must be a factor of 24, so $p=7,8,9,10,12,14,18,30$ with $q=28,16,12,10,8,7,6,5$. The areas ${\displaystyle \frac{pq}{2}}$ are 98,64,54,50,48,49,54,75, and the largest is 98, from intercepts 7 and 28.

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 to clear the reciprocals into a factored integer product and then choose divisors extremally. From $1\mathrm{/}x+1\mathrm{/}y=1\mathrm{/}12$, multiply by $12xy$ to get $12y+12x=xy$, so $xy-12x-12y=0$. Adding 144 completes the product: $(x-12)(y-12)=144$. If $x\le 12$ then $1\mathrm{/}x\ge 1\mathrm{/}12$, so the sum would exceed $1\mathrm{/}12$; thus $x>12$ and $y>12$, so both factors are positive divisors of 144. With $x<y$, write $x-12=d$ and $y-12=144\mathrm{/}d$ with $d<12$. To make $y$ smallest, make $144\mathrm{/}d$ smallest while staying above 12, so take the largest $d$ below 12 dividing 144, namely $d=9$. Then $y-12=16$ and $y=28$, realized by $x=21$ since $1\mathrm{/}21+1\mathrm{/}28=7\mathrm{/}84=1\mathrm{/}12$. Hence the smallest possible $y$ is 28.

The insight is double-counting memberships to bound extras combined with an extremal construction. Union is $32-3=29$ students with at least one sport. Total memberships are $18+16+14=48$, so extras beyond one per student are $48-29=19$. Each all-three student contributes 2 extras and each exactly-two student contributes 1, so with $t$ triples, $2t$ is at most 19, giving $t$ at most 9. Nine is attainable with 9 triple, 1 soccer-basketball double, plus 8 only-soccer, 6 only-basketball, and 5 only-tennis, which uses 29 union members and exactly 18, 16, and 14 memberships. Dividing extras by 3 or using only the smallest group misses the double-count.

The insight is to recover $xy$ from the shifted product instead of solving for $x$ and $y$. Expanding gives $(x+1)(y+1)=xy+x+y+1=20$, so with $x+y=8$ this is $xy+9=20$ and $xy=11$. Then the symmetric identity $x^2+y^2=(x+y{)}^2-2xy$ gives $64-22=42$. Solving for $x$ and $y$ individually would require $t^2-8t+11=0$, which needs the quadratic formula beyond this exam, so the symmetric route is the only short path.

The insight is to rewrite the fraction as an integer plus a proper remainder instead of testing $x$ values one by one. Write $4x+9=2(2x+1)+7$, so the fraction equals $2+{\displaystyle \frac{7}{2x+1}}$. For the whole to be an integer, ${\displaystyle \frac{7}{2x+1}}$ must be an integer, so $2x+1$ must be a divisor of 7. The divisors are $1,-1,7,-7$, giving $2x+1=1,-1,7,-7$, so $x=0,-1,3,-4$. That is four integers, and trying $x$ values directly would never end.

The insight is to square the sum of the legs instead of solving for each leg, combining the linear perimeter with the Pythagorean relation. Let the legs be $a$ and $b$. Then $a+b+9=20$, so $a+b=11$. By the Pythagorean theorem $a^2+b^2=81$. Squaring the sum gives $(a+b{)}^2=a^2+2ab+b^2$, so $121=81+2ab$. Thus $2ab=40$ and $ab=20$. The area is $ab\mathrm{/}2$, which equals 10 square inches. Trying to find $a$ and $b$ separately leads to an ugly quadratic with irrational roots, while the area follows at once.

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 $15+10-5=20$.

Think like a tutor: use the line itself. The line through $(0,u)$ and $(v,0)$ satisfies ${\displaystyle \frac{x}{v}}+{\displaystyle \frac{y}{u}}=1$, which is equal slopes in disguise. Substituting $(k,k)$ gives ${\displaystyle \frac{k}{v}}+{\displaystyle \frac{k}{u}}=1$. Keep the reciprocal together instead of clearing everything: ${\displaystyle \frac{k}{u}}=1-{\displaystyle \frac{k}{v}}={\displaystyle \frac{v-k}{v}}$. Invert once to get ${\displaystyle \frac{u}{k}}={\displaystyle \frac{v}{v-k}}$, so $u={\displaystyle \frac{vk}{v-k}}$. Clearing denominators first and chasing $u$ through each term is long and sign prone, while one common denominator plus one inversion is short. With $v=6$ and $k=2$, this gives $u=3$.

Insight: negate the compound inequality instead of solving for $x$, then filter by remainders and pair symmetrically. Since $y=3x-7$, observe $-3x+7=-y$, so $-50\le -y<10$. Multiplying by $-1$ reverses both signs to give $50\ge y>-10$, so $-10<y\le 50$. Because $y=3x-7$ with integer $x$, $y$ leaves remainder 2 upon division by 3. Among $-9$ through 50, those values are $-7$, $-4$, up to 50, twenty terms with first $-7$ and last 50. Pairing first with last gives 20 times 43 divided by 2, which is 430. Solving for $x$ first and evaluating each $y$ separately is much longer.

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 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.

Think like a tutor: add rates, not times. Pipe A fills ${\displaystyle \frac{1}{a}}$ tank per hour, Pipe B fills ${\displaystyle \frac{1}{b}}$, Pipe C fills ${\displaystyle \frac{1}{c}}$, and together they fill ${\displaystyle \frac{1}{t}}$ per hour, so ${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}={\displaystyle \frac{1}{t}}$. Combine the known parts first: ${\displaystyle \frac{1}{c}}={\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}={\displaystyle \frac{ab-at-bt}{abt}}$. Invert only at the end to get $c={\displaystyle \frac{abt}{ab-at-bt}}$. Multiplying by $abct$ at the start and then trying to collect $c$ from three terms is long and sign prone, but one denominator plus one inversion is quick. With $a=4$, $b=4$, and $t=1$, this gives $c=2$.

The insight is to read the absolute sum as distances on the number line and then impose divisibility by 4. The distance between 3 and $-5$ is 8. If $x$ lies between $-5$ and 3, there are 9 such integers, the two distances add to 8, which is divisible by 4, so all 9 work. If $x$ lies outside, write $d$ for the distance outside: for $x>3$, $d=x-3$ runs 1 through 17, and the sum is $8+2d$; for $x<-5$, $d=-5-x$ runs 1 through 15, and the sum is again $8+2d$. Now $8+2d$ is divisible by 4 exactly when $d$ is even. Among 1 through 17 there are 8 evens, among 1 through 15 there are 7 evens, giving $8+7=15$ outside values. Adding the middle 9 gives 24.

The insight is 180-degree rotational symmetry about the center plus a divisibility count for points on the diagonal. There are 7 choices for $x$ and 5 for $y$, so 35 equally likely points. The map $(x,y)$ to $(6-x,4-y)$ sends the rectangle to itself, fixes the line through $(0,0)$ and $(6,4)$, and swaps points strictly above with points strictly below, so off-line points split equally. On the line $y=2x\mathrm{/}3$ with integer bounds, $x$ must be a multiple of 3, giving only $(0,0)$, $(3,2)$, and $(6,4)$. Thus $35-3=32$ points split 16 and 16, probability $16\mathrm{/}35$, so $a+b=16+35=51$.

The insight is weighted extremal bounding with an explicit construction. Simple interest makes one dollar for one year earn 0.10 dollars, so with deposits $P1$ less than $P2$ less than $P3$ the interest condition is $P1$ plus $2P2$ plus $3P3$ equals 220. To have three deposits above 40 dollars the cheapest distinct choice is 41, 42, 43, giving 41 plus 84 plus 129 equals 254 which exceeds 220, so three is impossible. Two is possible with 1, 42, 45 since 1 plus 84 plus 135 equals 220 for 22 dollars interest, so the maximum is two.

Think like a tutor: never solve for the sides. Let the sides be $l$ and $w$. Perimeter gives a sum, $l+w={\displaystyle \frac{P}{2}}$, and Pythagoras gives $l^2+w^2=d^2$. The clever move is the square of a sum, $(l+w{)}^2=l^2+2lw+w^2$. Let $A=lw$. Then ${\displaystyle \frac{P^2}{4}}=d^2+2A$, so $2A={\displaystyle \frac{P^2}{4}}-d^2$ and $A={\displaystyle \frac{P^2-4d^2}{8}}$. Trying to find $l$ and $w$ separately forces a messy quadratic, but treating the product as one block finishes quickly. With 3,4,5, $P=14$ and $d=5$ give 12.

The insight is double-counting memberships combined with solving a small linear system for exactly-two and exactly-three. Union is $30-5=25$, so exactly-two plus exactly-three equals $25-9=16$. Total memberships are $18+16+14=48$, which also equal 9 times 1 plus exactly-two times 2 plus exactly-three times 3, so twice exactly-two plus three times exactly-three equals 39. Substituting exactly-two as 16 minus exactly-three gives 32 plus exactly-three equals 39, so exactly-three is 7. Counting each triple once or confusing none and exactly-one with triples misses the two-way count.