SSAT Upper Quantitative I challenge

The insight is to count total likes and cap how many can be explained without a triple overlap. Total likes are $85+80+75=240$. If no one liked all three, each student accounts for at most 2 likes, giving at most 200 likes, so at least 40 surplus likes must come from triple likers, each contributing one like beyond 2. Thus at least 40 students like all three, which is the bound. It is attainable with 40 triple likers and pairwise-only groups of 25, 20, and 15, since $40+25+20=85$ for soccer, $40+25+15=80$ for basketball, and $40+20+15=75$ for tennis, using exactly 100 students, which is the construction.

The insight is to replace distances along a fixed line by proportional coordinate differences, which avoids square roots, and then split into inside and outside cases. Along one line, distance is a constant multiple of horizontal change, so $\mathrm{\mid }x-2\mathrm{\mid }\mathrm{/}\mathrm{\mid }x-8\mathrm{\mid }=2\mathrm{/}1$. Hence $x-2=2(x-8)$ or $x-2=-2(x-8)$. The first gives $x=14$ outside the segment and the second gives $x=6$ inside it. Both positions lie on the line, so the sum is $6+14=20$.

The insight is inclusion-exclusion combined with the observation that the two quarters exactly cover the square. Each quarter has area $\pi (10{)}^2\mathrm{/}4=25\pi$, so the sum is $50\pi$. For any point in the square, the sum of the squared distances to opposite corners is at most 200, so at least one distance is at most 100, meaning every square point lies in at least one quarter. Hence the union is the 100-area square, and the overlap equals sum minus union, $50\pi -100$. Computing the lens directly is much longer.

The insight is to separate magnitudes from signs. Write $U=\mathrm{\mid }p\mathrm{\mid }$ and $V=\mathrm{\mid }q\mathrm{\mid }$, so $U+V=15$. If $p$ and $q$ have the same sign then $\mathrm{\mid }p+q\mathrm{\mid }=U+V=15$, which contradicts 11, so they must have opposite signs. Then $\mathrm{\mid }p+q\mathrm{\mid }=\mathrm{\mid }U-V\mathrm{\mid }=11$. Solving $U+V=15$ together with $\mathrm{\mid }U-V\mathrm{\mid }=11$ gives the magnitudes 13 and 2 in some order. Since the signs are opposite, the product is negative with magnitude 26, so $pq=-26$. A student who assumes both numbers are positive solves the magnitudes correctly but gets the sign wrong, while the sign elimination makes the computation short.

The insight is to apply the Pythagorean theorem twice to get the space diagonal and then balance the integer divisor triple of the volume. For edges $a$, $b$, and $c$ the base diagonal squares to $a^2+b^2$, and adding the height gives diagonal $\sqrt{a^2+b^2+c^2}$ with $abc=72$. To minimize the sum of squares the factors should be as equal as possible, near the cube root of 72. Checking balanced triples, $3\times 4\times 6=72$ gives $9+16+36=61$, while 2 × 6 × 6 gives 76, 3 × 3 × 8 gives 82, and every more spread triple is larger. Hence the smallest diagonal is $\sqrt{61}$.

The insight is to pair the outer and inner factors and complete a square instead of multiplying everything. Let $n=12$, so the product is $n(n+1)(n+2)(n+3)$. Group as $[n(n+3)][(n+1)(n+2)]=(n^2+3n)(n^2+3n+2)$. Put $m=n^2+3n=144+36=180$. Then the product is $m(m+2)=m^2+2m$, so adding 1 gives $m^2+2m+1=(m+1{)}^2$. Hence the root is $m+1=181$. Multiplying to 32761 and guessing the root by hand is the long failing route.

The insight is to pair the familiar triangular sum formula with the fact that consecutive integers share no prime factor, forcing both to be powers of two. Now $1+2+\cdots +n=n(n+1)\mathrm{/}2$, so requiring a power of two gives $n(n+1)=2^{k+1}$ for some $k$. Since $n$ and $n+1$ are consecutive, no prime can divide both, yet their product is a power of two, so the odd one of the pair must be 1. That forces $n=1$, giving $1=2^0$. No larger consecutive pair consists of two powers of two, since powers of two beyond 1 and 2 differ by more than one. Hence below 100 there is exactly one such $n$, while listing 100 sums and testing each is long.

The insight is to combine a modular residue restriction with the total-question bound to list the few possibilities and then extremize. Let $c$ be correct, $w$ wrong, and $b$ blank, so $c+w+b=30$ and $7c-3w=120$. Then $7c=120+3w$ is a multiple of 3, so $c$ must be a multiple of 3. Also $7c-120=3w>=0$ gives $c>=18$, while $c+w<=30$ caps large $c$. Trying multiples of 3 leaves only $c=18$ with $w=2$ and $b=10$, and $c=21$ with $w=9$ and $b=0$. Among these the greatest blank count is 10.

Let the sides be $l$ and $w$ with $l+w=21$. Then $l$ squared plus $w$ squared equals $c$ squared for integer $c$. Since $c$ is less than 21 and greater than 14, and twice the area $441-c$ squared must be even, $c$ is odd, so $c$ is 15, 17, or 19 giving area equal to 108, 76, or 40. Only 108 is a product of two positive integers summing to 21, namely 9 times 12, and 9 squared plus 12 squared equals 15 squared. The other two values have no factor pair summing to 21, so the area must be 108.

The insight is to square the Pythagorean relations and add all three by symmetry instead of solving for edges directly. Let the edges be $x$, $y$, $z$. Then $x^2+y^2=25$, $y^2+z^2=41$, $z^2+x^2=34$. Adding gives $2(x^2+y^2+z^2)=100$, so $x^2+y^2+z^2=50$. Subtracting each face equation isolates the opposite square: $z^2=25$, $x^2=9$, $y^2=16$. Hence the edges are 3, 4, 5 in some order, and the volume is $3\times 4\times 5=60$ cubic inches.

The insight is total balance for the mean combined with positional ordering for the median through a case split. The known sum is 70 plus 72 plus 76 plus 90 equals 308, so the mean is 308 plus $x$ over 5. If $x$ is below 70 the median is 72, giving 308 plus $x$ equals 360 for $x$ equals 52, which satisfies below 70. If the median is $x$ then $x$ equals 77, which is outside 72 to 76, and if the median is 76 then $x$ equals 72, which is outside its interval, so no other position works. Assuming the missing value is middle gives 77 and reporting the median gives 72, but only 52 makes mean and median both 72.

The insight is integer factorization combined with the extremal fact that the triple closest to a cube gives least surface. Factoring 72 gives triples including 3,3,8 with surface 114, 2,6,6 with surface 120, 2,4,9 with surface 124, and 2,3,12 with surface 132. The triple 3,4,6 has pairwise products 12, 18, and 24, summing to 54 and doubling to 108. No other integer triple is closer to a cube, so 108 square units is the least possible surface.

The insight is to read the slope as a reduced rise-over-run ratio and scale its Pythagorean triple instead of solving a quadratic, then compare the two products. A slope of $2\mathrm{/}3$ means the changes satisfy $dx=3k$ and $dy=2k$ for some scale factor $k$, so the distance is $\mathrm{\mid }k\mathrm{\mid }\sqrt{3^2+2^2}=\mathrm{\mid }k\mathrm{\mid }\sqrt{13}$. Setting this equal to $6\sqrt{13}$ gives $\mathrm{\mid }k\mathrm{\mid }=6$, so the two points are $(2+18,9+12)=(20,21)$ and $(2-18,9-12)=(-16,-3)$. Their products are 420 and 48, so the greatest possible value is 420. Solving $(x-2{)}^2+(2(x-2)\mathrm{/}3{)}^2=(6\sqrt{13}{)}^2$ directly is much longer.

The insight is to bound the smallest denominator by ordering. Since $x\le y\le z$, we have ${\displaystyle \frac{1}{x}}\ge {\displaystyle \frac{1}{y}}\ge {\displaystyle \frac{1}{z}}$, so $1={\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}\le {\displaystyle \frac{3}{x}}$, giving $x\le 3$. Hence $x$ is 2 or 3. If $x=3$ then ${\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}={\displaystyle \frac{2}{3}}$ forces $y=3$ and $z=3$. If $x=2$ then ${\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}={\displaystyle \frac{1}{2}}$ with $y\ge 2$ gives $y\le 4$, so $y=3$ yields $z=6$ and $y=4$ yields $z=4$. The attainable $z$ values are 3,4,6, greatest 6. Without the bound a solver faces an unbounded search, while the bound leaves only three short cases.

The insight is to combine the intercept form of a line with divisor optimization over factor pairs. Let the intercepts be $p$ on $x$ and $q$ on $y$, so $x\mathrm{/}p+y\mathrm{/}q=1$ and $6\mathrm{/}p+8\mathrm{/}q=1$ with integers $p$ larger than 6 and $q$ larger than 8. Clearing denominators gives $6q+8p=pq$, which rearranges to $(p-6)(q-8)=48$. Write $u=p-6$ and $v=q-8$ with $uv=48$, so the triangle area $pq\mathrm{/}2=(u+6)(v+8)\mathrm{/}2$. Checking the factor pairs of 48 gives 96 at $u=6$ and $v=8$ with $p=12$ and $q=16$, smaller than the neighboring feasible areas. Hence the least possible area is 96.

The insight is to expose the repetend with 999 and then count by division with remainder. Since 37 times 27 is 999, we have ${\displaystyle \frac{5}{37}}={\displaystyle \frac{135}{999}}=0.135135\dots$ with repeating block 135 summing to 9. One hundred digits contain 33 full blocks plus one extra digit because $100=33\cdot 3+1$. So the sum is $33\cdot 9+1=298$. Writing out one hundred digits by hand is the long failing route, while the period plus remainder makes it two multiplications.

The insight is that the positions of the 1s are triangular numbers, so the question becomes an inequality for $k$. The first 1 is at place 1, the next after one 0 is at place $3=1+2$, the next at $6=1+2+3$, and in general the $k$th 1 is at place $1+2+\cdots +k=k(k+1)\mathrm{/}2$. We need the largest $k$ with $k(k+1)\mathrm{/}2\le 100$. Since $13\times 14\mathrm{/}2=91$ and $14\times 15\mathrm{/}2=105>100$, exactly 13 ones occur by place 100, with the next 1 at place 105.

The key is grouping places by the digit length of the source integer and locating the target with division and remainder. Digits 1 through 9 use 9 places, so 141 places remain to reach 150. Each two-digit integer uses 2 places, and 141 divided by 2 gives 70 full integers with remainder 1. Starting at 10, seventy integers reach 79 and use 140 places, totaling 149 places with the first block. The next integer is 80, and remainder 1 asks for its first digit, which is 8. Writing 150 digits out would be very long, and forgetting the one-digit block shifts everything.

The insight is to combine the total-distance invariant with constant speed ratios. At the first meeting the two cars together have covered $D$, with the car from $P$ covering 40. At the second meeting they together have covered $3D$, because the first meeting accounts for one $D$ and the trips to the far towns plus the return meeting add two more $D$. Hence the car from $P$ has covered three times 40, namely 120, by then. From the $P$ side this same total equals going all the way to $Q$ and back to 20 from $P$, namely $2D-20$. So $2D-20=120$, giving $D=70$.

The insight is to track the juice fraction of the whole rather than volumes and then undo the repeated multiplication with a square root. The final juice fraction is $16\mathrm{/}(16+9)=16\mathrm{/}25$. Each removal keeps the fraction $(V-6)\mathrm{/}V$ of the juice, so after two identical steps the fraction is $((V-6)\mathrm{/}V{)}^2=16\mathrm{/}25$. Taking square roots gives $(V-6)\mathrm{/}V=4\mathrm{/}5$, so $V\mathrm{/}5=6$ and $V=30$.

The insight is to avoid solving for the legs and instead square their sum. Let the legs be $p$ and $q$. Then $p+q=46-20=26$ and $p^2+q^2=20^2=400$ by the Pythagorean theorem. Squaring gives $(p+q{)}^2=26^2=676=p^2+q^2+2pq$, so $2pq=676-400=276$. The triangle area is $pq\mathrm{/}2$, hence $276\mathrm{/}4=69$. A student who tries to find $p$ and $q$ individually faces an irrational quadratic, while the symmetric computation finishes in seconds.

The insight is to unify the two periods with a common multiple and then use monotonicity of a single ratio instead of tracking two growing populations. Write time as $t=6k$ with integer $k$, so the counts are $100\cdot 3^{2k}$ and $200\cdot 2^{3k}$. Cancelling gives first exceeds second exactly when $(3^2\mathrm{/}{2}^3{)}^k=(9\mathrm{/}8{)}^k>2$. Since $9\mathrm{/}8>1$, the left side increases with $k$, so it suffices to test the boundary. Now $9^5=59049$ and $2\cdot 8^5=65536$, so $k=5$ fails, while $9^6=531441$ and $2\cdot 8^6=524288$, so $k=6$ works. Hence the smallest such multiple is $6\cdot 6=36$ years, with short integer arithmetic.

The insight is to work backward from the given probability to the hidden number of reds. The two-draw probability gives $n(n-1)$ over $(n+4)(n+3)$ equals 1 over 3, so $3n^2-3n=n^2+7n+12$, hence $n^2-5n-6=0$ and $(n-6)(n+1)=0$, so $n=6$. The full bag then has 6 red and 4 blue. Exactly two red in three draws needs $15\cdot 4=60$ triples out of 120, so the probability is 60 over 120, which is 1 over 2.

The insight is to reduce the slope to lowest terms and then sum an arithmetic series. The change is 36 in $x$ and $-9$ in $y$, so the slope is $-9\mathrm{/}36=-1\mathrm{/}4$. Hence $y$ drops by 1 exactly when $x$ grows by 4, so integer points occur when $x-1$ is a multiple of 4. From 1 to 37 this gives $1,5,\dots ,37$, which is 10 points. Their $x$-coordinates form an arithmetic progression with first term 1, last term 37, and 10 terms, so the sum is $(1+37)\times 10\mathrm{/}2=190$. Testing every integer $x$ would take far longer.

The insight is to re-cut the surface: an outside path becomes a straight segment only after unfolding the two faces it crosses, so the problem is a choice among three unfoldings plus the Pythagorean theorem. Pairing the edges gives three straight candidates: $\sqrt{(2+3{)}^2+6^2}=\sqrt{61}$, $\sqrt{(2+6{)}^2+3^2}=\sqrt{73}$, and $\sqrt{(3+6{)}^2+2^2}=\sqrt{85}$. A straight segment is shorter than any bent one on the same unfolding, so the minimal surface path is the smallest of the three, $\sqrt{61}$. A student who stays in three dimensions tries bent paths or uses the interior diagonal 7, which is shorter but illegally passes through the box.