SSAT Upper · Competition-style problems

The insight is that reduction happens before the terminating test, so split on the common factor 11. A reduced fraction terminates exactly when its denominator uses only primes 2 and 5. If 11 does not divide $n$, then $n$ itself must have that form; through 60 these are 1,2,4,5,8,10,16,20,25,32,40,50, twelve numbers. If 11 divides $n$, write $n=11m$; then ${\displaystyle \frac{11}{n}}={\displaystyle \frac{1}{m}}$, so $m$ must use only 2 and 5 and satisfy $11m\le 60$, giving $m=1,2,4,5$ and hence $n=11,22,44,55$. That adds four new values, for $12+4=16$ total, including the integers from $n=1$ and $n=11$.

The insight is to read the sum as distance and then use parity. For $x$ between $-5$ and 2 the sum is $(2-x)+(x+5)=7$, the distance between 2 and $-5$. Outside, for $x$ below $-5$ it is $-2x-3$ and for $x$ above 2 it is $2x+3$, both odd for integer $x$ and larger than 7. Hence $k=7$ occurs for eight integer points, values below 7 never occur, even values above 7 never occur, and each odd $k$ above 7 occurs for exactly one point on each side. In $0\le k\le 20$ those are 9,11,13,15,17,19, six values. Missing the distance picture pushes a solver into testing every $k$ separately.

The insight is to sort the triple. Rename the numbers from smallest to largest as $x\le y\le z$. Then $\mathrm{\mid }a-b\mathrm{\mid }+\mathrm{\mid }b-c\mathrm{\mid }+\mathrm{\mid }c-a\mathrm{\mid }$ equals $(y-x)+(z-y)+(z-x)$, so the middle value cancels and the total is $2(z-x)$, twice the range. Since $a,b,c$ are integers between $-6$ and 6, the range $z-x$ is an integer from 0 to 12, so the sum can only be an even number $0,2,\dots ,24$. The largest is 24, attained for example at $-6,0,6$. The next attainable even value is 22, attained for example at $-6,0,5$ with $11+5+6=22$. A student who skips sorting is pushed into checking many triples, while sorting makes the arithmetic immediate.

The insight is to convert the repeating block to ninths and then turn denominator 33 into a gcd condition that can be counted. Since the two-digit block repeats every two places, $100N-N=99N$ equals the integer $AB=10a+b$, so $N=AB\mathrm{/}99$. In lowest terms the denominator is $99\mathrm{/}\gcd (AB,99)$. Setting this equal to 33 forces $\gcd (AB,99)=3$, so $AB$ is a multiple of 3 but of neither 9 nor 11. From 10 to 99 there are 30 multiples of 3; removing the 10 multiples of 9 leaves 20; removing the remaining multiples of 11, namely 33 and 66 with 99 already removed, leaves 18. Each gives a different $N$.

The insight is a case split on where the right angle sits, combining perpendicular slopes with Pythagorean distance. If the right angle is at $A$, then $AP$ must be vertical, so $x=0$, giving sides 6 and 3, not isosceles. If it is at $B$, then $x=6$, giving sides 6 and 3, again not isosceles. So any isosceles right triangle must be right-angled at $P$, with $AB$ as hypotenuse. Slopes from $P$ to $A$ and $B$ are 3 over $-x$ and 3 over $6-x$, whose product must be $-1$, so $x(6-x)=9$, or $(x-3{)}^2=0$. Hence $x=3$ is the only candidate. There $PA$ and $PB$ are each $\sqrt{9+9}=3\sqrt{2}$, and $18+18=36$, so it is an isosceles right triangle. Thus there is exactly one position.

The insight is a weighted-average shift with divisibility and bounding. In 2023 the overall pass fraction is 50 out of 80 equals 62.5 percent, with 50 percent and 75 percent by grade improving to 60 percent and 80 percent. With $n$ seventh-graders out of 100, the 2024 overall is 0.6 times $n$ plus 0.8 times 100 minus $n$ all over 100, which must be below 62.5, so $n$ is more than 87.5. Integer numbers of passers force $n$ to be a multiple of 5, so the smallest feasible $n$ is 90 with 54 plus 8 equals 62 passers for 62 percent. Ignoring multiples gives 88 and reversing the inequality gives 85.

The insight is that the diagonals share a midpoint, combined with a three-case split most students miss. In a parallelogram the midpoint of one diagonal equals the midpoint of the other, so if $D$ is opposite a known vertex, $D$ equals the sum of the other two vertices minus that opposite vertex. There are three choices for the opposite vertex: $A+B-C=(5,-4)$ with coordinate sum 1, $A+C-B=(-1,6)$ with sum 5, and $B+C-A=(9,10)$ with sum 19. Adding all possibilities gives $1+5+19=25$. Equivalently the total $x$ is $2+7+4=13$ and the total $y$ is $1+3+8=12$, for 25 altogether. A student who draws one parallelogram does three times the work yet finds only one of the three values.

The trick is to clear denominators first and let the given bounds certify the endpoints. Multiplying by 40 gives $56.4<40\sqrt{2}<56.8$ and $69.2<40\sqrt{3}<69.6$, so $k$ must satisfy $57\le k\le 69$, thirteen integers; indeed ${\displaystyle \frac{57}{40}}=1.425$ exceeds 1.42 and ${\displaystyle \frac{69}{40}}=1.725$ is below 1.73. Now sieve for lowest terms: ${\displaystyle \frac{k}{40}}$ is reduced exactly when $k$ is odd and not a multiple of 5. The odds from 57 through 69 are 57,59,61,63,65,67,69, seven numbers, and only 65 is a multiple of 5. Removing it leaves six fractions.

The trick is to work inside-out and watch parity pair the numbers. Start at the right: $2-1=1$, so $3-(2-1)=3-1=2$, then $4-2=2$, then $5-2=3$, then $6-3=3$. Every two layers increase the result by 1: with $V_k$ for the block from $k$ down to 1, $V_1=1$, $V_2=1$, $V_3=2$, $V_4=2$, and so on, so $V_{2m-1}=V_{2m}=m$. In particular the 60 numbers form 30 such pairs, giving 30. A student who subtracts left to right or who adds everything faces a long failing computation, while the pairing finishes in seconds.

The insight is volume subtraction combined with inclusion-exclusion for the overlapping tunnels. The uncut cube holds 343 cubic centimeters. Each straight tunnel has volume 28, so three tunnels total 84 if overlaps are ignored. Each pair of tunnels meets in an 8 cubic centimeter cube and all three meet in that same central 8 cube, so their union is 84 minus 24 plus 8, which is 68. Subtracting the union from the cube leaves 343 minus 68, which is 275 cubic centimeters.

The insight is to count the complement and then show the bound can be met by making the missing groups disjoint. The fractions missing each sport are $1\mathrm{/}4$, $1\mathrm{/}5$, and $1\mathrm{/}6$, whose sum is $15\mathrm{/}60+12\mathrm{/}60+10\mathrm{/}60=37\mathrm{/}60$. At most $37\mathrm{/}60$ of the class can miss at least one sport, so at least $1-37\mathrm{/}60=23\mathrm{/}60$ must play all three. This bound is attainable because $37\mathrm{/}60<1$: with 60 students, let 15 miss only soccer, 12 miss only basketball, and 10 miss only tennis; these groups are disjoint and use 37 students, leaving exactly 23 who play all three.

The insight is to bound the third side with the triangle inequality and then enforce the Pythagorean inequality for every longest side. Let the third side be $x$. Then $x+7>10$ and $7+10>x$, so $4\le x\le 16$. For an acute triangle the sum of the squares of the two shorter sides must exceed the square of the longest side. If $x\ge 10$ is longest, $7^2+10^2>x^2$ gives $x^2<149$, so $x\le 12$. If 10 is longest, $7^2+x^2>10^2$ gives $x^2>51$, so $x\ge 8$. Together $8\le x\le 12$, giving $8,9,10,11,12$, which is 5 values, and each indeed satisfies all three acute inequalities.

The insight is to rewrite the Pythagorean relation as a difference of squares and filter by parity. From $20^2+b^2=c^2$ we get $c^2-b^2=400$, so $(c-b)(c+b)=400$. Since their sum $2c$ is even, the two factors have the same parity, and since their product is even they must both be even. Writing $u=c-b$ and $v=c+b$ with $uv=400$ and $u<v$, the even pairs give $(2,200)$, $(4,100)$, $(8,50)$, and $(10,40)$, producing hypotenuses $(2+200)\mathrm{/}2=101$, $(4+100)\mathrm{/}2=52$, $(8+50)\mathrm{/}2=29$, and $(10+40)\mathrm{/}2=25$. Pairs with mixed parity give nonintegers and $(20,20)$ gives a degenerate segment, so they are excluded. The total is $101+52+29+25=207$.

The insight is that the altitude creates two smaller triangles similar to the whole, so the altitude is the geometric mean of the segments, and the integer difference then gives a quickly factored quadratic. Let the segments be $p$ and $p+7$. Similarity gives $12^2=p(p+7)$, so $p^2+7p-144=0$. This factors simply as $(p+16)(p-9)=0$, so the positive solution is $p=9$. The other segment is 16. The full hypotenuse is their sum, $9+16=25$. Checking, 9 times 16 is 144. Solving the three Pythagorean equations directly without the mean relation is much longer.

The insight is to cut open the box into a flat net so a surface route becomes a straight line, then minimize over the three ways to pair faces. A path that crosses two faces becomes the hypotenuse of a right triangle whose legs are the third edge and the sum of the other two. The three possibilities are $\sqrt{(4+5{)}^2+12^2}=\sqrt{225}=15$, $\sqrt{(4+12{)}^2+5^2}=\sqrt{281}$, and $\sqrt{(5+12{)}^2+4^2}=\sqrt{305}$. The first is smallest, so the shortest surface route has length 15. The interior diagonal $\sqrt{185}$ is shorter but leaves the surface and is not allowed.

The insight is that only the squarefree part matters: remove every squared factor and two numbers multiply to a square exactly when the remainders match. Here 1,4,9 leave 1; 2,8 leave 2; 3,12 leave 3; the rest are singletons. So equal-kernel pairs are 3 from the first group plus 1 plus 1, for 5 favorable pairs. There are 66 pairs from 12 numbers, so the probability is 5 over 66. Brute-force products would be long, but the kernel grouping makes it short.

The winning reframe is total minus excluded, with both parts paired. Pair $1+59,2+58,\dots$ to get the total 1 through 59 as ${\displaystyle \frac{59\cdot 60}{2}}=1770$. Multiples of 3 are $3(1+\cdots +19)=3\cdot 190=570$; multiples of 5 are $5(1+\cdots +11)=5\cdot 66=330$; multiples of both, hence of 15, are $15+30+45=90$. By inclusion-exclusion the excluded sum is $570+330-90=810$. Subtracting from the total gives $1770-810=960$. Listing and adding the nearly forty survivors would take far too long without a calculator.

The insight is that the unknown corner of the square lies on the hypotenuse, which fixes its side by similarity. If the side is $s$, the corner $(s,s)$ satisfies $s\mathrm{/}8+s\mathrm{/}6=1$, so $7s\mathrm{/}24=1$ and $s=24\mathrm{/}7$. Hence the square has area $576\mathrm{/}49$. The triangle has area 24. For a uniform point, probability is favorable area over total area, so $(576\mathrm{/}49)$ divided by 24 equals 24 over 49. Once the side is known, the arithmetic is one short division.

The insight is to treat the fold as a congruence that leaves a small right triangle whose sides satisfy the Pythagorean theorem and a linear relation. The hypotenuse is 15 since $9^2+12^2=225$. Keeping the acute end fixed lays 9 along 15, leaving $15-9=6$ exposed. Let the unknown distance along the side of length 12 be $x$. The folded-over segment also has length $x$ and stands perpendicular to the hypotenuse, so $x^2+6^2=(12-x{)}^2$. Expanding gives $x^2+36=144-24x+x^2$, so $24x=108$ and $x=9\mathrm{/}2$.

The insight is to count shared faces direction by direction instead of casework on corners and edges. Along one direction there are 3 · 3 rows each contributing 2 adjacencies, so 18 pairs; the same holds for each of the three directions, giving 54 adjacent pairs. Two cubes can be chosen in 351 ways. Thus the probability is 54 over 351, which reduces by 27 to 2 over 13.

The key insight is to turn strict fraction inequalities into positive integer gaps and then eliminate $p$ with a clever linear combination. From $2\mathrm{/}5<p\mathrm{/}q$ we get $5p-2q>0$, and since both sides are integers it is at least 1; similarly $p\mathrm{/}q<3\mathrm{/}7$ gives $3q-7p\ge 1$. Multiply the first gap by 7 and the second by 5 and add: $7(5p-2q)+5(3q-7p)=q\ge 7+5=12$. So any fraction strictly between must have denominator at least 12. This bound is sharp because $5\mathrm{/}12$ lies strictly between, since $2\mathrm{/}5=0.40<0.416\dots <0.428\dots =3\mathrm{/}7$. Hence the smallest possible denominator is 12.

The insight is to stop taking the root directly and instead trap $N$ between consecutive squares after factoring the large power. Write $2^{20}=(2^{10}{)}^2=1024^2$, so $N=1024^2+4096$. For any $k$, $(1024+k{)}^2=1024^2+2048k+k^2$. With $k=1$ this is $1048576+2048+1=1050625$. With $k=2$ this is $1048576+4096+4=1052676$. But $N=1048576+4096=1052672$, which is 2047 above the first square and 4 below the second. Hence $1025^2<N<1026^2$, so $\sqrt{N}$ lies strictly between 1025 and 1026 and the greatest integer below it is 1025.

The insight is the combined-distance invariant: at the first meeting the two walkers together have covered exactly one trail length, and at the second meeting they have covered exactly three trail lengths, so the speed ratio is preserved while distances triple. Let the trail be $L$. The north starter walks 240 first and $L+120$ by the second meeting, while the south starter walks $L-240$ first and $2L-120$ by the second. Equating ratios gives $(L+120)\mathrm{/}(2L-120)=240\mathrm{/}(L-240)$. Clearing gives $L^2-120L-28800=480L-28800$, so $L(L-600)=0$ and $L=600$. Without the three-length total, solvers set up separate unknowns for each speed and time and stall.

The insight is that the area formula depends on which side the line exits, so the slope must be bounded before solving and the extraneous root rejected. The whole area is 60, so the larger piece is 40 and the smaller 20. If $m$ is at most $6\mathrm{/}10$, the origin piece is $50m$; setting $50m=40$ gives $m=4\mathrm{/}5$, which violates the bound and is discarded. If $m$ exceeds $6\mathrm{/}10$, the origin piece is $60-18\mathrm{/}m$; setting $60-18\mathrm{/}m=40$ gives $18\mathrm{/}m=20$ and $m=9\mathrm{/}10$, which satisfies the bound. Hence only $9\mathrm{/}10$ works. Using one formula for all slopes keeps the invalid $4\mathrm{/}5$.

The insight is water volume conservation combined with a case check for partial versus full submersion. Water volume is 200. If the 6-cm cube were fully covered, depth would be $416\mathrm{/}100$, which is 4.16, less than 6, contradicting full cover. So the cube sticks out and water fills $100h$ minus $36h$, which equals 200. Thus $64h$ equals 200 and $h$ equals 3.125 centimeters.

The insight is to count face adjacencies by direction and then use favorable over total pairs. Along each axis there are 2 times 9, which is 18 adjacent pairs, so three directions give 54 favorable pairs. Two cubes chosen from 27 give 27 times 26 divided by 2, which is 351 total pairs. Hence the probability is $54\mathrm{/}351$, which reduces by 27 to $2\mathrm{/}13$.

The key is complementary counting with three-set inclusion exclusion finished by an explicit construction. Total likes count as 80 plus 75 plus 70 equals 225 percent while the union is 100 percent because everyone likes at least one fruit. Writing exactly-one plus exactly-two plus triple as 100 and exactly-one plus twice exactly-two plus three times triple as 225 gives exactly-two plus twice triple equals 125, so triple must be at least 25. The bound occurs with no exactly-one and 75 exactly-two split as 30 plus 25 plus 20 across pairs plus 25 triple, matching all three totals. Trying overlaps pair by pair without this relation becomes long casework.

The insight is that the diagonals of a parallelogram share a midpoint, so each fourth vertex is the sum of two given vertices minus the third. If the fourth vertex completes the diagonal between the first two given points, it equals $(1+5-9,4+10-2)=(-3,12)$ with sum 9; the other two completions give $(5,-4)$ with sum 1 and $(13,8)$ with sum 21. Adding the three possibilities gives $9+1+21=31$. Equivalently each given coordinate appears twice with a plus sign and once with a minus sign across the three cases, so the total is $(1+5+9)+(4+10+2)=31$ without listing all cases separately.