ISEE Upper · Competition-style problems · Part 4 of 8

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 to avoid solving for the sides and instead recover their sum with the identity $(a+b{)}^2=a^2+b^2+2ab$, then use the fact that any corresponding lengths share the same linear scale. Let the true sides be $a$ and $b$ in feet. Then $ab=1200$ and $a^2+b^2=50^2=2500$ by the Pythagorean theorem, so $(a+b{)}^2=2500+2400=4900$ and $a+b=70$, giving a true perimeter of 140 feet. The drawing diagonal of 2.5 cm corresponds to 50 feet, so the linear factor is $2.5\mathrm{/}50=1\mathrm{/}20$ cm per foot. Hence the drawing perimeter is $140\mathrm{/}20=7$ cm, a short computation once the sum is seen.

The insight is to translate divisor count into an equation for exponents and then use the size bound to pick the winner. For $N=2^x{3}^y$ the divisor count is $(x+1)(y+1)$, so $(x+1)(y+1)=12$ with $x,y\ge 1$ gives $(x,y)=(5,1),(3,2),(2,3),(1,5)$, producing $N=96,72,108,486$. Imposing $N<150$ leaves 96, 72, and 108, whose greatest is $108=2^2\cdot 3^3$. Factoring each candidate and counting divisors by hand faces four separate factorizations, while the exponent equation lists all possibilities at once.

The key is that the two groups overlap in the two middle numbers, so adding their totals double counts the median pair, and an even-count median is their average. The total of all eight is $8(20)=160$. The five smallest total $5(16)=80$ and the five largest total $5(26)=130$, together 210. The excess $210-160=50$ is the sum of the fourth and fifth numbers counted twice. For eight numbers the median is the average of those two, so $50\mathrm{/}2=25$. No listing is needed once the overlap is seen.

The insight is to subtract before adding, using cancellation of binomial products together with reordering of the sum. Expanding gives $f(x)=x^2+10x+16$ and $g(x)=x^2+10x+24$, so $f(x)-g(x)=-8$ for every $x$ because the $x^2$ and $x$ terms cancel. Hence the total difference is the sum of fifty copies of $-8$, namely 50 times $-8$ equals $-400$. A student who first sums $f(1)+\mathrm{.}\mathrm{.}\mathrm{.}+f(50)$ must handle $1^2+\mathrm{.}\mathrm{.}\mathrm{.}+50^2$ and $1+\mathrm{.}\mathrm{.}\mathrm{.}+50$, which is very long without a calculator, while subtracting first leaves one short multiplication.

The insight is factoring by grouping into distinct lines plus lattice counting with overlap correction. Write $2x^2+7xy+3y^2=2x^2+6xy+xy+3y^2=2x(x+3y)+y(x+3y)=(2x+y)(x+3y)$. Hence the equation means $2x+y=0$ or $x+3y=0$. With $-6\le x\le 6$ and $-6\le y\le 6$, the line $y=-2x$ forces $\mathrm{\mid }x\mathrm{\mid }\le 3$, giving 7 pairs, and the line $x=-3y$ forces $\mathrm{\mid }y\mathrm{\mid }\le 2$, giving 5 pairs. The pair $(0,0)$ lies on both lines, so inclusion-exclusion gives $7+5-1=11$ distinct pairs. Checking all 169 grid points one by one is the long failing route.

The idea is to work from the center. The diagonals of a square bisect each other, are equal, and are perpendicular. The center is the midpoint $(4,4)$, and the vector from the center to $(7,6)$ is $(3,2)$. Turning that vector a quarter turn gives $(-2,3)$ and $(2,-3)$, so the other two vertices are $(4-2,4+3)=(2,7)$ and $(4+2,4-3)=(6,1)$. Check: the distance from $(1,2)$ to $(2,7)$ is $\sqrt{1+25}=\sqrt{26}$, and from $(2,7)$ to $(7,6)$ it is $\sqrt{25+1}=\sqrt{26}$, as a square requires.

The key is to factor to expose the given relation so the cubic collapses to a constant without solving for $x$. Since $x^3-4x^2-7x=x(x^2-4x-7)$ and $x^2-4x=7$ means $x^2-4x-7=0$, the product is $x$ times zero, leaving 11. Solving $x^2-4x-7=0$ would need the quadratic formula with $\sqrt{44}$ and then cubing radicals, which is very long without a calculator, while factoring makes the answer immediate.

The insight is symmetry: the line $3x+5y=60$ is a diagonal of the rectangle with corners $(0,0)$ and $(20,12)$, and the map $(x,y)\to (20-x,12-y)$ swaps the two sides of that diagonal. Inside the rectangle there are $19\times 11=209$ points with integer coordinates. The points exactly on the line are those with $x$ a multiple of 5: $(5,9)$, $(10,6)$, $(15,3)$, which is 3 points. The other 206 points split evenly between the two sides, so the region $3x+5y<60$ holds ${\displaystyle \frac{206}{2}}=103$ of them.

The insight is an auxiliary-parallel re-cut plus a proportional split. Draw through $P$ a line parallel to $l$ and $m$; by alternate interior angles it splits the smaller angle $APB$ into exactly the two interior angles, so their sum is $100^{\circ }$, a fact that is not visible without the helper line. With ratio $2:3$, the parts are $2+3=5$ equal shares of $100^{\circ }$, so each share is $20^{\circ }$. The larger interior angle is $3\times 20=60^{\circ }$ and the smaller is $40^{\circ }$. Trying to relate the ratio to $180^{\circ }$ instead of $100^{\circ }$ fails.

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 idea is to locate the center by symmetry. The center lies on the perpendicular bisector of the side joining the two vertices, which is the line through the middle of the square. Tangency to the opposite side means the center is $r$ away from that side, so it is $8-r$ away from the side containing the two vertices. The right triangle with legs 4 (half the side) and $8-r$ has hypotenuse $r$: $(8-r{)}^2+16=r^2$, so $80-16r=0$ and $r=5$.

The insight is that sticker unit rates mislead once whole bags force waste, so the effective cost must include leftover ounces and the integer covering matters. The small bag costs 0.25 per ounce and the large bag costs 0.23 per ounce, so the large bag looks cheaper and 28 times 0.23 is 6.44, which is impossible because bags are whole. Checking whole-bag covers by number of large bags, zero larges needs five smalls for 7.50, one large needs three smalls for exactly 28 ounces at 2.30 plus 4.50 equals 6.80, two larges need two smalls for 7.60, and three larges cover 30 ounces for 6.90. The exact-fit mix beats the all-large cover, so the minimum is 6.80.

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 insight is that the order of operations forces exponents and division before subtraction, so each factor is genuinely 1 minus 1 over a square, and then each such factor splits by difference of squares into n minus 1 times n plus 1 over n squared. Writing the five factors as 1 times 3 over 2 times 2 times 2 times 4 over 3 times 3 times 3 times 5 over 4 times 4 times 4 times 6 over 5 times 5 times 5 times 7 over 6 times 6 lets almost every integer cancel diagonally. What remains is 1 over 6 times 7 over 2, which is 7 over 12, while multiplying the uncancelled numerators and denominators directly gives huge numbers that are hard to reduce without a calculator.

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.

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 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 an extremal bound plus an explicit construction to show the bound is sharp. The total must be $8\times 11=88$, a one-step equation. To have many numbers above 14, make everything as small as the rules allow: numbers above 14 cost at least $15,16,17,18,19$ and numbers below cost at least $1,2,3$. With five numbers above 14, the cheapest distinct list is $1,2,3,15,16,17,18,19$, whose sum is 91, already above 88, so five or more is impossible. With four, $1+2+3+4+15+16+17+18=76$, leaving 12 to add while staying distinct and above 14; $1,2,3,4,15,16,17,30$ totals 88 with four numbers above 14. So the maximum is 4.

The insight is to regroup the cost as a multiple of the item count plus a notebook surcharge and then check endpoints with the nonstrict bounds. With $p$ pens, the cost is $4n+3p=3(n+p)+n$, which is at most 35, while $n+p$ is at least 10. Hence $30+n\le 3(n+p)+n\le 35$, giving $n\le 5$, while $n\ge 0$ by definition. Every value occurs: $n$ notebooks together with $10-n$ pens uses exactly 10 items and costs $30+n$ dollars, which is at most 35 for each of $n=0,1,2,3,4,5$, so the attainable values are exactly 0 through 5.

The key is a cheapest-list bound plus a construction attaining it, since an angle has a positive complement exactly when it is acute under $90^{\circ }$. Four distinct angles with no complement need at least $90+91+92+93=366^{\circ }$, using $90^{\circ }$ as the smallest non-acute integer and distinctness to force the increase, which already exceeds $360^{\circ }$, so four is impossible. Three is attainable, for example $1^{\circ }+91^{\circ }+92^{\circ }+176^{\circ }=360^{\circ }$, with three distinct angles at least $90^{\circ }$ and one acute angle, all distinct. Hence the greatest possible number without a complement is three.

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 to cancel first and only then apply the 2-and-5 terminating test, splitting on whether 7 cancels. A fraction in lowest terms terminates exactly when its denominator has no prime besides 2 and 5. If 7 does not divide $n$, then $7\mathrm{/}n$ is already reduced, so $n$ itself must equal $2^a{5}^b$; below 30 these are 2, 4, 5, 8, 10, 16, 20, and 25, a total of 8. If 7 divides $n$, write $n=7m$ so $7\mathrm{/}n=1\mathrm{/}m$, and $m$ must equal $2^a{5}^b$ with $7m<30$; this gives $m=1$, 2, 4 and $n=7$, 14, 28. Long division on 28 values is long, while the cancel-first prime test is short, for a total of 8 plus 3 equals 11.

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.

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 read a route as a choice of positions for the right steps and then fix the double subtraction with inclusion-exclusion. A route uses 8 steps with 4 rights, so there are $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$ routes in total. Through $(1,1)$ there are $\left(\genfrac{}{}{0px}{}{2}{1}\right)\times \left(\genfrac{}{}{0px}{}{6}{3}\right)=2\times 20=40$ routes, and through $(2,3)$ there are $\left(\genfrac{}{}{0px}{}{5}{2}\right)\times \left(\genfrac{}{}{0px}{}{3}{1}\right)=10\times 3=30$ routes. Routes through both must visit $(1,1)$ then $(2,3)$, giving $\left(\genfrac{}{}{0px}{}{2}{1}\right)\times \left(\genfrac{}{}{0px}{}{3}{1}\right)\times \left(\genfrac{}{}{0px}{}{3}{1}\right)=2\times 3\times 3=18$ routes. Hence routes avoiding both are $70-40-30+18=18$.

The insight is to count the complement instead of listing numbers containing 7. There are $9\times 10\times 10=900$ three-digit numbers. Numbers with no 7 have 8 choices for the hundreds digit (1 to 9 except 7) and 9 choices each for the tens and ones (0 to 9 except 7), so $8\times 9\times 9=648$. Thus numbers with at least one 7 are $900-648=252$. Letting $252\mathrm{/}(252+n)=1\mathrm{/}3$ gives $756=252+n$, so $n=504$. Listing all 900 numbers would take far too long.

Bound width and height separately, then multiply, and show the bound is sharp. In one direction two length 8 intervals inside length 10 overlap in at least $8+8-10=6$, since pushing them apart leaves at most 2 of slack. The same holds vertically, so overlap width is at least 6 and height at least 6, giving area at least 36. Placing the squares in opposite corners makes the overlap exactly 6 by 6, achieving 36. Using only total areas gives $64+64-100=28$, which allows disconnected overlap shapes that axis-aligned rectangles cannot realize.

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 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 key insight is to simplify the right side and force the left side to share its squarefree part. Since $\sqrt{98}=7\sqrt{2}$, the equation becomes $\sqrt{m}+\sqrt{n}=7\sqrt{2}$, so each of $m$ and $n$ must be twice a perfect square, say $m=2a^2$ and $n=2b^2$ with $a+b=7$. The positive pairs for $(a,b)$ are $(1,6)$, $(2,5)$, and $(3,4)$, giving $(m,n)$ pairs $(2,72)$, $(8,50)$, and $(18,32)$. Only 18 and 32 are both less than 50, so the admissible unordered pair is 18 and 32, whose sum is 50. A solver who squares immediately faces messy integer hunting, while the squarefree view finishes quickly.

The insight is to divide first to isolate a fixed remainder and then count divisors instead of testing infinitely many $n$. Long division gives $n^2+3n+14=(n+2)(n+1)+12$, so the fraction equals $n+2+{\displaystyle \frac{12}{n+1}}$. Hence $n+1$ must be a positive divisor of 12 exceeding 1. The eligible divisors are $2,3,4,6,12$, giving $n=1,2,3,5,11$, five values. Testing $n$ one by one would never end, while the remainder makes it a short divisor list.

The insight is to locate the lone two from the even-count median straddle and then balance the mean to recover threes and fours. With 20 scores the median is the average of the 10th and 11th smallest. Nine scores are 1, so the 10th is at least 2. To average 2.5 the 10th must be 2 and the 11th must be 3, which forces exactly one 2, so $x=1$. Then $y+z=10$ from the total $9+1+y+z=20$. The total points are $20\times 2.2=44$, while $9\times 1+1\times 2=11$ are already accounted for, leaving $3y+4z=33$. Substituting $y=10-z$ gives $30+z=33$, so $z=3$ and $y=7$. Hence 7 of 20 scores are 3, giving ${\displaystyle \frac{7}{20}}$.