ISEE Upper · Competition-style problems · Part 6 of 8

The idea is to read the quadrants from the signs of the slope and the intercept. Missing Quadrant II while reaching Quadrants I and III requires a rising line, $m>0$, that crosses the $y$-axis below the origin, $b<0$. A line through the origin ($b=0$) touches only Quadrants I and III, and a horizontal line ($m=0$) misses Quadrant I. Now count with $m+\mathrm{\mid }b\mathrm{\mid }\le 4$: $m=1$ allows $b=-1,-2,-3$; $m=2$ allows $b=-1,-2$; $m=3$ allows $b=-1$. That is $3+2+1=6$ lines.

The insight is to add the perimeter equations and read off the border area as width times average perimeter without ever finding the length and width. Let the outer garden be $l$ by $w$ and the inner hole be $l-6$ by $w-6$ since 3 feet are removed from each side. Then the outer perimeter is $2l+2w$ and the inner is $2l+2w-24$, so their sum is $4l+4w-24=120$ and $l+w=36$. The path area is $lw-(l-6)(w-6)=6(l+w)-36=216-36=180$. Trying to find $l$ and $w$ separately is impossible from one sum, a failing route, while the combined quantity finishes in seconds.

The insight is the complement to 2, after which all deficits share a numerator. Write each fraction as 1 minus a unit fraction: ${\displaystyle \frac{4}{5}}=1-1\mathrm{/}5$ and ${\displaystyle \frac{13}{14}}=1-1\mathrm{/}14$, so the first sum is $2-(1\mathrm{/}5+1\mathrm{/}14)=2-19\mathrm{/}70$. Likewise the sums equal $2-19\mathrm{/}78$, $2-19\mathrm{/}88$, and $2-19\mathrm{/}90$ because $1\mathrm{/}6+1\mathrm{/}13=19\mathrm{/}78$, $1\mathrm{/}8+1\mathrm{/}11=19\mathrm{/}88$, and $1\mathrm{/}9+1\mathrm{/}10=19\mathrm{/}90$. With equal numerators, the fraction with the largest denominator is smallest, so $19\mathrm{/}90<19\mathrm{/}88<19\mathrm{/}78<19\mathrm{/}70$. Subtracting the smallest deficit leaves the largest total, so the last sum with deficit 19/90 is greatest. Direct addition needs four large common denominators, while the complement needs only one short unit sum each.

The insight is that napkins differ in size, so the per-napkin rate is the wrong unit and total usable area is the invariant that must be compared. Each napkin in the first pack is 96 square inches and 90 of them total 8640 square inches, while each napkin in the second pack is 72 square inches and 120 of them also total 8640 square inches, which is 60 square feet. The unit costs are therefore 5.40 divided by 60 equals 0.09 per square foot and 6.00 divided by 60 equals 0.10 per square foot. The first pack wins on area although it loses per napkin, 0.06 versus 0.05, so the cheaper area price is 0.09.

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 express the unknown angle from the interior sum and then trap it with two bounds. Let the unknown be $L$. Then $(n-2)180=151(n-1)+L$, so $L=29n-209$. Convex means $L<180$, so $29n<389$ and $n<13.5$. Largest means $L>151$, so $29n>360$ and $n>12.4$. The only integer in that open interval is $n=13$, giving $L=168$, which is indeed the largest and below 180. Testing values of $n$ one by one without both inequalities is much longer.

The insight is the surround-and-subtract re-cut with coordinate differences. The triangle fits in the box $0\le x\le 6$, $0\le y\le 5$ of area 30. The three right triangles outside it have legs 6 and 1, 4 and 4, 2 and 5, so areas 3, 8 and 5. Subtracting gives $30-16=14$. Trying a slanted base with the distance formula needs square roots and is very long without a calculator, while the box uses only halves of whole numbers.

The insight is to rearrange the intercept form into a factored constant and then count ordered divisors. Write intercepts as $a$ and $b$ so $4\mathrm{/}a+5\mathrm{/}b=1$, hence $4b+5a=ab$ and adding 20 gives $(a-4)(b-5)=20$. With $a>4$ and $b>5$ each positive divisor of 20 gives one line, and 20 has divisors 1,2,4,5,10,20 for 6 lines, realized by $(5,25)$, $(6,15)$, $(8,10)$, $(9,9)$, $(14,7)$, $(24,6)$.

The insight is the hidden-substitution reframe plus a squares-are-nonnegative bound. Let $y=x^2+2x$. Then the equation becomes $y^2+5y+6=0$, which factors as $(y+2)(y+3)=0$, so $y=-2$ or $y=-3$. Thus $x^2+2x+2=0$ or $x^2+2x+3=0$. Completing the square gives $x^2+2x+2=(x+1{)}^2+1$, which is at least 1 for every real $x$, and $x^2+2x+3=(x+1{)}^2+2$, which is at least 2. Neither can be zero, so there are no real solutions. Expanding to a quartic and searching for roots without the substitution is the long failing route.

First graph to locate the slanted edge. With $y=0$, $7x\le 154$ gives intercept $(22,0)$, and with $x=0$, $11y\le 154$ gives intercept $(0,14)$, so the slanted edge joins $(22,0)$ to $(0,14)$. Its rise is 14 over run $-22$, whose magnitudes reduce to 7 over 11 in lowest terms by dividing by 2. Thus moving from one lattice point to the next along the edge requires 11 steps in $x$ and 7 in $y$ to stay integral. Starting at $(22,0)$, one such step reaches $(11,7)$ and another reaches $(0,14)$, with no others in between. Hence there are 3 integer points on the edge, found without testing all 23 integer $x$-values.

The insight is to give every choice the same even power of ten so the question becomes which leftover integer is a square. Write each number as $k\times 10^4$: the values are $18\times 10^4$, $25\times 10^4$, $180\times 10^4$, and $250\times 10^4$. Since $10^4=(10^2{)}^2$ is already a perfect square, the whole product is a square exactly when $k$ is a square. Only $25=5^2$ qualifies, giving $(5\times 10^2{)}^2=500^2=250000=2.5\times 10^5$. Trying square roots of numbers near two million is slow without a calculator, while one common-factor rewrite reduces everything to recognizing 25.

The insight is to sort the numbers by their remainder when divided by 3. Remainder 0: $3,6,\dots ,18$, which is 6 numbers. Remainder 1: $1,4,\dots ,19$, which is 7 numbers. Remainder 2: $2,5,\dots ,20$, which is 7 numbers. A sum is a multiple of 3 when both numbers have remainder 0, giving $\left(\genfrac{}{}{0px}{}{6}{2}\right)=15$ pairs, or when one has remainder 1 and the other remainder 2, giving $7\cdot 7=49$ pairs. That is 64 of the $\left(\genfrac{}{}{0px}{}{20}{2}\right)=190$ pairs, so the probability is ${\displaystyle \frac{64}{190}}={\displaystyle \frac{32}{95}}$, slightly more than ${\displaystyle \frac{1}{3}}$.

The insight is that the sum of all 25 products factors by the distributive property into the sum from the first bag times the sum from the second bag, so no one needs to list products. The first sum is 30 with average 6, and the second sum is 60 with average 12. Hence the total of all products is 30 times 60, which is 1800, and dividing by 25 pairs gives 72, which also equals 6 times 12. Listing all products and averaging directly is long and error prone without a calculator, while the factored averages finish quickly.

The idea is that the two slanted lines meet at a single apex, $(3,7)$, and the triangle’s width grows in proportion to its distance from that apex, on either side. At height $k$ the lines are at $x=k-4$ and $x={\displaystyle \frac{13-k}{2}}$, a width of ${\displaystyle \frac{3}{2}}\mathrm{\mid }7-k\mathrm{\mid }$, and the height of the triangle is $\mathrm{\mid }7-k\mathrm{\mid }$. So the area is ${\displaystyle \frac{3}{4}}(7-k{)}^2=27$, giving $(7-k{)}^2=36$ and $k=1$ or $k=13$. Both give a triangle with base 9 and height 6, so the sum is 14.

The insight is to factor out the smallest power of 2 before asking about squares. Then $N=2^n(1+2^3+2^4)=2^n(1+8+16)=25\cdot 2^n=5^2\cdot 2^n$. A square times $5^2$ is a square exactly when $2^n$ is a square, and $2^n$ is a square exactly when $n$ is even, since prime exponents in a square are even. So $n$ must be even. Between 1 and 30 there are 15 even integers, and each indeed gives $N=(5\cdot 2^{n\mathrm{/}2}{)}^2$. Expanding powers directly faces rapidly growing values, while the factored form finishes the count immediately.

The key idea is that a line and its reflection over $y=x$ meet on the mirror line itself. Reflecting swaps $x$ and $y$, so the image is $x=2y+6$, or $y={\displaystyle \frac{x-6}{2}}$. The original line meets the $x$-axis at $(-3,0)$ and the image meets it at $(6,0)$. On $y=x$, the original line gives $x=2x+6$, so the two lines meet at $(-6,-6)$. The triangle has base $6-(-3)=9$ on the $x$-axis and height 6, so its area is ${\displaystyle \frac{1}{2}}\cdot 9\cdot 6=27$.

The insight is completing the square to get a difference of squares plus a same-parity divisor argument for squares. Write $n^2+14n+13=(n+7{)}^2-36$. If this equals $k^2$ with $k\ge 0$, then $(n+7{)}^2-k^2=36$, so $(n+7-k)(n+7+k)=36$. For positive $n$, both factors are positive and their sum $2(n+7)$ is even, so both factors are even. The even factor pairs of 36 are 2 and 18 and 6 and 6. The pair 2 and 18 gives $n+7=10$, so $n=3$ with $k=8$, and indeed $9+42+13=64$. The pair 6 and 6 gives $n=-1$, not positive. Mixed-parity pairs would give half-integers and are impossible. Hence the smallest positive $n$ is 3, while blind testing of values would be unfocused.

The insight is the straight-angle total combined with prime parity. Let the three angles be $p$, $q$ and $r$ with $q=p+36$ and $p+q+r=180$, so $2p+r=144$. Since 144 and $2p$ are even, $r$ must be even, and the only even prime is 2, so $r=2$. Then $p+q=178$ and $q-p=36$, giving $2q=214$ and $q=107$, with $p=71$. All three values are prime, so the triple works. Without seeing that $r$ must be 2, a solver must test many prime pairs that sum to 180, which is long without a calculator.

The idea is that perpendicular lines have slopes whose product is $-1$. Segment $AB$ has slope ${\displaystyle \frac{2}{6}}={\displaystyle \frac{1}{3}}$, so line $BC$ has slope $-3$: $y-2=-3(x-6)$. At $x=0$ this gives $y=20$, so $C=(0,20)$. Now take side $AC$, which lies on the $y$-axis and has length 20, as the base; the height to it is the horizontal distance from $B$ to the $y$-axis, which is 6. The area is ${\displaystyle \frac{1}{2}}\cdot 20\cdot 6=60$.

The key is to re-base percents that refer to different wholes and then balance the two counts. Let the numbers of boys and girls be $x$ and $y$. Walkers total $0.20x+0.80y$, and girls who walk total $0.80y$. Since half the walkers are girls, $0.80y=0.50(0.20x+0.80y)$, so $0.80y=0.10x+0.40y$ and $0.40y=0.10x$, giving $x=4y$. Hence girls are one out of five students, which is $20\mathrm{\%}$. A student who treats every percent as a share of the same whole is pushed into equating the walker share with the school share directly.

The key insight is to count the complement with inclusion-exclusion instead of enumerating many cell pairs. There are $\left(\genfrac{}{}{0px}{}{12}{2}\right)=66$ equally likely pairs. Same-grade pairs total $3\left(\genfrac{}{}{0px}{}{4}{2}\right)=18$, same-fruit pairs total $2\left(\genfrac{}{}{0px}{}{6}{2}\right)=30$, and pairs sharing both total $6\left(\genfrac{}{}{0px}{}{2}{2}\right)=6$. By inclusion-exclusion, pairs sharing a grade or a fruit total $18+30-6=42$, so pairs differing in both total $66-42=24$. The probability is $24\mathrm{/}66=4\mathrm{/}11$. Direct enumeration over many cell pairs is much longer, while $36\mathrm{/}66$ and $48\mathrm{/}66$ ignore one condition.

The insight is that the product is largest when $2x$ and $3y$ are about equal, which here means $2x=3y=15$, or $x=7.5$ and $y=5$, with product 37.5. Those are not integers, so check the integer points near that balance, using the largest $x$ allowed for each $y$. For $y=3$: $x\le 10$, product 30. For $y=4$: $x\le 9$, product 36. For $y=5$: $x\le 7$, product 35. For $y=6$: $x\le 6$, product 36. For $y=7$: $x\ge 7$ would need $14+21\le 30$, which fails. The greatest product is 36.

The key combines the area-product invariant with a perimeter comparison. Let the square side be $s$. After lengthening one direction by $25\mathrm{\%}$, that side is $5s\mathrm{/}4$. To keep area $s^2$, the other side must be $s^2\mathrm{/}(5s\mathrm{/}4)=4s\mathrm{/}5$, a $20\mathrm{\%}$ decrease found by division rather than subtraction. The new perimeter is $2(5s\mathrm{/}4+4s\mathrm{/}5)=2(41s\mathrm{/}20)=41s\mathrm{/}10=4.10s$ versus $4s$, so the increase is $0.10s\mathrm{/}4s=1\mathrm{/}40$, which is $2.5\mathrm{\%}$. A student who subtracts percents is pushed into a longer route that confuses linear and multiplicative change.

The trick is to count games by pairs and then use the total points invariant. Five teams give 5 times 4 divided by 2 equals 10 games. All decisive games would give 30 points, while the table totals 8 plus 5 plus 5 plus 3 plus 3 equals 24. Each tie replaces 3 points with 2 points, so each tie costs exactly 1 point. The shortfall 30 minus 24 equals 6 must be the number of ties, achieved by records such as 2 wins and 2 ties for 8 points with matching opponents.

The insight is periodicity of a full block plus prime-factor containment. Since 36 is a multiple of 4, 6, 9, 12, 18, and 36, any 36 consecutive integers contain 9, 6, 4, 3, 2, and 1 of those multiples respectively. A multiple of 4 and 9 has factors 2 squared and 3 squared, so it is a multiple of 6, meaning the 4-and-9-only region is empty and the triple count equals the 4-and-9 count 1. Hence exactly-two equals $(3-1)+(2-1)+0=3$.

The insight is inclusion-exclusion for three overlapping multiple sets plus careful handling of a strict inequality with a negative coefficient. Multiples of 2 contribute 50, of 3 contribute 33, of 5 contribute 20, for 103 total. Subtract multiples of $6,10,15$, which are $16,10,6$, leaving 71, then add back multiples of 30, which are 3, giving $C=74$. Then $-2n>-148$ flips on division by $-2$ to $n<74$. The greatest integer below 74 is 73, since 74 makes both sides equal.

The insight is to group digits by remainder modulo 3 and then permute within each feasible remainder pattern. The remainders are 0 for 3,6,9, 1 for 1,4, and 2 for 2. A three-digit sum is a multiple of 3 only for $0+0+0$ and $0+1+2$. The first gives 1 digit set with 6 orders, and the second gives $3\times 2\times 1=6$ digit sets with 6 orders each, for 36 numbers. Adding the 6 from the first pattern gives 42 numbers, while checking all 120 ordered triples would take far too long.

The insight is to add the six orthogonal views instead of counting cubes. The top and bottom each see the 3 by 4 footprint, giving 12 and 12. The front and back each see $4+3+1=8$. The left and right each see $3+2+1=6$. Adding gives $12+12+8+8+6+6=52$. Counting cube by cube forces tracking dozens of hidden glued faces and usually misses the bottom or double counts the steps.

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 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 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 write the common ratio in lowest terms and use a divisibility invariant, then enumerate divisors with a bound. Let the terms be $a$, $ar$ and $a{r}^2$ with $r=p\mathrm{/}q$ in lowest terms and $q>0$. Then $a$ must supply $q^2$ in its factorization, so write $a=k{q}^2$ with $k$ a positive integer. The three terms become $k{q}^2$, $kpq$ and $k{p}^2$ and their sum is $k(q^2+pq+p^2)=91$, where 91 is 7 times 13. Since $p$ and $q$ are positive coprime integers, $q^2+pq+p^2$ must be a divisor of 91, so 7, 13 or 91. Checking gives $1+2+4=7$ with $k=13$ for the triple 13,26,52, and $1+3+9=13$ with $k=7$ for the triple 7,21,63, and $1+9+81=91$ with $k=1$ for the triple 1,9,81 after factoring $p^2+p-90=0$. The first triple has middle term 26, which exceeds 24, while the other two have middle terms 21 and 9, which are below 24. So different admissible triples order differently.

The insight is to combine exponential growth with modular periodicity instead of computing the huge term. The 100th term is $5\cdot 2^{99}$, far too large for hand division. Work modulo 11: powers of 2 cycle $1,2,4,8,5,10,9,7,3,6$ and back to 1, a period of 10. Since $99=90+9$, $2^{99}$ leaves the same remainder as $2^9=512$, and $512=11\cdot 46+6$, so it is 6. Then $5\cdot 6=30$ leaves remainder 8 upon division by 11. Listing all 100 remainders would be long, but the cycle makes it short.

Because the quadrilateral is convex, its diagonals cross at an interior point $O$. Apply the strict triangle inequality in each of the four small triangles with vertex $O$: side $AB$ is shorter than $AO$ plus $BO$, side $BC$ is shorter than $BO$ plus $CO$, side $CD$ is shorter than $CO$ plus $DO$, and side $DA$ is shorter than $DO$ plus $AO$. Adding all four strict inequalities counts each piece $AO$, $BO$, $CO$ and $DO$ twice, so the perimeter $P$ is shorter than twice the sum $d1+d2$. Dividing gives $d1+d2$ larger than $P\mathrm{/}2$ for every convex quadrilateral.

The key is mean centering plus the fact that a square is never negative. The numbers 3,7,8,10 have mean 7, so shift by writing $x=7+t$. Then $f=(t-4{)}^2+t^2+(t+1{)}^2+(t+3{)}^2=4t^2+26$ after expanding, since the linear terms cancel. Equivalently $f(x)=4(x-7{)}^2+26$. Because $(x-7{)}^2$ is at least 0, the smallest possible value is 26, reached at $x=7$, without expanding four squares and completing the square the long way.

The trick is to write area through the fraction of a full circle. Let the radius be $r$ and the arc be $20-2r$. The sector is the fraction $(20-2r)\mathrm{/}2\pi r$ of the whole circle $\pi r^2$, which simplifies to $r(10-r)=25-(r-5{)}^2$, so no sector exceeds 25. This bound is attainable with $r=5$ and arc 10, which is shorter than the full circumference $10\pi$ and gives area $5\cdot 5=25$. But $r=8$ with arc 4 is also valid since 4 is shorter than $16\pi$, and it gives area $8\cdot 2=16$, strictly smaller. Since one admissible sector matches the second quantity and another is smaller, the comparison is not fixed.

The insight is to combine the arithmetic-series sum with the polygon angle-sum formula to get a quadratic, then use convexity to discard the extraneous root. Let $n$ be the number of sides. The angle sum from the sequence is $120n+5n(n-1)\mathrm{/}2$ and from geometry is $(n-2)180$. Hence $120n+5n(n-1)\mathrm{/}2=180n-360$, so $n(n-1)=24n-144$, that is $n^2-25n+144=0$, which factors as $(n-9)(n-16)=0$. Thus $n$ is 9 or 16. If $n$ were 16, the largest angle would be $120+75=195$ degrees, exceeding 180 degrees and contradicting convexity. For $n=9$ the largest angle is 160 degrees, which is convex, so the polygon has 9 sides.