1997 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $a$ and $b$ are digits for which $\begin{array}{r} 2a\\[-2pt] {}\times b3\\ \hline 69\\ 92\phantom0\\ \hline 989 \end{array}$ then $a+b=$

$3$

$4$

$7$

$9$

$12$

Concepts:long multiplication digits

Difficulty rating: 920

Small Hint:

Use the first partial product to determine the two-digit top factor

Big Hint:

The second partial product is the top factor multiplied by $b$

Solution:

The first partial product says $3(20+a)=69,$ so $a=3.$ The second says $23b=92,$ so $b=4.$ Thus $a+b=7,$ and the correct answer is C.

2.

The adjacent sides of the decagon shown meet at right angles. What is its perimeter?

$22$

$32$

$34$

$44$

$50$

Concepts:perimeter rectilinear polygons

Difficulty rating: 1020

Small Hint:

The unlabeled rightward horizontal lengths together equal the bottom width

Big Hint:

Find the total vertical extent by combining the labeled height and the upper step

Solution:

The total of the rightward horizontal sides is $12,$ so all horizontal sides total $24.$ The full height is $8+2=10,$ so the vertical sides total $20.$ Therefore the perimeter is $24+20=44,$ and the correct answer is D.

3.

If $x,$ $y,$ and $z$ are real numbers such that $\begin{aligned} (x-3)^2+(y-4)^2\\ {}+(z-5)^2&=0, \end{aligned}$ then $x+y+z=$

$-12$

$0$

$8$

$12$

$50$

Concepts:nonnegative squares equations

Difficulty rating: 1060

Small Hint:

Each squared real quantity is nonnegative

Big Hint:

A sum of nonnegative terms is zero only when every term is zero

Solution:

All three squares are nonnegative, so each must be $0.$ Hence $x=3,$ $y=4,$ $z=5,$ and $x+y+z=12.$ The correct answer is D.

4.

If $a$ is $50\%$ larger than $c,$ and $b$ is $25\%$ larger than $c,$ then $a$ is what percent larger than $b?$

$20\%$

$25\%$

$50\%$

$100\%$

$200\%$

Concepts:percent change ratios

Difficulty rating: 920

Small Hint:

Write both $a$ and $b$ as multiples of $c$

Big Hint:

The requested percentage uses $b$ , not $c$ , as its base

Solution:

We have $a=1.5c$ and $b=1.25c.$ Thus $\frac{a}{b}=\frac{1.5}{1.25}=1.2,$ so $a$ is $20\%$ larger than $b.$ The correct answer is A.

5.

A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles?

$35.2$

$76$

$80$

$84$

$86$

Concepts:similar dimensions perimeter

Difficulty rating: 1260

Small Hint:

Let the short and long sides of a small rectangle be $x$ and $y$

Big Hint:

The total width is both $3x$ across the top and $2y$ across the bottom

Solution:

Let a small rectangle have sides $x$ and $y.$ The common width gives $3x=2y,$ while the large rectangle has dimensions $3x$ by $x+y.$ Its perimeter is $2(4x+y)=176.$ Since $y=\frac{3x}{2},$ this is $11x=176,$ so $x=16$ and $y=24.$ One small perimeter is $2(16+24)=80,$ making C correct.

6.

Consider the sequence $1,-2,3,-4,5,-6,\ldots,$ whose $n$ th term is $(-1)^{n+1}n.$ What is the average of the first $200$ terms of the sequence?

$-1$

$-0.5$

$0$

$0.5$

$1$

Concepts:sequences arithmetic mean

Difficulty rating: 1160

Small Hint:

Group consecutive terms into odd-even pairs

Big Hint:

Each pair has the same sum, and there are $100$ pairs

Solution:

Each pair $(2k-1)-2k$ has sum $-1.$ The $100$ pairs therefore total $-100,$ and their $200$ -term average is $-\frac{100}{200}=-0.5.$ The correct answer is B.

7.

The sum of seven integers is $-1.$ What is the maximum number of the seven integers that can be larger than $13?$

$1$

$4$

$5$

$6$

$7$

Concepts:integers extremal reasoning

Difficulty rating: 920

Small Hint:

All seven cannot exceed $13$ because their sum would be positive

Big Hint:

There is no lower bound on the remaining integer

Solution:

If all seven integers exceeded $13,$ their sum would be at least $98.$ Six can exceed $13$ : take six copies of $14$ and a seventh integer of $-85.$ Thus the maximum is $6,$ and the correct answer is D.

8.

Mientka Publishing Company prices its best seller Where’s Walter? as follows: $C(n)= \begin{cases} 12n, & 1\le n\le24,\\ 11n, & 25\le n\le48,\\ 10n, & 49\le n, \end{cases}$ where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ books cost less than $24$ books. For how many values of $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?

$3$

$4$

$5$

$6$

$8$

Concepts:piecewise functions inequalities

Difficulty rating: 1360

Small Hint:

Only the two price-break points can make a larger order cheaper

Big Hint:

Compare $C(n)$ with $C(25)$ near the first break and with $C(49)$ near the second

Solution:

At the first break, $C(25)=275$ is below $C(23)=276$ and $C(24)=288,$ giving $n=23$ and $n=24.$ At the second, $C(49)=490$ is below $C(n)=11n$ for $n=45,$ $n=46,$ $n=47,$ $n=48.$ No other $n$ works because each piece increases. There are $2+4=6,$ so D is correct.

9.

In the figure, $ABCD$ is a $2$ by $2$ square, $E$ is the midpoint of $\overline{AD},$ and $F$ is on $\overline{BE}.$ If $\overline{CF}$ is perpendicular to $\overline{BE},$ then the area of quadrilateral $CDEF$ is

$2$

$3-\frac{\sqrt3}{2}$

$\frac{11}{5}$

$\sqrt5$

$\frac94$

Concepts:coordinate geometry area

Difficulty rating: 1630

Small Hint:

Place $B=(0,0),$ $C=(2,0),$ $A=(0,2),$ and $E=(1,2)$

Big Hint:

Find $F$ as the intersection of $BE$ with the line through $C$ perpendicular to $BE$

Solution:

With the coordinates in the first hint, $BE$ has equation $y=2x,$ and the perpendicular through $C$ is $y=-\frac{(x-2)}{2}.$ Their intersection is $F=(\frac{2}{5},\frac{4}{5}).$ The square has area $4,$ while triangles $AEF$ and $BCF$ have areas $\frac{3}{5}$ and $\frac{4}{5},$ respectively. Hence $[CDEF]=4-\frac{3}{5}$ ${}-\frac{4}{5}=\frac{11}{5},$ so C is correct.

10.

Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the $4$ replaced by $3$ and the other die has the $3$ replaced by $4.$ When these dice are rolled, what is the probability that the sum is odd?

$\frac13$

$\frac49$

$\frac12$

$\frac59$

$\frac{11}{18}$

Concepts:probability parity

Difficulty rating: 1330

Small Hint:

An odd sum requires one odd result and one even result

Big Hint:

Count odd-labeled faces on each modified die, including repeated labels

Solution:

The first modified die has $4$ odd faces and $2$ even faces; the second has $2$ odd and $4$ even. Thus the probability is $(\frac{4}{6})(\frac{4}{6})+(\frac{2}{6})(\frac{2}{6})$ $=\frac{20}{36}=\frac{5}{9}.$ The correct answer is D.

11.

In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23,$ $14,$ $11,$ and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18,$ what is the least number of points she could have scored in the tenth game?

$26$

$27$

$28$

$29$

$30$

Concepts:averages inequalities

Difficulty rating: 1440

Small Hint:

Let $S$ be her point total in the first five games

Big Hint:

Use the nine-game comparison to maximize $S$ , then apply the strict ten-game average bound

Solution:

Games $6$ – $9$ total $68.$ The condition $\frac{S+68}{9}>\frac{S}{5}$ gives $S<85,$ so the greatest integral $S$ is $84.$ A ten-game average above $18$ requires a total at least $181,$ hence the tenth score is at least $181-(84+68)=29.$ This is attainable, so D is correct.

12.

If $m$ and $b$ are real numbers and $mb\gt0,$ then the line whose equation is $y=mx+b$ cannot contain the point

$(0,1997)$

$(0,-1997)$

$(19,97)$

$(19,-97)$

$(1997,0)$

Concepts:linear equations signs

Difficulty rating: 1280

Small Hint:

The condition $mb\gt0$ means the slope and vertical intercept have the same sign

Big Hint:

Substitute each point; a positive $x$ -intercept forces the slope and intercept to have opposite signs

Solution:

If $(1997,0)$ were on the line, then $0=1997m+b,$ so $b=-1997m$ and $mb=-1997m^2\lt0,$ a contradiction. Each other point can occur for suitable same-sign $m,b.$ Thus the correct answer is E.

13.

How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of $N$ is a perfect square?

$4$

$5$

$6$

$7$

$8$

Concepts:digits perfect squares

Difficulty rating: 1540

Small Hint:

Write $N=10x+y$ and add its reversal

Big Hint:

Determine when $11(x+y)$ , with $1\le x\le9$ and $0\le y\le9$ , can be square

Solution:

The sum is $11(x+y).$ Since $1\le x+y\le18,$ this is square only when $x+y=11,$ giving $121.$ The tens digit $x$ can be any of $2,3,\ldots,9,$ with $y=11-x.$ There are $8$ integers, so E is correct.

14.

The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1.$ If the populations in the years $1994,$ $1995,$ and $1997$ were $39,$ $60,$ and $123,$ respectively, then the population in $1996$ was

$81$

$84$

$87$

$90$

$102$

Concepts:recurrences proportionality

Difficulty rating: 1570

Small Hint:

Let $x$ be the $1996$ population and $k$ the constant of proportionality

Big Hint:

Translate the years $1994$ – $1996$ and $1995$ – $1997$ into two equations using the same $k$

Solution:

The rule gives $x-39=60k$ and $123-60=kx.$ Eliminating $k$ yields $x(x-39)=3780,$ or $(x-84)(x+45)=0.$ The population is positive, so $x=84,$ and B is correct.

15.

Medians $\overline{BD}$ and $\overline{CE}$ of triangle $ABC$ are perpendicular, $BD=8,$ and $CE=12.$ The area of triangle $ABC$ is

$24$

$32$

$48$

$64$

$96$

Concepts:medians perpendicular diagonals area

Difficulty rating: 1630

Small Hint:

View $BD$ and $CE$ as the diagonals of quadrilateral $BCDE$

Big Hint:

Triangle $ADE$ is similar to $ABC$ with scale factor $\frac{1}{2}$

Solution:

Quadrilateral $BCDE$ has perpendicular diagonals $BD=8$ and $CE=12,$ so its area is $\frac12(8)(12)=48.$ Since $D,E$ are midpoints, triangle $ADE$ has one fourth the area of $ABC,$ making $BCDE$ three fourths of it. Thus $[ABC]=\frac{48\cdot4}{3}=64,$ so D is correct.

16.

The three row sums and the three column sums of the array $\begin{bmatrix} 4&9&2\\ 8&1&6\\ 3&5&7 \end{bmatrix}$ are the same. What is the least number of entries that must be altered to make all six sums different from one another?

$1$

$2$

$3$

$4$

$5$

Concepts:arrays invariants construction

Difficulty rating: 1860

Small Hint:

With only three altered entries, examine the rows and columns containing no alteration

Big Hint:

For an upper bound, try changing four entries so that their row and column effects are all distinct

Solution:

With at most three alterations, either two of the six lines are unchanged, or some altered entry is the only alteration in both its row and its column. In the first case those two sums remain equal; in the second, that entry changes its row sum and column sum by the same amount, so those two sums remain equal. Four suffice: replace $4,$ $1,$ $2,$ $6$ by $5,$ $3,$ $7,$ $9,$ respectively. The resulting row sums are $21,$ $20,$ $15$ and column sums are $16,$ $17,$ $23.$ Hence the minimum is $4,$ and D is correct.

17.

A line $x=k$ intersects the graph of $y=\log_5x$ and the graph of $y=\log_5(x+4).$ The distance between the points of intersection is $0.5.$ Given that $k=a+\sqrt b,$ where $a$ and $b$ are integers, what is $a+b?$

$6$

$7$

$8$

$9$

$10$

Concepts:logarithms radical equations

Difficulty rating: 1590

Small Hint:

Because both points have $x$ -coordinate $k$ , their distance is the difference of their logarithms

Big Hint:

Combine the logarithms and exponentiate base $5$

Solution:

The vertical distance is $\begin{aligned} \log_5(k+4)-\log_5k &=\frac12,\\ \frac{k+4}{k}&=\sqrt5. \end{aligned}$ Thus $\frac{k+4}{k}=\sqrt5,$ so $k=\frac{4}{\sqrt5-1}=1+\sqrt5.$ Therefore $a+b=1+5=6,$ and A is correct.

18.

A list of integers has mode $32$ and mean $22.$ The smallest number in the list is $10.$ The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10,$ the mean and median of the new list would be $24$ and $m+10,$ respectively. If $m$ were instead replaced by $m-8,$ the median of the new list would be $m-4.$ What is $m?$

$16$

$17$

$18$

$19$

$20$

Concepts:mean median mode

Difficulty rating: 2010

Small Hint:

A total increase of $10$ raises the mean by $2$ , determining the list length

Big Hint:

Order the five entries and use the two stated median changes to identify the entries beside $m$

Solution:

The list has $\frac{10}{2}=5$ entries. Write them $10\le a\le m\le b\le c.$ Replacing $m$ by $m+10$ makes that value the median, so $b\ge m+10$ and $c\ge m+10;$ because the mode is $32,$ we must have $b=c=32.$ The original total is $110,$ giving $a+m=36.$ Replacing $m$ by $m-8$ makes the median $a=m-4,$ so $2m-4=36$ and $m=20.$ Thus E is correct.

19.

A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle $ABC$ as shown, where $AB=1.$ To the nearest hundredth, what is the radius of the circle?

$2.18$

$2.24$

$2.31$

$2.37$

$2.41$

Concepts:tangent circles right triangles

Difficulty rating: 2120

Small Hint:

If the circle radius is $r$ , then $O=(r,r)$ in the displayed coordinates

Big Hint:

Write the hypotenuse line of the $30^\circ$ - $60^\circ$ - $90^\circ$ triangle and set its distance from $O$ equal to $r$

Solution:

Take $A=(0,0),$ $B=(1,0),$ and $C=(0,\sqrt3).$ The hypotenuse is $\sqrt3x+y=\sqrt3,$ and the circle center is $O=(r,r).$ Tangency gives $\frac{\left|(\sqrt3+1)r-\sqrt3\right|}{2}=r.$ The pictured external circle has $(\sqrt3+1)r-\sqrt3=2r,$ so $r=\frac{\sqrt3}{\sqrt3-1}$ $=\frac{3+\sqrt3}{2}\approx2.37.$ Thus D is correct.

20.

Which one of the following integers can be expressed as the sum of $100$ consecutive positive integers?

$1{,}627{,}384{,}950$

$2{,}345{,}678{,}910$

$3{,}579{,}111{,}300$

$4{,}692{,}581{,}470$

$5{,}815{,}937{,}260$

Concepts:arithmetic sequences divisibility

Difficulty rating: 1410

Small Hint:

Write the terms as $a+1,a+2,\ldots,a+100$

Big Hint:

Their sum must be congruent to $50$ modulo $100$

Solution:

The sum is $100a+(1+\cdots+100)$ $=100a+5050,$ so it ends in $50.$ Only $1,627,384,950$ has that property, and it gives the positive integer $a=\frac{1,627,384,950-5050}{100}.$ Thus A is correct.

21.

For any positive integer $n,$ let $f(n)= \begin{cases} \log_8n, & \log_8n\in\mathbb{Q},\\ 0, & \log_8n\notin\mathbb{Q}. \end{cases}$ What is $\displaystyle\sum_{n=1}^{1997}f(n)?$

$\log_8 2047$

$6$

$\frac{55}{3}$

$\frac{58}{3}$

$585$

Concepts:logarithms prime factorization sums

Difficulty rating: 1860

Small Hint:

Determine which integers can be rational powers of $8=2^3$

Big Hint:

List the powers of $2$ not exceeding $1997$ and sum their base- $8$ logarithms

Solution:

For integer $n,$ $\log_8n$ is rational exactly when $n$ is a power of $2.$ The relevant values are $n=2^k$ for $0\le k\le10,$ and $f(2^k)=\frac{k}{3}.$ Therefore the sum is $\frac{0+1+\cdots+10}{3}=\frac{55}{3},$ so C is correct.

22.

Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $\$56.$ The absolute difference between the amounts Ashley and Betty had to spend was $\$19.$ The absolute difference between the amounts Betty and Carlos had was $\$7,$ between Carlos and Dick was $\$5,$ between Dick and Elgin was $\$4,$ and between Elgin and Ashley was $\$11.$ How much did Elgin have?

$\$6$

$\$7$

$\$8$

$\$9$

$\$10$

Concepts:absolute differences signed cases

Difficulty rating: 2010

Small Hint:

Assign a sign to each successive difference around the five-person cycle

Big Hint:

The signed differences must total zero before the five amounts can be summed

Solution:

The signed differences around the cycle have magnitudes $19,$ $7,$ $5,$ $4,$ $11$ and sum $0.$ Thus one side of the sign split must total half of $46,$ namely $23.$ The only split is $19+4=7+5+11.$ In one orientation, the amounts are $A=E+11,$ $B=E-8,$ $C=E-1,$ and $D=E+4.$ Their total is $5E+6,$ so $5E+6=56$ and $E=10.$ Reversing every sign would give $5E-6=56,$ not an integral $E.$ Thus Elgin had $\$10$ and E is correct.

23.

In the figure, polygons $A,$ $E,$ and $F$ are isosceles right triangles; $B,$ $C,$ and $D$ are squares with sides of length $1;$ and $G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is

$\frac12$

$\frac23$

$\frac34$

$\frac56$

$\frac43$

Concepts:nets polyhedron volume

Difficulty rating: 2120

Small Hint:

Recognize the three unit-square faces as faces meeting at a corner of a unit cube

Big Hint:

The triangular faces cap the solid obtained by slicing one corner from that cube

Solution:

The net forms a unit cube with one corner cut off. The removed corner is a triangular pyramid with three mutually perpendicular unit edges, so its volume is $\frac13\cdot\frac12\cdot1=\frac{1}{6}.$ The remaining polyhedron has volume $1-\frac{1}{6}=\frac{5}{6},$ and the correct answer is D.

24.

A rising number, such as $34689,$ is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom95=126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97$ th number in the list does not contain the digit

$4$

$5$

$6$

$7$

$8$

Concepts:combinations lexicographic order

Difficulty rating: 2120

Small Hint:

Count how many rising numbers begin with $1$ , then with $23$

Big Hint:

After those blocks, list the first few numbers beginning with $24$

Solution:

There are $\binom84=70$ beginning with $1.$ Among those beginning with $2,$ the first $\binom63=20$ begin with $23,$ occupying positions $71$ – $90.$ Thus the $97$ th overall is the seventh beginning with $24:$ $\begin{gathered} 24567,24568,24569,24578,\\ 24579,24589,24678. \end{gathered}$ The number $24678$ omits $5,$ so B is correct.

25.

Let $ABCD$ be a parallelogram and let $\overrightarrow{AA'},$ $\overrightarrow{BB'},$ $\overrightarrow{CC'},$ and $\overrightarrow{DD'}$ be parallel rays in space on the same side of the plane determined by $ABCD.$ If $AA'=10,$ $BB'=8,$ $CC'=18,$ $DD'=22,$ and $M$ and $N$ are the midpoints of $A'C'$ and $B'D',$ respectively, then $MN=$

$0$

$1$

$2$

$3$

$4$

Concepts:vectors three-dimensional geometry

Difficulty rating: 2010

Small Hint:

Use vectors and the parallelogram identity $A+C=B+D$

Big Hint:

The planar components of the two midpoints coincide; compare only their ray-direction components

Solution:

Choose the common ray direction as a unit vector $u.$ Then $\begin{aligned} M&=\frac{A+C}{2}+14u,\\ N&=\frac{B+D}{2}+15u. \end{aligned}$ Since $A+C=B+D$ for a parallelogram, $N-M=u,$ so $MN=1.$ The correct answer is B.

26.

Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B,$ angle $APB$ is twice angle $ACB,$ and $\overline{AC}$ intersects $\overline{BP}$ at point $D.$ If $PB=3$ and $PD=2,$ then $AD\cdot CD=$

$5$

$6$

$7$

$8$

$9$

Concepts:circles power of a point

Difficulty rating: 2170

Small Hint:

Draw the circle centered at $P$ through $A$ and $B$

Big Hint:

The central-angle condition puts $C$ on that circle, so apply intersecting chords at $D$

Solution:

Because $PA=PB,$ draw their circle with center $P.$ The condition $\angle APB=2\angle ACB$ is the central-inscribed angle relation, so $C$ lies on the same circle. Along line $PB,$ the other circle intersection is $E$ with $PE=3.$ Power of $D$ gives $\begin{aligned} AD\cdot CD&=DE\cdot DB\\ &=(3-2)(3+2)=5. \end{aligned}$ Thus A is correct.

27.

Consider those functions $f$ that satisfy $f(x+4)+f(x-4)=f(x)$ for all real $x.$ Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p.$

$8$

$12$

$16$

$24$

$32$

Concepts:functional equations recurrences periodicity

Difficulty rating: 2180

Small Hint:

For fixed $x$ , study the sequence $u_n=f(x+4n)$

Big Hint:

Use $u_{n+1}+u_{n-1}=u_n$ repeatedly to find when every initial pair returns

Solution:

For $u_n=f(x+4n),$ the equation is $u_{n+1}=u_n-u_{n-1}.$ Starting from $u_0,u_1,$ the sequence is $\begin{gathered} u_0,u_1,u_1-u_0,-u_0,\\ -u_1,u_0-u_1,u_0,u_1,\ldots, \end{gathered}$ so every such function has period $6\cdot4=24.$ This is least because $f(x)=\sin(\frac{\pi x}{12})$ satisfies the equation and has fundamental period $24.$ Hence D is correct.

28.

How many ordered triples of integers $(a,b,c)$ satisfy $\begin{aligned} |a+b|+c&=19,\\ ab+|c|&=97? \end{aligned}$

$0$

$4$

$6$

$10$

$12$

Concepts:Diophantine equations absolute values factorization

Difficulty rating: 2290

Small Hint:

Set $s=|a+b|$ , so $c=19-s$ , and split according to the sign of $c$

Big Hint:

When $c\lt0$ , rewrite the resulting equations as products equal to $117$

Solution:

If $c\ge0,$ substitution and factoring lead to no pair consistent with the required sign of $a+b.$ Thus $c\lt0,$ so $s\gt19$ and $ab+s=116.$ If $a+b=s,$ then $(a+1)(b+1)=117,$ producing the unordered pairs $\{0,116\},$ $\{2,38\},$ $\{8,12\}.$ If $a+b=-s,$ then $(a-1)(b-1)=117,$ producing $\{-116,0\},$ $\{-38,-2\},$ $\{-12,-8\}.$ Each unordered pair has two orders, giving $6\cdot2=12$ triples. The correct answer is E.

29.

Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7.$ For example, $\frac{700}{99}=7.07=7.070707\ldots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?

$7$

$8$

$9$

$10$

$1$ cannot be represented as a sum of finitely many special numbers

Concepts:decimal expansions place value construction

Difficulty rating: 2410

Small Hint:

If $a_k$ summands have $7$ in decimal place $k$ , divide the sum by $7$

Big Hint:

Compare the resulting digit counts with the repeating decimal for $\frac{1}{7}$ , then seek a six-digit repeating construction

Solution:

Suppose $1$ is a sum of $n$ special numbers, and let $a_k$ count summands having a $7$ in the $k$ th decimal place. Dividing by $7$ gives $\begin{aligned} \frac17&=\frac{a_1}{10}+\frac{a_2}{10^2}+\cdots\\ &=0.\overline{142857}. \end{aligned}$ For $n\le9,$ each $a_k$ is a digit, so $a_1,a_2,\ldots=1,4,2,8,5,7,\ldots;$ hence $n\ge8.$ Eight suffice because the repeating special decimals represented by $\begin{aligned} 700700+2(070707)\\ {}+2(077777)\\ {}+3(000777)&=999999. \end{aligned}$ Their sum is $1.$ Therefore the minimum is $8,$ and B is correct.

30.

For positive integers $n,$ denote by $D(n)$ the number of pairs of different adjacent digits in the binary (base two) representation of $n.$ For example, $D(3)=D(11_2)=0,$ $D(21)=D(10101_2)=4,$ and $D(97)=D(1100001_2)=2.$ For how many positive integers $n$ less than or equal to $97$ does $D(n)=2?$

$16$

$20$

$26$

$30$

$35$

Concepts:binary representation combinations

Difficulty rating: 2290

Small Hint:

A valid binary numeral consists of a block of $1$ s, then $0$ s, then $1$ s

Big Hint:

Count by bit length through six bits, then handle the seven-bit cutoff $97=1100001_2$ separately

Solution:

A $d$ -bit numeral with exactly two changes has form $1^a0^b1^c$ with positive $a,$ $b,$ $c,$ giving $\binom{d-1}{2}$ choices. For $d=3,$ $d=4,$ $d=5,$ $d=6,$ the total is $1+3+6+10=20.$ Among seven-bit numbers at most $1100001_2=97,$ the five forms beginning with one $1$ all work, and the only form beginning with at least two $1$ s is $1100001_2$ itself. Thus there are $20+6=26,$ and C is correct.