1998 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

Each of the sides of five congruent rectangles is labeled with an integer, as shown. These five rectangles are placed, without rotating or reflecting, in positions $I$ through $V$ so that the labels on coincident sides are equal. Which of the rectangles is in position $I?$

$A$

$B$

$C$

$D$

$E$

Concepts:logic edge matching

Difficulty rating: 1430

Small Hint:

The rectangle in position $II$ must match another rectangle on both vertical sides

Big Hint:

After finding position $II$ , match its left label to the right label of position $I$

Solution:

Position $II$ needs both its left and right labels to occur as right and left labels, respectively, on two other rectangles. Only $D,$ with side labels $7$ and $4,$ permits both matches: $E$ has right label $7,$ and $A$ has left label $4.$ Hence position $I$ contains $E,$ so the correct answer is E.

2.

Letters $A,$ $B,$ $C,$ and $D$ represent four different digits selected from $0,$ $1,$ $2,$ $\ldots,$ $9.$ If $\frac{A+B}{C+D}$ is an integer that is as large as possible, what is the value of $A+B?$

$13$

$14$

$15$

$16$

$17$

Concepts:digits optimization ratios

Difficulty rating: 1100

Small Hint:

Make the denominator as small as possible using distinct digits

Big Hint:

The two smallest digits and the two largest digits are disjoint pairs

Solution:

The least positive denominator is $0+1=1,$ and the greatest numerator using two other digits is $8+9=17.$ Their ratio is the integer $17,$ which is plainly maximal. Thus $A+B=17,$ and E is correct.

3.

If $a,$ $b,$ and $c$ are digits for which $\begin{array}{r} 7a2\\ {}-48b\\ \hline c73 \end{array}$ then $a+b+c=$

$14$

$15$

$16$

$17$

$18$

Concepts:subtraction place value digits

Difficulty rating: 1020

Small Hint:

The units column must borrow from the tens column

Big Hint:

Continue the borrowing through the tens and hundreds columns

Solution:

The units column gives $12-b=3,$ so $b=9.$ In the tens column, $a-1$ must borrow, and $a-1+10-8=7,$ giving $a=6.$ The hundreds column then gives $7-1-4=c=2.$ Therefore $a+b+c=6+9+2=17,$ so D is correct.

4.

Define $[a,b,c]$ to mean $\frac{a+b}{c},$ where $c\ne0.$ What is the value of $[[60,30,90],[2,1,3],[10,5,15]]?$

$0$

$0.5$

$1$

$1.5$

$2$

Concepts:defined operations substitution

Difficulty rating: 1070

Small Hint:

Evaluate each of the three inner brackets first

Big Hint:

All three inner brackets have the same value

Solution:

Each inner value is $1:$ $\frac{90}{90}=1,\frac{3}{3}=1,$ and $\frac{15}{15}=1.$ Thus the outer expression is $[1,1,1]=\frac{1+1}{1}=2.$ The correct answer is E.

5.

If $2^{1998}-2^{1997}$ ${}-2^{1996}+2^{1995}$ $=k\cdot2^{1995},$ what is the value of $k?$

$1$

$2$

$3$

$4$

$5$

Concepts:exponents factoring

Difficulty rating: 1160

Small Hint:

Factor out the smallest power of $2$

Big Hint:

The remaining coefficient is $2^3-2^2-2+1$

Solution:

Factoring gives $2^{1995}(8-4-2+1)=3\cdot2^{1995}.$ Hence $k=3,$ so C is correct.

6.

If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is

$8$

$15$

$17$

$47$

$93$

Concepts:factorization optimization

Difficulty rating: 1310

Small Hint:

Factor $1998$ and look for divisors near its square root

Big Hint:

The closest factor pair is formed from $37$ and the remaining factor

Solution:

Since $1998=2\cdot3^3\cdot37,$ its factor pair nearest $\sqrt{1998}$ is $37\cdot54.$ Their difference is $54-37=17,$ so C is correct.

7.

If $N\gt1,$ then $\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}=$

$N^{\frac{1}{27}}$

$N^{\frac{1}{9}}$

$N^{\frac{1}{3}}$

$N^{\frac{13}{27}}$

$N$

Concepts:radicals rational exponents

Difficulty rating: 1280

Small Hint:

Start with the innermost cube root and convert it to an exponent

Big Hint:

At each outer level, add $1$ to the exponent and divide by $3$

Solution:

The innermost radical is $N^{\frac{1}{3}}.$ The next is $(N\cdot N^{\frac{1}{3}})^{\frac{1}{3}}=N^{\frac{4}{9}}.$ The outermost is $(N\cdot N^{\frac{4}{9}})^{\frac{1}{3}}=N^{\frac{13}{27}}.$ Thus D is correct.

8.

A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x,$ the length of the longer parallel side of each trapezoid.

$\frac35$

$\frac23$

$\frac34$

$\frac56$

$\frac78$

Concepts:trapezoid area equal areas

Difficulty rating: 1540

Small Hint:

Each of the three regions has area $\frac{1}{3}$

Big Hint:

A trapezoid has height $\frac{1}{2}$ and parallel sides $x$ and $\frac{1}{2}$

Solution:

Each trapezoid has area $\frac{1}{3}.$ Its height is $\frac{1}{2},$ and its parallel sides have lengths $x$ and $\frac{1}{2}.$ Therefore $\frac12\cdot\frac12\left(x+\frac12\right)=\frac13,$ which gives $x=\frac{5}{6}.$ Thus D is correct.

9.

A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?

$24$

$27$

$30$

$33$

$36$

Concepts:weighted averages percentages

Difficulty rating: 1180

Small Hint:

The two halves of the remainder are each $35\%$ of the audience

Big Hint:

Weight the heard times $60,0,20,$ and $40$ by their audience fractions

Solution:

The average is $0.20(60)+0.10(0)$ $+0.35(20)+0.35(40)$ $=12+7+14=33$ minutes. Thus D is correct.

10.

A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimeter of each of the congruent rectangles is $14.$ What is the area of the large square?

$49$

$64$

$100$

$121$

$196$

Concepts:rectangle dimensions square area

Difficulty rating: 1200

Small Hint:

Let the sides of one rectangle be $x$ and $y$

Big Hint:

The side length of the large square is $x+y$

Solution:

If a rectangle has sides $x,y,$ then $2x+2y=14,$ so $x+y=7.$ The large square has side $x+y,$ hence area $7^2=49.$ The correct answer is A.

11.

Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R?$

$1$

$2$

$4$

$5$

$6$

Concepts:circles rectangles counting

Difficulty rating: 1310

Small Hint:

There are six unordered pairs of rectangle vertices

Big Hint:

The two diagonals determine the same circle

Solution:

The four sides determine four distinct diameter circles. The two diagonals are equal and share a midpoint, so they determine the same fifth circle. Hence there are $5,$ and D is correct.

12.

How many different prime numbers are factors of $N$ if $\log_2(\log_3(\log_5(\log_7N)))=11?$

$1$

$2$

$3$

$4$

$7$

Concepts:nested logarithms prime factors

Difficulty rating: 1360

Small Hint:

Undo the logarithms one at a time, starting with base $2$

Big Hint:

The final expression for $N$ is a power of $7$

Solution:

Successively exponentiating gives $N=7^{\,5^{\,3^{\,2^{11}}}}.$ Thus $7$ is the only prime factor of $N,$ so the correct answer is A.

13.

Walter rolls four standard six-sided dice and finds that the product of the numbers on the upper faces is $144.$ Which of the following could not be the sum of the upper four faces?

$14$

$15$

$16$

$17$

$18$

Concepts:factorization casework

Difficulty rating: 1700

Small Hint:

Factor $144=2^4\cdot3^2$ using four factors from $1$ through $6$

Big Hint:

Separate the cases with zero, one, or two faces showing $6$

Solution:

Valid rolls include $(4,4,3,3)$ with sum $14,$ $(6,4,3,2)$ with sum $15,$ $(6,6,2,2)$ with sum $16,$ and $(6,6,4,1)$ with sum $17.$ If two $6$ s occur, the other two faces have product $4$ and sum at most $5,$ giving total at most $17;$ the cases with fewer $6$ s also cannot total $18$ while retaining product $144.$ Thus $18$ is impossible, and E is correct.

14.

A parabola has vertex at $(4,-5)$ and has two $x$ -intercepts, one positive and one negative. If this parabola is the graph of $y=ax^2+bx+c,$ which of $a,$ $b,$ and $c$ must be positive?

only $a$

only $b$

only $c$

$a$ and $b$ only

none

Concepts:quadratic functions signs of coefficients

Difficulty rating: 1360

Small Hint:

The vertex lies below two intercepts, so determine the opening direction

Big Hint:

Use the signs of the roots’ product and of $-\frac{b}{2a}=4$

Solution:

The parabola opens upward, so $a\gt0.$ Its roots have opposite signs, so $\frac{c}{a}\lt0$ and hence $c\lt0.$ Also $-\frac{b}{2a}=4\gt0,$ so $b\lt0.$ Therefore only $a$ must be positive, and A is correct.

15.

A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?

$\sqrt3$

$2$

$\sqrt6$

$3$

$6$

Concepts:regular polygons area ratios

Difficulty rating: 1360

Small Hint:

Partition the regular hexagon into six equilateral triangles

Big Hint:

Equilateral-triangle area is proportional to the square of its side

Solution:

A regular hexagon of side $s$ is six equilateral triangles of side $s.$ An equal-area equilateral triangle of side $t$ therefore satisfies $t^2=6s^2,$ so $\frac{t}{s}=\sqrt6.$ The correct answer is C.

16.

The figure shown is the union of a circle and two semicircles of diameters $a$ and $b,$ all of whose centers are collinear. The ratio of the area of the shaded region to that of the unshaded region is

$\sqrt{\frac ab}$

$\frac ab$

$\frac{a^2}{b^2}$

$\frac{a+b}{2b}$

$\frac{a^2+2ab}{b^2+2ab}$

Concepts:circle area algebraic simplification

Difficulty rating: 1800

Small Hint:

Express each region using the large semicircle and one small semicircle

Big Hint:

Both resulting areas factor a common multiple of $a+b$

Solution:

The shaded area is $\begin{aligned} &\frac{\pi}{2}\left(\frac{a+b}{2}\right)^2 +\frac{\pi}{2}\left(\frac a2\right)^2\\ &\qquad-\frac{\pi}{2}\left(\frac b2\right)^2 =\frac{\pi a(a+b)}4. \end{aligned}$ Similarly, the unshaded area is $\frac{\pi b(a+b)}{4}.$ Their ratio is $\frac{a}{b},$ so B is correct.

17.

Let $f(x)$ be a function with the two properties:

(a) for any two real numbers $x$ and $y,$ $f(x+y)=x+f(y),$ and

(b) $f(0)=2.$

What is the value of $f(1998)?$

$0$

$2$

$1996$

$1998$

$2000$

Concepts:functional equations substitution

Difficulty rating: 1240

Small Hint:

Set $y=0$ in the functional equation

Big Hint:

The two properties determine $f(x)$ directly for every real $x$

Solution:

Taking $y=0$ gives $f(x)=x+f(0)=x+2.$ Hence $f(1998)=2000,$ so E is correct.

18.

A right circular cone of volume $A,$ a right circular cylinder of volume $M,$ and a sphere of volume $C$ all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then

$A-M+C=0$

$A+M=C$

$2A=M+C$

$A^2-M^2+C^2=0$

$2A+2M=3C$

Concepts:solid geometry volume formulas

Difficulty rating: 1360

Small Hint:

Let the common radius be $r$ , so the cone and cylinder height is $2r$

Big Hint:

Write all three volumes as multiples of $\pi r^3$

Solution:

The volumes are $A=\frac23\pi r^3,$ $M=2\pi r^3,$ and $C=\frac43\pi r^3.$ Therefore $A-M+C$ $=(\frac{2}{3}-2+\frac{4}{3})\pi r^3=0,$ so A is correct.

19.

How many triangles have area $10$ and vertices at $(-5,0),$ $(5,0),$ and $(5\cos\theta,5\sin\theta)$ for some angle $\theta?$

$0$

$2$

$4$

$6$

$8$

Concepts:coordinate geometry trigonometry area

Difficulty rating: 1570

Small Hint:

The fixed base has length $10$ , and the third vertex has height $|5\sin\theta|$

Big Hint:

Count the distinct points on the circle satisfying the resulting sine equation

Solution:

The area is $\frac12(10)|5\sin\theta|=25|\sin\theta|.$ Thus $|\sin\theta|=\frac{2}{5}.$ There are four corresponding points on the circle, and each gives a distinct triangle. Hence the answer is $4,$ making C correct.

20.

Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that

(a) the numbers are all different,

(b) they sum to $13,$ and

(c) they are in increasing order, left to right.

First, Casey looks at the number on the leftmost card and says, “I don’t have enough information to determine the other two numbers.” Then Tracy looks at the number on the rightmost card and says, “I don’t have enough information to determine the other two numbers.” Finally, Stacy looks at the number on the middle card and says, “I don’t have enough information to determine the other two numbers.” Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?

$2$

$3$

$4$

$5$

There is not enough information to determine the number.

Concepts:logic ordered triples

Difficulty rating: 2190

Small Hint:

List all increasing positive triples with sum $13$

Big Hint:

Eliminate triples in the order of the three statements, using what each speaker has learned

Solution:

The eight triples are $\begin{gathered} (1,2,10),(1,3,9),\\ (1,4,8),(1,5,7),\\ (2,3,8),(2,4,7),\\ (2,5,6),(3,4,6). \end{gathered}$ Casey’s statement excludes $(3,4,6).$ With that known, Tracy’s statement excludes rightmost values $10,9,$ and $6.$ The remaining triples are $(1,4,8),(2,3,8)$ and $(1,5,7),(2,4,7).$ A middle value $3$ or $5$ would now identify the triple, so Stacy’s statement forces the middle value $4.$ Thus C is correct.

21.

In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?

$\frac dh$

$0$

$\frac{d^2}{h}$

$\frac{h^2}{d}$

$\frac{d^2}{h-d}$

Concepts:rates algebra

Difficulty rating: 1630

Small Hint:

Express Windy’s speed as a fraction of Sunny’s speed using the first race

Big Hint:

In the second race Sunny travels $h+d$ meters before finishing

Solution:

If Sunny’s speed is $r,$ Windy’s is $\frac{r(h-d)}{h}.$ Sunny needs time $\frac{h+d}{r}$ in the second race, during which Windy runs $\frac{r(h-d)}h\cdot\frac{h+d}{r} =h-\frac{d^2}{h}.$ Sunny finishes at position $h,$ so the lead is $\frac{d^2}{h}.$ Thus C is correct.

22.

What is the value of the expression $\begin{aligned} &\frac1{\log_2 100!}+\frac1{\log_3 100!}\\ &\quad+\frac1{\log_4 100!}+\cdots\\ &\qquad+\frac1{\log_{100}100!}? \end{aligned}$

$0.01$

$0.1$

$1$

$2$

$10$

Concepts:logarithms sums

Difficulty rating: 1630

Small Hint:

Use the reciprocal identity $\frac{1}{\log_bx}=\log_xb$

Big Hint:

Combine the resulting logarithms of $2,3,\ldots,100$

Solution:

Each term equals $\log_{100!}b.$ Therefore the sum is $\begin{aligned} &\log_{100!}(2\cdot3\cdots100)\\ &\qquad=\log_{100!}(100!)=1. \end{aligned}$ Thus C is correct.

23.

The graphs of $x^2+y^2=4+12x+6y$ and $x^2+y^2=k+4x+12y$ intersect when $k$ satisfies $a\le k\le b,$ and for no other values of $k.$ Find $b-a.$

$5$

$68$

$104$

$140$

$144$

Concepts:circles completing the square intersection

Difficulty rating: 1860

Small Hint:

Complete the square to find both circle centers and radii

Big Hint:

Two circles intersect exactly when the center distance lies between the difference and sum of their radii

Solution:

The circles are $(x-6)^2+(y-3)^2=49,$ $(x-2)^2+(y-6)^2=k+40.$ Their centers are $5$ units apart. With second radius $r=\sqrt{k+40},$ intersection requires $|7-r|\le5\le7+r,$ or $2\le r\le12.$ Thus $-36\le k\le104,$ and $b-a=140.$ The correct answer is D.

24.

Call a $7$ -digit telephone number $d_1d_2d_3$ - $d_4d_5d_6d_7$ memorable if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,$ $1,$ $2,$ $\ldots,$ $9,$ the number of different memorable telephone numbers is

$19{,}810$

$19{,}910$

$19{,}990$

$20{,}000$

$20{,}100$

Concepts:inclusion-exclusion digit strings

Difficulty rating: 1800

Small Hint:

Count numbers satisfying each of the two matching conditions separately

Big Hint:

If both matches hold, all seven digits are forced by one repeated digit

Solution:

Each matching condition gives $10^4=10,000$ numbers: choose the three prefix digits and the one unconstrained remaining digit. If both hold, then $d_1d_2d_3=d_4d_5d_6=d_5d_6d_7,$ forcing all digits equal, so there are $10$ overlaps. Inclusion-exclusion gives $10,000+10,000-10=19,990,$ so C is correct.

25.

A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0),$ and $(7,3)$ is matched with $(m,n).$ Find $m+n.$

$6.7$

$6.8$

$6.9$

$7.0$

$8.0$

Concepts:reflections coordinate geometry

Difficulty rating: 2010

Small Hint:

The crease is the perpendicular bisector of the segment joining $(0,2)$ and $(4,0)$

Big Hint:

Find the line $y=2x-3$ , then reflect $(7,3)$ across it

Solution:

The crease passes through $(2,1)$ with slope $2,$ so it is $y=2x-3.$ The perpendicular through $(7,3)$ has slope $-\frac{1}{2}$ and meets the crease at $(\frac{19}{5},\frac{23}{5}).$ This intersection is the midpoint of $(7,3)$ and $(m,n),$ giving $m=\frac{3}{5},\ n=\frac{31}{5}.$ Hence $m+n=\frac{34}{5}=6.8,$ so B is correct.

26.

In quadrilateral $ABCD,$ it is given that $\angle A=120^\circ,$ angles $B$ and $D$ are right angles, $AB=13,$ and $AD=46.$ Then $AC=$

$60$

$62$

$64$

$65$

$72$

Concepts:cyclic quadrilaterals law of cosines circumradius

Difficulty rating: 2010

Small Hint:

The two right angles make $ABCD$ cyclic with $AC$ as a diameter

Big Hint:

Find $BD$ in triangle $ABD$ , then use the extended law of sines

Solution:

Because $\angle B+\angle D=180^\circ,$ the quadrilateral is cyclic, and $AC$ is its diameter. In triangle $ABD,$ $\begin{aligned} BD^2&=13^2+46^2\\ &\quad-2(13)(46)\cos120^\circ\\ &=2883=(31\sqrt3)^2. \end{aligned}$ By the extended law of sines, $AC=\frac{BD}{\sin120^\circ}$ $=\frac{31\sqrt3}{\frac{\sqrt3}{2}}=62.$ Thus B is correct.

27.

A $9\times9\times9$ cube is composed of twenty-seven $3\times3\times3$ cubes. The big cube is “tunneled” as follows: First, the six $3\times3\times3$ cubes which make up the center of each face as well as the center $3\times3\times3$ cube are removed as shown. Second, each of the twenty remaining $3\times3\times3$ cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The surface area of the final figure is

$384$

$729$

$864$

$1024$

$1056$

Concepts:surface area three-dimensional counting

Difficulty rating: 2290

Small Hint:

After the first stage, classify the twenty remaining large subcubes as corner or edge cubes

Big Hint:

For each second-stage tunnel, subtract exposed center squares and add the newly exposed tunnel walls

Solution:

After the first stage, $8$ corner subcubes contribute $27$ exposed units each, and $12$ edge subcubes contribute $36$ each. Tunneling a corner subcube removes $3$ exposed unit squares and adds $24$ tunnel-wall squares; tunneling an edge subcube removes $4$ and also adds $24.$ Hence the final area is $\begin{aligned} &8(27-3+24)\\ &\qquad+12(36-4+24)\\ &=384+672=1056. \end{aligned}$ Thus E is correct.

28.

In triangle $ABC,$ angle $C$ is a right angle and $CB\gt CA.$ Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB.$ If $\frac{AC}{AD}=\frac{2}{3},$ then $\frac{CD}{BD}=\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

$10$

$14$

$18$

$22$

$26$

Concepts:trigonometry angle multiplication

Difficulty rating: 2290

Small Hint:

Let $\angle DAB=\alpha$ , so $\angle CAB=3\alpha$ and $\cos2\alpha=\frac{AC}{AD}$

Big Hint:

Set $AC=1$ and express $CD$ and $CB$ using $\tan2\alpha$ and $\tan3\alpha$

Solution:

Let $\angle DAB=\alpha.$ Since $\frac{AC}{AD}=\cos2\alpha=\frac{2}{3},$ the identity for $\cos2\alpha$ gives $\tan\alpha=\frac{1}{\sqrt5}.$ Taking $AC=1,$ $\begin{aligned} CD&=\tan2\alpha=\frac{\sqrt5}{2},\\ CB&=\tan3\alpha=\frac{7\sqrt5}{5}. \end{aligned}$ Thus $BD=CB-CD=\frac{9\sqrt5}{10},$ and $\frac{CD}{BD}=\frac{5}{9}.$ Therefore $m+n=14,$ so B is correct.

29.

A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to

$4.0$

$4.2$

$4.5$

$5.0$

$5.6$

Concepts:lattice points extremal geometry

Difficulty rating: 2520

Small Hint:

Three collinear interior lattice points force a fourth, so use a smallest noncollinear lattice triangle

Big Hint:

At a maximal square, opposite sides are pinned by nearby lattice points; compare their separation

Solution:

It suffices to enclose the three noncollinear lattice points $(0,0),(0,1),(1,0).$ In a maximal placement, two opposite sides are pinned by neighboring lattice points; the greatest possible separation is the distance $\sqrt5$ between parallel lines through $(1,1)$ and $(0,-1).$ Hence the area is at most $5.$ This is attained by the square bounded by $\begin{aligned} y&=2x+2,\\ y&=2x-3,\\ 2y+x&=3,\\ 2y+x&=-2. \end{aligned}$ Its area is $5$ and exactly the three stated lattice points lie inside. Thus D is correct.

30.

For each positive integer $n,$ let $a_n=\frac{(n+9)!}{(n-1)!}.$ Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is

$1$

$3$

$5$

$7$

$9$

Concepts:prime valuations trailing digits modular arithmetic

Difficulty rating: 2630

Small Hint:

Write $a_n=n(n+1)\cdots(n+9)$ and compare its powers of $2$ and $5$

Big Hint:

An odd rightmost nonzero digit first becomes possible when the ten-term block contains $5^7$

Solution:

The five even terms in any ten consecutive integers contribute at least $2^8.$ Thus the rightmost nonzero digit can be odd only when the block contains at least eight factors of $5,$ first possible when it contains $5^7=78125.$ For $n=5^7-9,$ the block has $v_2=9$ and $v_5=8,$ so the digit remains even. For $n=5^7-8=78117,$ both valuations are $8.$ Cancelling $2^8 5^8$ and multiplying the remaining odd unit digits gives $\begin{aligned} &7\cdot9\cdot1\cdot3\cdot9\cdot3\\ &\qquad\cdot1\cdot1\cdot3 \equiv9\pmod{10}. \end{aligned}$ Hence the first odd rightmost nonzero digit is $9,$ and E is correct.