What is the difference between the sum of the first
2003 even counting numbers and the sum of the first
2003 odd counting numbers?
(A) 0
(B) 1
(C) 2
(D) 2003
(E) 4006
2003 AMC 10 A, Problem #1—
2003 AMC 12 A, Problem #1—
“Each even counting number is one more than the
preceding odd counting number.”
Solution
Answer (D): Each even counting number, beginning with 2, is one
more than the preceding odd counting number. Therefore the difference is
(1)(2003) = 2003.
Difficulty: Medium
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
Number Theory > Integers
Members of the Rockham Soccer League buy socks
and T–shirts. Socks cost $4 per pair and each T–
shirt costs $5 more than a pair of socks. Each member
needs one pair of socks and a shirt for home games
and another pair of socks and a shirt for away games.
If the total cost is $2366, how many members are in
the League?
(A) 77
(B) 91
(C) 143
(D) 182
(E) 286
2003 AMC 10 A, Problem #2—
2003 AMC 12 A, Problem #2—
“Each member will have two pairs of socks, and two
shirts.”
Solution
Answer (B): The cost for each member is the price of two pairs of socks,
$8, and two shirts, $18, for a total of $26. So there are 2366/26 = 91
members.
Difficulty: Easy
NCTM Standard: Algebra Standard for Grades 9–12: Use mathematical models to represent and
understand quantitative relationships.
Mathworld.com Classification:
Number Theory > Arithmetic > General Arithmetic
A solid box is 15 cm by 10 cm by 8 cm. A new solid
is formed by removing a cube 3 cm on a side from
each corner of this box. What percent of the original
volume is removed?
(A) 4.5
(B) 9
(C) 12
(D) 18
(E) 24
2003 AMC 10 A, Problem #3—
2003 AMC 12 A, Problem #3—
“The total volume of the eight removed cubes is 8×33 =
216 cubic centimeters.”
Solution
Answer (D): The total volume of the eight removed cubes is 8 × 33 =
216 cubic centimeters, and the volume of the original box is 15 × 10 ×
8¡ = 1200
cubic centimeters. Therefore the volume has been reduced by
¢
216
(100%)
= 18%.
1200
Difficulty: Medium-hard
NCTM Standard: Geometry Standard for Grades 9–12: Analyze properties and determine
attributes of two- and three-dimensional objects.
Mathworld.com Classification:
Geometry > Solid Geometry > Polyhedra > Cubes
It takes Mary 30 minutes to walk uphill 1 km from
her home to school, but it takes her only 10 minutes
to walk from school to home along the same route.
What is her average speed, in km/hr, for the round
trip?
(A) 3
(B) 3.125
(C) 3.5
(D) 4
(E) 4.5
2003 AMC 10 A, Problem #4—
2003 AMC 12 A, Problem #4—
“Mary walks a total of 2 km in 40 minutes.”
Solution
Answer (A): Mary walks a total of 2 km in 40 minutes. Because 40
minutes is 2/3 hr, her average speed, in km/hr, is 2/(2/3) = 3.
Difficulty: Medium
NCTM Standard: Algebra Standard for Grades 9–12: Identify essential quantitative relationships
in a situation and determine the class or classes of functions that might model the relationships.
Mathworld.com Classification:
Number Theory > Arithmetic > General Arithmetic
Let d and e denote the solutions of 2x2 + 3x − 5 = 0.
What is the value of (d − 1)(e − 1)?
(A) −
5
2
(B) 0
(C) 3
(D) 5
(E) 6
2003 AMC 10 A, Problem #5—
“Solve for d and e .”
Solution
Answer (B): Since
0 = 2x2 + 3x − 5 = (2x + 5)(x − 1) we have d = −
5
and e = 1.
2
So (d − 1)(e − 1) = 0.
OR
If x = d and x = e are the roots of the quadratic equation ax2 +bx+c = 0,
then
c
b
de =
and d + e = − .
a
a
For our equation this implies that
µ ¶
3
5
(d − 1)(e − 1) = de − (d + e) + 1 = − − −
+ 1 = 0.
2
2
Difficulty: Medium
NCTM Standard: Algebra Standard for Grades 9–12: Represent and analyze mathematical
situations and structures using algebraic symbols.
Mathworld.com Classification:
Calculus and Analysis > Roots
Define x♥y to be |x − y| for all real numbers x and
y. Which of the following statements is not true?
(A) x♥y = y♥x for all x and y
(B) 2(x♥y) = (2x)♥(2y) for all x and y
(D) x♥x = 0 for all x
(C) x♥0 = x for all x
(E) x♥y > 0 if x 6= y
2003 AMC 10 A, Problem #6—
2003 AMC 12 A, Problem #6—
“Test each choice.”
Solution
Answer (C): For example, −1♥0 = | − 1 − 0| = 1 6= −1. All the other
statements are true:
(A) x♥y = |x − y| = | − (y − x)| = |y − x| = y♥x for all x and y.
(B) 2(x♥y) = 2|x − y| = |2x − 2y| = (2x)♥(2y) for all x and y.
(D) x♥x = |x − x| = 0 for all x.
(E) x♥y = |x − y| > 0 if x 6= y.
Difficulty: Medium-hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand meanings of
operations and how they relate to one another.
Mathworld.com Classification:
Recreational Mathematics > Cryptograms > Cryptarithmetic
How many non-congruent triangles with perimeter 7
have integer side lengths?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
2003 AMC 10 A, Problem #7—
2003 AMC 12 A, Problem #7—
“The longest side cannot be greater than 3 .”
Solution
Answer (B): The longest side cannot be greater than 3, since otherwise
the remaining two sides would not be long enough to form a triangle. The
only possible triangles have side lengths 1–3–3 or 2–2–3.
Difficulty: Medium-hard
NCTM Standard: Geometry Standard for Grades 9–12: Analyze properties and determine
attributes of two- and three-dimensional objects.
Mathworld.com Classification:
Geometry > Plane Geometry > Triangles
What is the probability that a randomly drawn positive
factor of 60 is less than 7?
(A)
1
10
(B)
1
6
(C)
1
4
(D)
1
3
(E)
1
2
2003 AMC 10 A, Problem #8—
2003 AMC 12 A, Problem #8—
“Write out all factors of 60 .”
Solution
Answer (E): The factors of 60 are
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
Six of the twelve factors are less than 7, so the probability is 1/2.
Difficulty: Medium-easy
NCTM Standard: Data Analysis and Probability Standard for Grades 9–12: Understand and
apply basic concepts of probability.
Mathworld.com Classification:
Probability and Statistics > Probability
Simplify
(A)
√
x
s r
q
3
√
3
3
x x x x.
(B)
√
3
x2
(C)
√
27
x2
(D)
√
54
x
(E)
√
81
x80
2003 AMC 10 A, Problem #9—
“Write it out in exponential form.”
Solution
Answer (A): We have
s r
q
3
√
1 1 1 1
3
3
x x x x = (x(x(x · x 2 ) 3 ) 3 ) 3
1
3
1
1
= (x(x(x 2 ) 3 ) 3 ) 3
1
1
1
= (x(x · x 2 ) 3 ) 3
3
1
1
1
1
3
1
1
= (x(x 2 ) 3 ) 3 = (x · x 2 ) 3 = (x 2 ) 3 = x 2 =
√
x.
Difficulty: Hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand meanings of
operations and how they relate to one another.
Mathworld.com Classification:
Calculus and Analysis > Special Functions > Radicals > Cube Root
The polygon enclosed by the solid lines in the figure
consists of 4 congruent squares joined edge-to-edge.
One more congruent square is attached to an edge at
one of the nine positions indicated. How many of the
nine resulting polygons can be folded to form a cube
with one face missing?
6
7
5
8
4
9
2
3
1
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
2003 AMC 10 A, Problem #10—
2003 AMC 12 A, Problem #13—
“Construct the existing squares to a cube with faces
missing.”
Solution
Answer (E): If the polygon is folded before the fifth square is attached,
then edges a and a0 must be joined, as must b and b0 . The fifth face of
the cube can be attached at any of the six remaining edges.
b'
a'
a,a'
b
a
b,b'
Difficulty: Medium-hard
NCTM Standard: Geometry Standard for Grades 9–12: Explore relationships (including
congruence and similarity) among classes of two- and three-dimensional geometric objects, make
and test conjectures about them, and solve problems involving them.
Mathworld.com Classification:
Geometry > Solid Geometry > Polyhedra > Cubes
The sum of the two 5-digit numbers AM C10 and
AM C12 is 123422. What is A + M + C?
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
2003 AMC 10 A, Problem #11—
2003 AMC 12 A, Problem #5—
“AM C10 = AM C×100+10, AM C12 = AM C×100+12 .”
Solution
Answer (E): Since the last two digits of AM C10 and AM C12 sum to
22, we have
AM C + AM C = 2(AM C) = 1234.
Hence AM C = 617, so A = 6, M = 1, C = 7, and A + M + C =
6 + 1 + 7 = 14.
Difficulty: Medium-easy
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
Recreational Mathematics > Cryptograms > Alphametic
A point (x, y) is randomly picked from inside the
rectangle with vertices (0, 0), (4, 0), (4, 1), and (0, 1).
What is the probability that x < y ?
(A)
1
8
(B)
1
4
(C)
3
8
(D)
1
2
(E)
3
4
2003 AMC 10 A, Problem #12—
“The point (x, y) satisfies x < y if and only if it belongs
to the triangle bounded by the lines x = y, y = 1, and
x = 0 .”
Solution
Answer (A): The point (x, y) satisfies x < y if and only if it belongs
to the shaded triangle bounded by the lines x = y, y = 1, and x = 0, the
area of which is 1/2. The ratio of the area of the triangle to the area of
the rectangle is 1/2
= 18 .
4
y
y=x
x
Difficulty: Medium-hard
NCTM Standard: Geometry Standard for Grades 9–12: Use Cartesian coordinates and other
coordinate systems, such as navigational, polar, or spherical systems, to analyze geometric situations.
Mathworld.com Classification:
Recreational Mathematics > Cryptograms > Alphametic
The sum of three numbers is 20. The first is 4 times
the sum of the other two. The second is seven times
the third. What is the product of all three?
(A) 28
(B) 40
(C) 100
(D) 400
(E) 800
2003 AMC 10 A, Problem #13—
“Use an equation to represent the relations.”
Solution
Answer (A): Let a, b, and c be the three numbers. Replace a by four
times the sum of the other two to get
4(b + c) + b + c = 20,
so
b + c = 4.
Then replace b with 7c to get
7c + c = 4,
1
so c = .
2
The other two numbers are b = 7/2 and a = 16, and the product of the
three is 16 · 7/2 · 1/2 = 28.
OR
Let the first, second, and third numbers be x, 7x, and 32x, respectively.
Then 40x = 20 so x = 21 and the product is
µ ¶
1
3
(32)(7)x = (32)(7)
= 28.
8
Difficulty: Medium
NCTM Standard: Algebra Standard for Grades 9–12: Use mathematical models to represent and
understand quantitative relationships.
Mathworld.com Classification:
Number Theory > Arithmetic > General Arithmetic
Let n be the largest integer that is the product of
exactly 3 distinct prime numbers, d, e and 10d + e,
where d and e are single digits. What is the sum of
the digits of n?
(A) 12
(B) 15
(C) 18
(D) 21
(E) 24
2003 AMC 10 A, Problem #14—
“Try different single digits primes for d and e .”
Solution
Answer (A): The largest single-digit primes are 5 and 7, but neither 75
nor 57 is prime. Using 3, 7, and 73 gives 1533, whose digits have a sum
of 12.
Difficulty: Hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
History and Terminology > Notation > Prime
What is the probability that an integer in the set
{1, 2, 3, . . . , 100} is divisible by 2 and not divisible
by 3 ?
(A)
1
6
(B)
33
100
(C)
17
50
(D)
1
2
(E)
18
25
2003 AMC 10 A, Problem #15—
“There are b 100
6 c = 16 that are divisible by both 2 and
3 .”
Solution
Answer (C): Of the 100
= 50 integers that are divisible by 2, there are
2
100
b 6 c = 16 that are divisible by both 2 and 3. So there are 50 − 16 = 34
that are divisible by 2 and not by 3, and 34/100 = 17/50.
Difficulty: Medium
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
Number Theory > Divisors > Divisible
What is the units digit of 132003 ?
(A) 1
(B) 3
(C) 7
(D) 8
(E) 9
2003 AMC 10 A, Problem #16—
“Powers of 13 have the same units digit as the
corresponding powers of 3 .”
Solution
Answer (C): Powers of 13 have the same units digit as the corresponding
powers of 3; and
31 = 3,
32 = 9,
33 = 27,
34 = 81,
and 35 = 243.
Since the units digit of 31 is the same as the units digit of 35 , units digits
of powers of 3 cycle through 3, 9, 7, and 1. Hence the units digit of 32000
is 1, so the units digit of 32003 is 7. The same is true of the units digit of
132003 .
Difficulty: Medium-hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Compute fluently and make
reasonable estimates.
Mathworld.com Classification:
Calculus and Analysis > Special Functions > Powers
The number of inches in the perimeter of an
equilateral triangle equals the number of square inches
in the area of its circumscribed circle. What is the
radius, in inches, of the circle?
√
3 2
(A)
π
√
3 3
(B)
π
(C)
√
3
(D)
6
π
(E)
√
3π
2003 AMC 10 A, Problem #17—
“Construct an equation with radius of the circle to
represent the relation above.”
Solution
Answer (B): Let the triangle have vertices A, B, and C, let O be the
center of the circle, and let D be the midpoint of BC. Triangle COD
is a 30–60–90 degree triangle. If r √is the radius of the circle, then the
sides
are r, r/2, and r 3/2. The perimeter of 4ABC is
³ √ of´ 4COD
√
√
r 3
= 3r 3, and the area of the circle is πr2 . Thus 3r 3 = πr2 ,
6 2
√
and r = (3 3)/π.
A
r
C
O
r 2
r=√3 2 D
B
Difficulty: Hard
NCTM Standard: Geometry Standard for Grades 9–12: Analyze characteristics and properties of
two- and three-dimensional geometric shapes and develop mathematical arguments about geometric
relationships.
Mathworld.com Classification:
Geometry > Plane Geometry > Triangles > Special Triangles > Equilateral Triangles
Geometry > Plane Geometry > Circles
What is the sum of the reciprocals of the roots of the
equation
2003
1
x + 1 + = 0?
2004
x
(A) −
2004
2003
(B) −1
(C)
2003
2004
(D) 1
(E)
2004
2003
2003 AMC 10 A, Problem #18—
“Reform the equation to quadratic equation.”
Solution
Answer (B): Let a = 2003/2004. The given equation is equivalent to
ax2 + x + 1 = 0.
If the roots of this equation are denoted r and s, then
rs =
so
1
a
1
and r + s = − ,
a
1 1
r+s
+ =
= −1.
r s
rs
OR
If x is replaced by 1/y, then the roots of the resulting equation are the
reciprocals of the roots of the original equation. The new equation is
2003
2003
+ 1 + y = 0 which is equivalent to y 2 + y +
= 0.
2004y
2004
The sum of the roots of this equation is the opposite of the y-coefficient,
which is −1.
Difficulty: Hard
NCTM Standard: Algebra Standard for Grades 9–12: Understand the meaning of equivalent forms
of expressions, equations, inequalities, and relations.
Mathworld.com Classification:
Algebra > Algebraic Equations > Quadratic Equation
A semicircle of diameter 1 sits at the top of a semicircle of diameter 2,
as shown. The shaded area inside the smaller semicircle and outside the
larger semicircle is called a lune. Determine the area of this lune.
1
(A) π −
6
√
3
(D)
+
4
√
3
4
1
π
24
√
3
1
(B)
− π
4
12
√
3
1
(E)
+ π
4
12
√
(C)
3
1
− π
4
24
2003 AMC 10 A, Problem #19—
2003 AMC 12 A, Problem #15—
“Area of the lune equals to the area of the region determined by the
triangle topped by the semicircle of diameter 1 minus the area of the
sector of the larger semicircle.”
Solution
Answer (C): First note that the area of the region determined by the
triangle topped by the semicircle of diameter 1 is
√
µ ¶2 √
1
3 1
3 1
1
·
+ π
=
+ π.
2 2
2
2
4
8
The area of the lune results from subtracting from this the area of the
sector of the larger semicircle,
1
1
π(1)2 = π.
6
6
So the area of the lune is
√
√
3 1
1
3
1
+ π− π=
− π.
4
8
6
4
24
Note that the answer does not depend on the position of the lune on the
semicircle.
Difficulty: Hard
NCTM Standard: Geometry Standard for Grades 9–12: Analyze characteristics and properties of
two- and three-dimensional geometric shapes and develop mathematical arguments about geometric
relationships.
Mathworld.com Classification:
Geometry > Plane Geometry > Arcs > Lune
A base-10 three-digit number n is selected at random.
Which of the following is closest to the probability
that the base-9 representation and the base-11
representation of n are both three-digit numerals?
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6
(E) 0.7
2003 AMC 10 A, Problem #20—
“The largest base-9 three-digit number is 93 − 1 = 728
and the smallest base-11 three-digit number is 112 =
121 .”
Solution
Answer (E): The largest base-9 three-digit number is 93 − 1 = 728
and the smallest base-11 three-digit number is 112 = 121. There are
608 integers that satisfy 121 ≤ n ≤ 728, and 900 three-digit numbers
altogether, so the probability is 608/900 ≈ 0.7.
Difficulty: Hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
Number Theory > Arithmetic > Number Bases
Pat is to select six cookies from a tray containing only chocolate chip,
oatmeal, and peanut butter cookies. There are at least six of each of these
three kinds of cookies on the tray. How many different assortments of six
cookies can be selected?
(A) 22
(B) 25
(C) 27
(D) 28
(E) 729
2003 AMC 10 A, Problem #21—
“The numbers of the three types of cookies must have a sum of six.”
Solution
Answer (D): The numbers of the three types of cookies must have a
sum of six. Possible sets of whole numbers whose sum is six are
0, 0, 6; 0, 1, 5; 0, 2, 4; 0, 3, 3; 1, 1, 4; 1, 2, 3; and 2, 2, 2.
Every ordering of each of these sets determines a different assortment of
cookies. There are 3 orders for each of the sets
0, 0, 6; 0, 3, 3; and 1, 1, 4.
There are 6 orders for each of the sets
0, 1, 5; 0, 2, 4; and 1, 2, 3.
There is only one order for 2, 2, 2. Therefore the total number of
assortments of six cookies is 3 · 3 + 3 · 6 + 1 = 28.
OR
Construct eight slots, six to place the cookies in and two to divide the
cookies by type. Let the number of chocolate chip cookies be the number
of slots to the left of the first divider, the number of oatmeal cookies be
the number of slots between the two dividers, and the number of peanut
butter cookies be the number of slots to the right of the second divider.
For example, 111 | 11 | 1 represents three chocolate chip¡ cookies,
two
¢
8
oatmeal cookies, and one peanut butter cookie. There are 2 = 28 ways
to place the two dividers, so there are 28 ways to select the six cookies.
Difficulty: Hard
NCTM Standard: Algebra Standard for Grades 9–12: Understand patterns, relations, and
functions.
Mathworld.com Classification:
Discrete Mathematics > Combinatorics > Binomial Coefficients
In rectangle ABCD, we have AB = 8, BC = 9, H is on BC with
BH = 6, E is on AD with DE = 4, line EC intersects line AH at G,
and F is on line AD with GF ⊥ AF . Find the length GF .
G
H
6
C
B
8
F
A
D
(A) 16
(B) 20
(C) 24
4
E
(D) 28
(E) 30
2003 AMC 10 A, Problem #22—
“Triangles GCH and GEA are similar; Triangles GF E and CDE are
similar.”
Solution
Answer (B): We have EA = 5 and CH = 3. Triangles GCH and GEA are similar, so
3
GC
=
GE
5
and
CE
GE − GC
3
2
=
=1− = .
GE
GE
5
5
Triangles GF E and CDE are similar, so
GF
CE
5
=
=
8
GE
2
and F G = 20.
OR
Place the figure in the coordinate plane with the origin at D, DA on the positive x-axis, and DC on
the positive y-axis. Then H = (3, 8) and A = (9, 0), so line AG has the equation
4
y = − x + 12.
3
Also, C = (0, 8) and E = (4, 0), so line EG has the equation
y = −2x + 8.
The lines intersect at (−6, 20), so F G = 20.
Difficulty: Medium-hard
NCTM Standard: Geometry Standard for Grades 9–12: Explore relationships (including
congruence and similarity) among classes of two- and three-dimensional geometric objects, make
and test conjectures about them, and solve problems involving them.
Mathworld.com Classification:
Geometry > Plane Geometry > Geometric Similarity > Similar
A large equilateral triangle is constructed by using toothpicks to create
rows of small equilateral triangles. For example, in the figure we have 3
rows of small congruent equilateral triangles, with 5 small triangles in the
base row. How many toothpicks would be needed to construct a large
equilateral triangle if the base row of the triangle consists of 2003 small
equilateral triangles?
(A) 1,004,004
(B) 1,005,006
(D) 3,015,018
(E) 6,021,018
(C) 1,507,509
2003 AMC 10 A, Problem #23—
“Each succeeding row will require one less set of 3 toothpicks.”
Solution
Answer (C): The base row of the large equilateral triangle has 1001 triangles pointing downward
and 1002 pointing upward. This base row requires 3(1002) toothpicks since the downward pointing
triangles require no additional toothpicks. Each succeeding row will require one less set of 3 toothpicks,
so the total number of toothpicks required is
3(1002 + 1001 + 1000 + · · · + 2 + 1) = 3 ·
1002 · 1003
= 1, 507, 509.
2
OR
Create a table:
Number of Rows
1
2
3
..
.
n
Number of Triangles
in Base Row
1
3
5
..
.
2n − 1
Number of Toothpicks
in All Rows
3
3+6
3+6+9
..
.
3(1 + 2 + · · · + n)
Thus
2003 = 2n − 1
so
n = 1002.
The number of toothpicks is
3(1 + 2 + · · · + 1002) = 3
(1002)(1003)
= 1, 507, 509.
2
Difficulty: Hard
NCTM Standard: Algebra Standard for Grades 9–12: Understand patterns, relations, and
functions.
Mathworld.com Classification:
Discrete Mathematics > Graph Theory > Simple Graphs > Planar Graphs > Matchstick Graph
Sally has five red cards numbered 1 through 5 and four
blue cards numbered 3 through 6. She stacks the cards
so that the colors alternate and so that the number
on each red card divides evenly into the number on
each neighboring blue card. What is the sum of the
numbers on the middle three cards?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
2003 AMC 10 A, Problem #24—
2003 AMC 12 A, Problem #12—
“Establish the whole stack of cards.”
Solution
Answer (E): Let R1, . . ., R5 and B3, . . ., B6 denote the numbers on the
red and blue cards, respectively. Note that R4 and R5 divide evenly into
only B4 and B5, respectively. Thus the stack must be R4, B4, . . ., B5,
R5, or the reverse. Since R2 divides evenly into only B4 and B6, we must
have R4, B4, R2, B6, . . ., B5, R5, or the reverse. Since R3 divides evenly
into only B3 and B6, the stack must be R4, B4, R2, B6, R3, B3, R1, B5,
R5, or the reverse. In either case, the sum of the middle three cards is 12.
Difficulty: Medium
NCTM Standard: Algebra Standard for Grades 9–12: Use mathematical models to represent and
understand quantitative relationships.
Mathworld.com Classification:
Discrete Mathematics > Combinatorics > Permutations > Arrangement
Let n be a 5-digit number, and let q and r be
the quotient and remainder, respectively, when n is
divided by 100. For how many values of n is q + r
divisible by 11 ?
(A) 8180
(B) 8181
(C) 8182
(D) 9000
(E) 9090
2003 AMC 10 A, Problem #25—
2003 AMC 12 A, Problem #18—
“n = 100q + r = q + r + 99q .”
Solution
Answer (B): Note that n = 100q+r = q+r+99q. Hence q+r is divisible
by 11 if and only if n is divisible by 11. Since 10, 000 ≤ n ≤ 99, 999, there
are
º ¹
º
¹
9999
99999
−
= 9090 − 909 = 8181
11
11
such numbers.
Difficulty: Hard
NCTM Standard: Number and Operations Standard for Grades 9–12: Understand numbers, ways
of representing numbers, relationships among numbers, and number systems.
Mathworld.com Classification:
Number Theory > Arithmetic > Multiplication and Division > Division