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2001 AMC 10 Problem 13

2001 AMC 10 Problem 13AMC 10 Step by Step

Name: Date: October 28, 2026

Logic Puzzles · difficulty 3 of 5 · about 4 min · position P11-15

A telephone number has the form $\text{ABC-DEF-GHIJ}$ , where each letter represents a different digit. The digits in each part of the number are in decreasing order; that is, $A > B > C$ , $D > E > F$ , and $G > H > I > J$ . Furthermore, $D$ , $E$ , and $F$ are consecutive even digits; $G$ , $H$ , $I$ , and $J$ are consecutive odd digits; and $A + B + C = 9$ . Find $A$ .

A) $4$
B) $5$
C) $6$
D) $7$
E) $8$

Answer

Key insight

Ten letters and ten digits means every digit 0 to 9 is used once. Pick the odd run and the even run; A, B, C are the leftovers.

Solution

Count the letters: $A$ through $J$ is ten letters, and they are all different digits. There are only ten digits, $0$ to $9$ , so every digit is used exactly once. So $A$ , $B$ , $C$ are whatever three digits are left after choosing $DEF$ and $GHIJ$ .
Start with the part that has the fewest choices. $GHIJ$ is four odd digits in a row, going down. The odd digits are $9, 7, 5, 3, 1$ , so $GHIJ$ is $9753$ or $7531$ .
$DEF$ is three even digits in a row, going down. The even digits are $8, 6, 4, 2, 0$ , so $DEF$ is $864$ , $642$ or $420$ .
That is only $2 \times 3 = 6$ cases. For each, list the three digits left over and add them. We need a sum of $9$ .

$GHIJ$	$DEF$	Left over	Sum
$7531$	$864$	$0, 2, 9$	$11$
$7531$	$642$	$0, 8, 9$	$17$
$7531$	$420$	$6, 8, 9$	$23$
$9753$	$864$	$0, 1, 2$	$3$
$9753$	$642$	$0, 1, 8$	$9$
$9753$	$420$	$1, 6, 8$	$15$

Only one row works: the left-over digits are $0, 1, 8$ . In decreasing order, $ABC = 810$ .
The full number is $810\text{-}642\text{-}9753$ , and $A = 8$ .

The answer is $\boxed{\textbf{(E)}\ 8}$ .

Why this works

The key fact is hidden in "each letter represents a different digit": with ten letters, all ten digits appear. That turns the problem around. Instead of searching for $A$ , $B$ , $C$ directly, you pick the parts with very few options (the runs of odd and even digits), and $A$ , $B$ , $C$ are forced. When a problem uses every item exactly once, think about what is left over.

The trap

Forgetting that 0 is a digit, so 420 is a possible DEF and 810 a possible ABC, or not noticing that all ten digits must be used.

Concise solution

There are ten letters and ten digits, all different, so every digit $0$ through $9$ is used exactly once. That means $A, B, C$ are simply the three digits not used by $DEF$ and $GHIJ$ .

Four consecutive odd digits in decreasing order: $GHIJ = 9753$ or $7531$ . Three consecutive even digits: $DEF = 864$ , $642$ or $420$ .

Case $GHIJ = 7531$ : the leftover pool is $\{0, 2, 4, 6, 8, 9\}$ . Removing $864$ leaves $\{0, 2, 9\}$ (sum $11$ ); removing $642$ leaves $\{0, 8, 9\}$ (sum $17$ ); removing $420$ leaves $\{6, 8, 9\}$ (sum $23$ ). No sum is $9$ .

Case $GHIJ = 9753$ : the pool is $\{0, 1, 2, 4, 6, 8\}$ . Removing $864$ leaves $\{0, 1, 2\}$ (sum $3$ ); removing $642$ leaves $\{0, 1, 8\}$ (sum $9$ ); removing $420$ leaves $\{1, 6, 8\}$ (sum $15$ ).

The only match is $ABC = 810$ , $DEF = 642$ , $GHIJ = 9753$ , so $A = 8$ .

The answer is $\boxed{\textbf{(E)}\ 8}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic

Techniques

Split into exhaustive cases and handle each