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2001 AMC 10 Problem 13

2001 AMC 10 Problem 13AMC 10 Step by Step

Name: Date: September 13, 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

All ten digits are used, so ABC is whatever the odd run GHIJ and even run DEF leave behind; only 9753 and 642 leave digits summing to 9.

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 $\boxed{\textbf{(E)}\ 8}$ .

Why this works

The hidden constraint "ten distinct letters" turns a search into a complement: fix the rigid parts (runs of consecutive odd or even digits, which have very few options) and the flexible part is forced. Six cases, each a three-digit sum, is a small and safe enumeration.

The trap

Forgetting that 0 is a digit (so 420 is a valid DEF and 810 a valid ABC), or not using the fact that all ten digits must appear.

Common mistakes

Forgetting that 0 is a digit (so 420 is a valid DEF and 810 a valid ABC), or not using the fact that all ten digits must appear.
Trying to build $ABC$ directly from digits summing to $9$ and checking compatibility afterwards, which invites overlooked overlaps.

Techniques

Split into exhaustive cases and handle each