2001 AMC 12 Problem 6

Below is the professionally curated solution for Problem 6 of the 2001 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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Concepts:digitslogical deductioncasework

Difficulty rating: 1420

6.

A telephone number has the form ABCDEFGHIJ,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,A \gt B \gt C, D>E>F,D \gt E \gt F, and G>H>I>J.G \gt H \gt I \gt J. Furthermore, D,D, E,E, and FF are consecutive even digits; G,G, H,H, I,I, and JJ are consecutive odd digits; and A+B+C=9.A + B + C = 9. Find A.A.

44

55

66

77

88

Solution:

The four consecutive decreasing odd digits GHIJGHIJ are either 97539753 or 7531,7531, leaving one odd digit (11 or 99) for ABC.ABC.

Since A+B+C=9A + B + C = 9 and the other two digits of ABCABC are even, the odd digit must be 11 (a 99 would force the two even digits to sum to 00). So the two even digits sum to 8.8.

The three consecutive decreasing even digits DEFDEF are 864,864, 642,642, or 420,420, leaving the even pairs {2,0},\{2, 0\}, {8,0},\{8, 0\}, or {8,6}\{8, 6\} for ABC.ABC. Only {8,0}\{8, 0\} sums to 8,8, so ABC=810ABC = 810 and A=8.A = 8.

Thus, the correct answer is E.

Problem 6 in Other Years