Let a, b, c and d be integers with a<2b, b<3c, and c<4d. If d<100, the largest possible value for a is

1989 AMC 12 Problem 11

Attempt Problem 11 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

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11.

Let

a,

b,

c

and

d

be integers with

a\lt2b,

b\lt3c,

and

c\lt4d.

d\lt100,

the largest possible value for

a

2367

2375

2391

2399

2400

Maximize the integers from

d

backward to

a

Each strict inequality lowers the greatest allowable integer by

1

The largest possible values are

d=99,\ c=395,\ b=1184,

and finally

a=2367.

Each value satisfies its strict inequality, so this maximum is attainable.

Thus the correct answer is A.