1972 AMC 12 Problem 33

Attempt Problem 33 of the 1972 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 1972 AMC 12 solutions, or check the answer key.

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33.

The minimum value of the quotient of a (base ten) number of three different nonzero digits divided by the sum of its digits is:

9.79.7

10.110.1

10.510.5

10.910.9

20.520.5

Answer: C
Concepts:digitsoptimizationinequality
Difficulty rating: 2140
Small Hint:

Let the hundreds, tens, and units digits be H,H, T,T, and U,U, and consider which position should contain the largest digit

Big Hint:

At a minimum U=9U=9; then maximize TT and minimize HH subject to distinct nonzero digits

Solution:

Let QQ denote the quotient. Then Q=1+99H+9TH+T+U. Q=1+\frac{99H+9T}{H+T+U}. Interchanging UU with a larger digit in either earlier position decreases the quotient, so the units digit must be the largest. Increasing that units digit lowers a quotient greater than 1,1, so U=9.U=9. Then T+11HT+H+9=1+10H9T+H+9. \frac{T+11H}{T+H+9} =1+\frac{10H-9}{T+H+9}. This is minimized by taking the largest available T=8T=8 and then the smallest H=1.H=1. The number is 189,189, and 1891+8+9=18918=10.5. \frac{189}{1+8+9}=\frac{189}{18}=10.5.

Therefore, the correct answer is C.

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