1998 AMC 12 Problem 2

Attempt Problem 2 of the 1998 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 1998 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

Letters A,A, B,B, C,C, and DD represent four different digits selected from 0,0, 1,1, 2,2, …,\ldots, 9.9. If A+BC+D\frac{A+B}{C+D} is an integer that is as large as possible, what is the value of A+B?A+B?
1313
1414
1515
1616
1717
Answer: E
Concepts:digitsoptimizationratios
Small hint:
Make the denominator as small as possible using distinct digits
Big hint:
The two smallest digits and the two largest digits are disjoint pairs
Solution:
The least positive denominator is 0+1=1,0+1=1, and the greatest numerator using two other digits is 8+9=17.8+9=17. Their ratio is the integer 17,17, which is plainly maximal. Thus A+B=17,A+B=17, and E is correct.
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