2016 AMC 10B Problem 6

Attempt Problem 6 of the 2016 AMC 10B 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 2016 AMC 10B solutions, or check the answer key.

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6.

Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number S.S. What is the smallest possible value for the sum of the digits of S?S?

 1 \ 1

 4 \ 4

 5 \ 5

 15 \ 15

 21 \ 21

Answer: B
Concepts:digitsextremal argument
Difficulty rating: 960
Solution:

To minimize the sum of the addends, use the two smallest nonzero hundreds digits, 11 and 22, and then the four smallest remaining digits, 0,3,4,50,3,4,5. Placing the smaller digits in the tens places shows that every possible sum is at least 104+235=339104+235=339. On the other hand, any three-digit number with digit sum less than 44 is at most 300300. Therefore the digit sum of SS is at least 44.

The example 157+243=400157+243=400 uses six distinct digits and has digit sum 44. Thus the smallest possible digit sum is 44.

Thus, the correct answer is B.

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