# 2016 AMC 10B Problems/Problem 6

## Problem

Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$. What is the smallest possible value for the sum of the digits of $S$? $\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

## Solution

Since we know that the sum of two three digit positive integers cannot start with a one, we can get rid of option A. Since $4$ is the next smallest integer, we can try to make an integer that starts with $4$ and ends with $00$ We can find that $143+257$ fits the parameters(along with many others), in which all the digits are distinct, and indeed sum to $400$, summing the digits we reach the correct answer choice; $\textbf{(B)}\ 4$.

## See Also

 2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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