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

Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$ . The hundreds digits of these numbers must be at least $1$ and $2$ , so $a\ge 102$ and $b\ge 203$ , which means $S\ge 305$ , so the digits of $S$ must sum to at least $4$ , in which case $S$ would have to be either $310$ or $400$ . But $b$ is too big for $310$ , so we consider the possibility $S=400$ .

Say $a=100+p$ and $b=200+q$ ; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$ .There are many solutions, but $p=3$ and $q=97$ give $103+297=400$ which proves that $\textbf{(B)}\ 4$ is attainable.

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2016 AMC 10B Problems/Problem 6

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