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Difference between revisions of "2016 AMC 10B Problems/Problem 6"
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==Solution==
 
==Solution==
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Let the two three-digit numbers she added be <math>a</math> and <math>b</math> with <math>a+b=S</math> and <math>a<b</math>. The hundreds digits of these numbers must be at least <math>1</math> and <math>2</math>, so <math>a\ge 102</math> and <math>b\ge 203</math>, which means <math>S\ge 305</math>, so the digits of <math>S</math> must sum to at least <math>4</math>, in which case <math>S</math> would have to be either <math>310</math> or <math>400</math>. But <math>b</math> is too big for <math>310</math>, so we consider the possibility <math>S=400</math>.
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Let the two three-digit numbers she added be <math>a</math> and <math>b</math> with <math>a+b=S</math> and <math>a<b</math>. The hundreds digits of these numbers must be at least <math>1</math> and <math>2</math>, so <math>a\ge 100</math> and <math>b\ge 200</math>.
  
 
Say <math>a=100+p</math> and <math>b=200+q</math>; then we just need <math>p+q=100</math> with <math>p</math> and <math>q</math> having different digits which aren't <math>1</math> or <math>2</math>.There are many solutions, but <math>p=3</math> and <math>q=97</math> give <math>103+297=400</math> which proves that <math>\textbf{(B)}\ 4</math> is attainable.
 
Say <math>a=100+p</math> and <math>b=200+q</math>; then we just need <math>p+q=100</math> with <math>p</math> and <math>q</math> having different digits which aren't <math>1</math> or <math>2</math>.There are many solutions, but <math>p=3</math> and <math>q=97</math> give <math>103+297=400</math> which proves that <math>\textbf{(B)}\ 4</math> is attainable.

Revision as of 13:42, 20 November 2016

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 $a\ge 100$ and $b\ge 200$.

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.

Solution 2

Ok, so for this problem, to find the $3$-digit integer with the smallest sum of digits, one should make the units and tens digit add to $0$. To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $7$ works best for the top number which makes the bottom digit $3$. The tens digits need to add to $9$ because of the $1$ that needs to be carried from the addition of the units digits. We see that $5$ and $4$ work the best as we can't use $6$ and $3$. Finally, we use $2$ and $1$ for our hundreds place digits.

Adding the numbers $257$ and $143$ ,

We get $400$ which means our answer is $\textbf{(B)}\ 4$.

See Also

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