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Difference between revisions of "2016 AMC 10B Problems/Problem 6"

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==Solution==
 
==Solution==
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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 <math>4</math> is the next smallest integer, we can try to make an integer that starts with <math>4</math> and ends with <math>00</math> We can find that <math>143+257</math> fits the parameters(along with many others), in which all the digits are distinct, and indeed sum to <math>400</math>, summing the digits we reach the correct answer choice; <math>\textbf{(B)}\ 4</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 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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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.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:30, 21 February 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 $a\ge 102$ 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.

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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