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Difference between revisions of "2016 AMC 10B Problems/Problem 6"
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Adding the numbers <math>257</math> and <math>143</math>, we get <math>400</math> which means our answer is <math>\boxed{\textbf{(B)}\ 4}</math>.
 
Adding the numbers <math>257</math> and <math>143</math>, we get <math>400</math> which means our answer is <math>\boxed{\textbf{(B)}\ 4}</math>.
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/6cwlGbTpp2Y
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~Education, the Study of Everything
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==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
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== Video Solution ==
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https://youtu.be/HISL2-N5NVg?t=1046
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~ pi_is_3.14
  
 
==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}}

Latest revision as of 11:47, 2 July 2023

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 1

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 $\boxed{\textbf{(B)}\ 4}$ is attainable.

Solution 2

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 $\boxed{\textbf{(B)}\ 4}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/6cwlGbTpp2Y

~Education, the Study of Everything


Video Solution

https://youtu.be/6sr62OMX8p4

~savannahsolver

Video Solution

https://youtu.be/HISL2-N5NVg?t=1046

~ pi_is_3.14

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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