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

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==Solution==
 
==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 <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>.
 
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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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 12:29, 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

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 (ProblemsAnswer KeyResources)
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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