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

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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.
 
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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==Solution 2==
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Ok, so for this problem, to find the <math>3</math>-digit integer with the smallest sum of digits, one should make the units and tens digit add to <math>0</math>. 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. <math>7</math> works best for the top number which makes the bottom digit <math>3</math>. The tens digits need to add to <math>9</math> because of the <math>1</math> that needs to be carried from the addition of the units digits. We see that <math>5</math> and <math>4</math> work the best as we can't use <math>6</math> and <math>3</math>. Finally, we use <math>2</math> and <math>1</math> for our hundreds place digits.
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Adding the numbers <math>257</math> and <math>143</math> ,
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We get <math>400</math> which means our answer is  <math>\textbf{(B)}\ 4</math>.
  
 
==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 14:31, 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.

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 (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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