Difference between revisions of "2016 AMC 10B Problems/Problem 6"
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | 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 | + | 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>\boxed{\textbf{(B)}\ 4}</math> is attainable. |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/6sr62OMX8p4 | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/HISL2-N5NVg?t=1046 | ||
+ | |||
+ | ~ 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 23:18, 27 January 2021
Problem
Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number . What is the smallest possible value for the sum of the digits of ?
Solution 1
Let the two three-digit numbers she added be and with and . The hundreds digits of these numbers must be at least and , so and .
Say and ; then we just need with and having different digits which aren't or .There are many solutions, but and give which proves that is attainable.
Solution 2
For this problem, to find the -digit integer with the smallest sum of digits, one should make the units and tens digit add to . 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. works best for the top number which makes the bottom digit . The tens digits need to add to because of the that needs to be carried from the addition of the units digits. We see that and work the best as we can't use and . Finally, we use and for our hundreds place digits.
Adding the numbers and , we get which means our answer is .
Video Solution
~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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