Difference between revisions of "2016 AMC 10B Problems/Problem 6"
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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 . What is the smallest possible value for the sum of the digits of ?
Solution
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 , which means , so the digits of must sum to at least , in which case would have to be either or . But is too big for , so we consider the possibility .
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.
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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