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>. | ||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=B|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 11:29, 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 
. What is the smallest possible value for the sum of the digits of ?
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 
is the next smallest integer, we can try to make an integer that starts with and ends with 
We can find that 
fits the parameters(along with many others), in which all the digits are distinct, and indeed sum to 
, summing the digits we reach the correct answer choice; 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
