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 100</math> and <math>b\ge 200</math>. | 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== | ==Solution 2== | ||
Revision as of 21:26, 4 February 2017
Contents
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
Ok, so 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 
.
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. 
