Difference between revisions of "1990 AJHSME Problems/Problem 1"
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Latest revision as of 17:12, 29 October 2016
Problem
What is the smallest sum of two 
-digit numbers that can be obtained by placing each of the six digits 
in one of the six boxes in this addition problem?
![[asy] unitsize(12); draw((0,0)--(10,0)); draw((-1.5,1.5)--(-1.5,2.5)); draw((-1,2)--(-2,2)); draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle); draw((4,1)--(6,1)--(6,3)--(4,3)--cycle); draw((4,4)--(6,4)--(6,6)--(4,6)--cycle); draw((7,1)--(9,1)--(9,3)--(7,3)--cycle); draw((7,4)--(9,4)--(9,6)--(7,6)--cycle); [/asy]](http://latex.artofproblemsolving.com/9/1/4/914e5e0471d3dc0feff40587010f980154414e03.png)
Solution
Let the two three-digit numbers be 
and 
. Their sum is equal to 
.
To minimize this, we need to minimize the contribution of the 
factor, so we let 
and 
. Similarly, we let 
, 
, and then 
and 
. The sum is ![\[100(9)+10(13)+(17)=1047 \rightarrow \boxed{\text{C}}\]](http://latex.artofproblemsolving.com/e/a/f/eaf7d8be906cc6243dc6997ef53f64f39876d975.png)
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
