Difference between revisions of "1996 AJHSME Problems/Problem 6"
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From there, compute the <math>3</math> ways you can do the two operations: | From there, compute the <math>3</math> ways you can do the two operations: | ||
| − | <math>(3+5)7 = 8\cdot 7 = | + | <math>(3+5)7 = 8\cdot 7 = 56</math> |
<math>(3 + 7)5 = 10\cdot 5 = 50</math> | <math>(3 + 7)5 = 10\cdot 5 = 50</math> | ||
Latest revision as of 12:55, 23 October 2016
Problem
What is the smallest result that can be obtained from the following process?
- Choose three different numbers from the set
. - Add two of these numbers.
- Multiply their sum by the third number.
.
Solution
Since we want the smallest possible result, and we are only adding and multiplying positive numbers over 
, we can "prune" the set to the three smallest numbers 
. Using bigger numbers will create bigger sums and bigger products.
From there, compute the 
ways you can do the two operations:
The smallest number is 36, giving an answer of 
See also
| 1996 AJHSME (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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
