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 = 54</math>
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<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>
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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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 $\{3,5,7,11,13,17\}$.
  • Add two of these numbers.
  • Multiply their sum by the third number.

$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 50 \qquad \text{(E)}\ 56$

Solution

Since we want the smallest possible result, and we are only adding and multiplying positive numbers over $1$, we can "prune" the set to the three smallest numbers $\{3,5,7\}$. Using bigger numbers will create bigger sums and bigger products.

From there, compute the $3$ ways you can do the two operations:

$(3+5)7 = 8\cdot 7 = 56$

$(3 + 7)5 = 10\cdot 5 = 50$

$(7+5)3 = 12\cdot 3 = 36$

The smallest number is 36, giving an answer of $\boxed{C}$

See also

1996 AJHSME (ProblemsAnswer KeyResources)
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

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