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Difference between revisions of "1990 AJHSME Problems/Problem 21"

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==Solution==
 
==Solution==
  
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We just use the definition to find the first number is <math>1/4 \rightarrow \boxed{\text{B}}</math>.
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If we do <math>64/16</math> we get <math>4</math>, then we do <math>16/4</math> giving <math>4</math> again, then <math>4/4</math> is <math>1</math> and <math>4/1</math> is <math>4</math> and finally <math>1/4</math> gives <math>1/4</math> as the answer, which is <math>\boxed{\text{B}}</math>
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==Solution 2==
 
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If we do <math>64/16</math> we get <math>4</math>, then we do <math>16/4</math> giving <math>4</math> again, then <math>4/4</math> is <math>1</math> and <math>4/1</math> is <math>4</math> and finally <math>1/4</math> gives <math>1/4</math> as the answer. Which is <math>\boxed{\text{B}}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 12:18, 16 October 2023

Problem

A list of $8$ numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three are shown: \[\text{\underline{\hspace{3 mm}?\hspace{3 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}16\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}64\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{1 mm}1024\hspace{1 mm}}}\] $\text{(A)}\ \frac{1}{64} \qquad \text{(B)}\ \frac{1}{4} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 4$

Solution

If we do $64/16$ we get $4$, then we do $16/4$ giving $4$ again, then $4/4$ is $1$ and $4/1$ is $4$ and finally $1/4$ gives $1/4$ as the answer, which is $\boxed{\text{B}}$

See Also

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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