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

(Created page with '==Problem== A list of <math>8</math> 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 fi…')
 
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We just use the definition to find the first number is <math>1/4 \rightarrow \boxed{\text{B}}</math>.
 
We just use the definition to find the first number is <math>1/4 \rightarrow \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==

Revision as of 14:47, 22 October 2018

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

We just use the definition to find the first number is $1/4 \rightarrow \boxed{\text{B}}$.



Solution 2

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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