Difference between revisions of "1990 AJHSME Problems/Problem 21"
(removed a vague solution) |
m (→Solution) |
||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | 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 | + | 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 
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:
![\[\text{\underline{\hspace{3 mm}?\hspace{3 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm} \underline{\hspace{7 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}16\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{2 mm}64\hspace{2 mm}}\hspace{1 mm},\hspace{1 mm}\underline{\hspace{1 mm}1024\hspace{1 mm}}}\]](http://latex.artofproblemsolving.com/7/6/2/76215b8f7b52ed31b0c274ef270d019ed2fd160e.png)
Solution
If we do 
we get 
, then we do 
giving again, then 
is 
and 
is and finally 
gives as the answer, which is 
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 | ||
