Difference between revisions of "1990 AJHSME Problems/Problem 21"
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==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 13:18, 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:
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 |