Question 574553: Heeding the advice to “Go west, young man,” Harold and Horace greedy set out for Las Vegas to make their fortunes. Luck was with the greedy brothers, for each doubled his money every hour. Harold, who began with $20, gambled for 20 hours and then sent his fortune home. The next day, Horace began with $40 and gambled until his fortune matched his brother’s. For how many hours did Horace Greedy gamble that day.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Harold Greedy started with $20. After $1 hour, he had $40. It took him 19 more hours to go from $40 to the amount/fortune he wanted. I'll call it X, but I do not have to calculate it.
Horace, started with $40, and increased his fortune at the same rate as his brother had done.
So he got from $40 to X in  hours, just like his brother had done before.
My question is, was the official solution to the problem unnecessarily more complicated than that?
JUST FOR FUN (and to apply a little of the stuff you're learning in class)
The brothers' fortunes were increasing in geometric sequence (or geometric progression, depending on book/teacher/country).
After n hours Harold had
$
After 20 hours he had
X=$
I still do not have to calculate it
Horace, starting with $40, would have
$ after t hours
When Horace's fortune matched his brother's X=$, we would have
 -->  -->  -->  -->  -->  --> 
OK. I'm curious. I'll calculate it after all.
X=$ =$20,971,520
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