There are 4 entrances to a casino.......as you enter each entrance it costs you $1.00 ........and when you exit the entrance you entered it costs you another $1.00....you enter 4 times and exit 4 times.......if each time you enter the Casino you lose half of what you have with you.....and upon entering and exiting ......you have exactly $1.00 left to exit with ......how much did you start out with ???
Your riddle doesn't give correct information. The road to the answer seems to me broken. You better check it again. I'm sure, In Sha Allah, I will guess the answer, then.
the answer is $45, we could use the formulas:
1. ((a-1)/2)-1=b ..... at the first exit ....("a" means the money we bring at the first time before entering the casino)
("a-1" means the money minus $1 after entering the casino)
(" (a-1)/2 " means after entering the casino we lose half of the money)
(" ( (a-1)/2 ) -1 " means after losing half of the money we exit the casino and cost $1) ("b" means the amount of money left after entering, loosing,and exiting the first time)

2. ((b-1)/2)-1=c ..... at the second exit ... (almost the same explanation as the first one)
3. ((d-1)/2)-1=e ..... at the third exit
4. ((e-1)/2)-1=0 ..... at the forth exit...(after exiting for the forth time we have no money left)

so after doing the math we will get a=$45 is the amount of money before entering the casino for the first time

siaufa
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the answer

$256 but in terrible at maths so it's probabaly wromg
the answer is
$61
$2.00, if upon entering the first time it cost you $1.00 and you lost 1/2 of what you started with. You must have won money each time you were in the casino,
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$5.00. You enter 4 times but by only 1, 2, or 3 of the entrances, and exit by the one or ones you haven't entered by, thereby not paying anything to exit except the very last time you exit. This costs you $4.00 to enter and nothing to exit, leaving $1.00 in your pocket.

The 1st digit is 1/2 of the 4th

The 1st 2 digits are 1/2 of the last 2

The 2nd and 3rd digits are the same number