Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has $36 when they begin and $36 when they end, what is the total amount that all three friends have?

$252

Step-by-step explanation:

This is quite a neat question, with no fixed equation. Given that each person gives the other two enough money to double their cash, if Toy had 36 dollars beginning, and 36 at the end - presumably the cash of each person, ( their starting and original ) should be the same as well. Respectively each should be a multiple of 36 dollars.

Jan's " give away " = Ami + 36, Jan - 108, Toy + 72

Toy's " give away " = Ami + 72, Jan + 36, Toy - 108

Therefore, we can conclude that Ami = 144 at the start, presuming he gave away 108 dollars, with a remaining 36. Jan, having 144 dollars ( after having his 72 dollars doubled by Ami ) gives 36 to Ami to double his amount, and 72 to double Toy's doubled amount, remaining with 36 dollars. Now Ami has 72 dollars, Jan has 36, and Toy has 144. Then, Toy double's Ami and Jan's amount, giving away 72 and 36 dollars, remaining with 36 dollars himself. Therefore, Ami has 144 dollars, Jan has 72 dollars, and Toy has 36 dollars both at the beginning and end.

144 + 72 + 36 = 252 dollars ( in total )

answer is 252

Step-by-step explanation:

Good luck :)

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58