Can u solve this?!
Question: a jewel theif takes a bunch of diamonds,
runs into 1st security guard and gives him half plus one,
then another guard gives him half of what he had left plus one,
then runs into a 3rd guard and gives him half plus one again.
the theif is left with one diamond. How many did he start with?
please work this out. Thanks
Answers: a jewel theif takes a bunch of diamonds,
runs into 1st security guard and gives him half plus one,
then another guard gives him half of what he had left plus one,
then runs into a 3rd guard and gives him half plus one again.
the theif is left with one diamond. How many did he start with?
please work this out. Thanks
Start at the end and work forward....
He has one, so he must have had 4 when he ran into the 3rd guard, giving him half (2) plus 1. Leaving him with 1.
So having 4, he must have had 10 when he ran into the 2nd guard, giving him half (5) plus 1 (6) leaving him with 4.
So having 10, he must have had 22 when he ran into the 1st guard, giving him half (11) plus 1 (12) leaving him with 10.
Now do it forwards.
1st guard
22 / 2 = 11 + 1 = 12
22 - 12 = 10
2nd guard
10 / 2 = 5 + 1 = 6
10 - 6 = 4
3rd guard
4 / 2 = 2 + 1 = 3
4 - 3 = 1
1 diamond left.
19
-10.5
9.5
9.5
-5.75
3.85
3.85
-2.425
1.425
no
Start with 22
11+1 10
5+1 4
2+1 1
Since no diamond is cut in 2 Integers must be used. Half has to be the remainder plus one. Going back thru the guards.