brain teasers (more)?

Question

1.The famous playboy Hystrix Tardigradus explained to a pretty lady his system for playing roulette: in each round, i always bet 1/2 of the money i have at the time on red. Yestrday i counted and i had won as many rounds as i lost.
that night, did he win lose or break even?
2. Professor Zizoloziz always adds the 5 digits on a bus transfer.Yesterday, he rose rout 62 w/ friend. As soon as he got the tickets, consecutivly #ed, he added the sum of all 10 digits was exactly 62. His logical friend asked him if the sum of the #s on either of the tickets was by any chance 35. Professor Z answered and his friend then knew the #s on the tickets.
What were the #s on the 2 tickets?

PLEASE SHOW UR WORK

thanxx.

Answer 1

1) If you have x, and win, you end up with 1.5x. If you lose, you get 0.5x. So every win/loss combines to give 1.5*0.5x, or 0.75x. So every win/loss pair he loses 25%; overall he loses.

2) Lets say the first ticket had digits abcde.
If e isn't 9, the second is abcd(e+1), so the total sum is 2(a+b+c+d+e)+1 which is odd - impossible.
So e = 9.

Now, if d isn't 9, then we have:
abcd9 and abc(d+1)0.
The total sum is 2(a+b+c+d) + 10 = 62, so a+b+c+d = 26, and any such a,b,c,d work.
But that makes the total sum of the first ticket 35. So, if Professor Z answered 'yes', there would be multiple solutions (eg 99179 and 99180, 66689 and 66690, etc).

Since his friend could work out the numbers, Professor Z must have answered 'No', and therefore d = 9.

Now, if c is not 9, we get abc99 and ab(c+1)00, which again gives a contradiction of an odd sum being 62. So c = 9 as well.

If b is not 9, we get ab999 and a(b+1)000. Then 2(a+b) + 28 = 62, so a+b = 17.
The only possibility (with b not 9) is a=9 and b=8. So one solution is 98999 and 99000.

Is it possible for b to be 9? No, because then the numbers are a9999 and (a+1)0000, which gives the same old even/odd contradiction.

So the numbers are 98999 and 99000.

Answer 2

1. Every time he wins, he multiplies his bank by 1.5. Every time he loses, he multiplies his bank by 0.5. Since 1.5 times 0.5 = 0.75, every win/loss reduces his bank by 25 percent.

So he lpses. More specifically, if he wins n times and loses n times, then he has 100*(0.75)^n percent of his original bank left.

Concrete example: Suppose he has $64. He wins (after betting $32), and his bank is now $32 (what he didn't bet) + $32 (what he bet) + $32 (what he won on the bet) = $96. If he bets half of that ($48) and loses, his bank is now $48 ($96 minus the $48 he lost on the bet).

Th enet result of one win and one loss is $48, which is 25% less than the $64 he started with.

Also, the order is irrelevant (try dong it where he loses then wins: he still winds up with $48).


2. There is insufficient information. All ou can conclude is that the first ticket's # ended in 9, and the next ticket's # ended in 0. You know this as follows.

Suppose the first ticket did NOT end in 9. Then adding 1 to the ticket number would increase the digital sum by only 1 (sicne there is no carrying involvedin the addition). So, if the first ticket's digital sum is n, then the next sum is n + 1. But the sum of n and n + 1 is 2n + 1, an odd number. Since 62 is even, the last digit of the first ticket's number must be 9.

Moreover, one of the ticket's digital sum was 35, it must be the first. Here's why. Supose the digital sum of the first ticket is n, AND, the last digit is 9. Then adding 1 to the number will lower the one's place by 9 (it goes from 9 to 0), AND raise the ten's palce by 1 (from carrying). So the net change in the digital sum is -9+1 = -8. Since the digital sum decrease by 8, the first sum was 35, and the second is 35 - 8 = 27. the sum of 35 and 27 is 62, as stated. Note that when 1 was added, the 10's digit could NOT have also been 9 to start with, because this would have resulted in - in the one's place, - 9 in the tens place, and +1 in the undreds place, for a next change of -9-9+1=-17. Since 35 - 17 = 18, and since 35 + 18 isnt 62, there was no carry in the tens place as well, i.e. the original tens place in the first ticket number was not 9.

This only narrows down the ones digits.

Ticket #1 = ABCD9
Ticket #2 = ABCE0, where E = D + 1.

All I know is (A + B + C + D + 9) + (A + B + C + E) = 62, or
(A + B + C + D + 9) + (A + B + C + D + 1) = 62, or
2A + 2B + 2C + 2D + 10 = 62, or
2A + 2B + 2C + 2D = 52, or
A + B + C + D = 26

Hmmm, how can you add four different numbers and get 26?
9 + 8 + 7 + 2
9 + 8 + 6 + 3
9 + 8 + 5 + 4
9 + 7 + 6 + 4
8 + 7 + 6 + 5

So A, B, C, D can be any permutation of any of the five rows listed, with the condition that D isnt 9.

So the tickets could have been 98729 and 98730, or 97829 and 97830, or... etc.


And this doesnt even take into consideration that one of the first four digitst could be repeated, like 97559.


Sorry, #2 does not have a unique solution.

Answer 3

1. Lets say he had 200 bux, 100 of that stays on the side. If he plays 10 times he wins 200 per time and loses. Thus if he wins 5 times and loses 5 times he is up 1000 and down 500 giving him a profit of 500 which is GREATER than he started out with, discarding the 100 he set aside.
2. I dont understand the question, please clarify

Answer 4

1: Every time he lost, he multipled his money by 0.5, because he lost half of it. Every time he won, he multiplied his money by 1.5, because he gained half of it. Since he won as often as he lost, each win can be said to have been offset by a loss. That would have the overall effect of multiplying his money by 0.75. Every time he won a round, he lost another and multiplied his money by another 0.75. Therefore he lost money.

Answer 5

tricky thing. try searching using google or bing. this may help!