Puzzle: There are two urns with one ball each. Each of subsequent n-2 balls is placed into one of these urns, with probability proportional to the number of balls already in that urn. What is the expected number of balls in the smaller sized urn? Source: P. Winkler's Puzzles book. (Chapter: Probability). Solution: Highlight the part between the * symbols for the answer. * This problem can be reformulated as the following problem. Suppose I have a stack of black cards and one red card. Initially I take red card in my hand. Now I add black cards randomly between any two cards (so, initially its either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent. No...
answer: 4
ReplyDeletesum all the digits of every exponent of 2 till you get a single digit. Observe that the pattern is 1,2,4,8,7,5,1,2,4,8,7,5(repeats in cycles of 6).
Thus the single digit for 2^29 will be 5. Now since we know that it consists of 9 distinct digits, try to find the digit which when removed from the number 0123456789 gives the single digit sum as 5.
:) yo asad!! fundoo!!
ReplyDeletethanx..
Correct Answer!
thanx.
ReplyDeleteDude u r doing a fundoo job of maintaining ur blog and updating the puzzles regularly.
Anyways try to post a proof for the solution I have given, like why is the cycle length 6.
thanx Asad!
ReplyDeleteActually there exists a simpler solution:
2^29 mod 9 = (8^9)*4 = -4 = 5
So, (sum of the digits of 2^29) mod 9 = 5 {This is true since 10 mod 9 = 100 mod 9 = 1}
Let the missing number be k
(Sum from 0 to 9) - k mod 9 = 5
k mod 9 = 40
k = 4
:)
I am happy! :D
I dont know abt the circle fundaa
ReplyDeletebut it is done by the mod method
it is simpler.
nice question, blows u at first
Proof of Asad's solution:
ReplyDelete2^6 mod 9 = 64 mod 9 = 1
So, the cycle repeats after every 6 steps
Effectively, Asad is also doing exactly the same thing!!
"sum all the digits of every exponent of 2 till you get a single digit. Observe that the pattern is 1,2,4,8,7,5,1,2,4,8,7,5(repeats in cycles of 6)."
i.e. find 2^i mod 9 for all i and see that since 2^6 mod 9 =1, the sequence repeats on length 6.
"Thus the single digit for 2^29 will be 5. Now since we know that it consists of 9 distinct digits, try to find the digit which when removed from the number 0123456789 gives the single digit sum as 5."
i.e. since 2^29 mod 9 = 5, find the sum of digits such that sum of digits mod 9 = 5, i.e. 45-k = 5 mod 9, i.e. k =4 :)