Source: Sent to me by Prateek Chandra Jha (IIT Bombay) Problem: This problem is inspired by the Cheryl's Birthday Puzzle ( FB Post , Guardian Link ). Paul, Sam and Dean are assigned the task of figuring out two numbers. They get the following information: Both numbers are integers between (including) 1 and 1000 Both numbers may also be identical. Paul is told the product of the two numbers, Sam the sum and Dean the difference. After receiving their number, the following conversation takes place: Paul: I do not know the two numbers. Sam: You did not have to tell me that, I already knew that. Paul: Then I now know the two numbers. Sam: I also know them. Dean: I do not know the two numbers. I can only guess one which may probably be correct but I am not sure. Paul: I know which one you are assuming but it is incorrect. Dean: Ok, I also know the two numbers. What are the two numbers? Disclaimer: Its not a puzzle for 14-15 year olds like Cheryl's
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 :)