2^29, expressed in base 10, is a ninedigit number. All nine of its digits are different. Find the digit that is missing without explicitly calculating 2^29.
This question was posed to me by Dinesh Dharme (CSE, IITB). Simple but interesting puzzle!!
Update (11/12/09): Solution in comments!!
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Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this art...

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tos...
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. 45k = 5 mod 9, i.e. k =4 :)