### Missing Digit

2^29, expressed in base 10, is a nine-digit 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!!

This question was posed to me by Dinesh Dharme (CSE, IITB). Simple but interesting puzzle!!

Update (11/12/09): Solution in comments!!

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 :)