The number 12345678901234567890123…890 is four thousand digits long. First, you remove all of the digits in odd numbered places starting at the leftmost place. Next, you remove the digits in the odd numbered places in the remaining 2000 digits. You perform the same operation until no digits remain. What digit is the last to be removed?
Answers
Answer:
Hey buddy here's ur answer !
Step-by-step explanation:
A simpler problem, 100-digit number
Stage
Digits
Number Pattern
0
100
12345678901234567890…
1
50
2468024680… pattern 1
2
25
4826048260… pattern 2
3
12
864208642086 pattern 3
4
6
628406 pattern 4
5
3
246 cycle back to pattern 1
6
1
4 cycle back to pattern 2
In stage 0, called the initial stage, the number is 100 digits long.
In stage 1, the number is 50 digits long and consists of the repeating pattern 24680…
In stage 2, the number is 25 digits long and consists of the repeating pattern 48260…
In stage 3, the number is 12 digits long and consists of the repeating pattern 86420…
In stage 4, the number is 6 digits long and consists of the repeating pattern 62840…
In stage 5, the number is 3 digits long and consists of 3 digits of the repeating pattern 24680 of stage 1.
In stage 6, called the final stage, the number is 1 digit long and consists of 1 digit of the repeating pattern 48260 of stage 2.
Therefore, 4 will be the last digit to be removed.
Answer: After the first removal, the 2nd, 4th, 6th, 8th, 10th, 12th, 14th, 16th,… will be left. After the second
removal, 4th, 8th, 12th, 16th, 20th, 24th, digits will be left. After the third removal, 8th, 16th, 24th, 32nd, digits
will be left… In short, after the nth removal, the first digit would be (2n
)th digit in the original number.
To reduce 4000 digits to a single digit, we need to perform the halving operation 11 times. Therefore, the
digit left would be (211)th digit or 2048th digit which is 8