Sequence!

Logic Level 3

$1 + 11 + 101 + 1001 + 10001 + \cdots+ 1\underbrace{000\ldots00}_{\text{fifty 0's}}1$

Find the sum of the digits when the number above is written as a single integer.

The answer is 58.

3 solutions

Zeeshan Ali
Feb 5, 2016

$1 + 11 + 101 + 1001 + 10001 + \cdots+ 1\underbrace{000\ldots00}_{\text{fifty 0's}}1$

                        5
                          1
    +                   1 1
    +                 1 0 1
    +               1 0 0 1
    +             1 0 0 0 1
    +           1 0 0 0 0 1
                          .
                          .
                          .
    +   1 0 0 0 . . 0 0 0 1 (50 Zeros)
    _______________________

    =   1 1 1 1 . . 1 1 6 2 (50 Ones)
    _______________________

Therefore Sum of digits = 50(1)+6+2 = $\boxed{58}$

Eli Ross Staff
Feb 4, 2016

The number will be $\underbrace{111\ldots1}_{49\text{ ones}}63,$ giving a digit sum of $49 + 6 + 3 = 58.$

Can someone explain why?

We see that the first zero in the sequence is in the third number (101). Thus the last number provided with the 50 zeros will be the 53rd number in the sequence. We also see that only the least significant digit increases with successive sums while the others stay 1. We then sum the first three numbers (1+11+101) and add 50. The last two digits are then 63. If you're adding 50 '10...01' to 163, it means you have 48 more '1' (50-3+1). The last number in the sum will have 49 ones + '63'. Then 1*49+6+3 = 58.

Giovanni GuessWhat - 5 years, 4 months ago

NO, in will be 50 ones 62

Hasmik Garyaka - 2 years, 6 months ago

Dong kwan Yoo
Nov 6, 2019

$1+11+101+1001+10001+ \dots + 100 \dots 001$

$= 1+ 10+ 100 + 1000+ \dots + 100 \dots 00 + 1+1+1+ \dots + 1+1$

$= 111 \dots 111 + 51$

So sum of digits is $52 + 6 = 58$

Sequence!

The answer is 58.

3 solutions

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