Big number

Number of 4 digit components (years): 2017 - 999 = 1018

Number of 3 digit components (years): 999 - 99 = 900

Number of 2 digit components (years): 99 - 9 = 90

Number of 1 digit components (years): 9 - 0 = 9

Hence, the total number of digits:

$4×1018+3×900+2×90+1×9 = \boxed {6961}$

Let's start out generous and assign 4 digits for each number from 0 through 2017. (Why do I include zero? It makes for easier math!)

Now the first 1000 numbers miss their thousands digits, so subtract those.

And the first 100 numbers also miss their hundreds digit, so subtract those.

The first 10 numbers miss their tens digit; again, subtract.

And the number zero isn't even there, so subtract 1.

$4\times 2018 - 1000 - 100 - 10 - 1 = 8072 - 1111 = \boxed{6961}.$

The given number list is 1, 2, 3, 4, . . . , 2017. It can be classified as follows to get the total digits.

Total digits = (9 x 1) + (90 x 2) + (900 x 3) + (1018 x 4) =6961

2017 - 999 = 1018

999 - 99 = 900

99 - 9 = 90

Now the equation is=1018 * 4 + 900 * 3 + 90 * 2 + 9 = 6961

The sum of digits of the single digit numbers is odd. The sum of digits of the multidigits number is even. So the total is odd. 6961 is the only odd number of the four choices.

There are even numbers of two-digit, three-digit, and four-digit numbers, but the number of one-digit numbers is odd. Therefore the answer is odd

Here is the "Brute force" way using JustBasic:

  for number = 1 to 2017

  number$ = str$(number)

   lenght = len(number$)

     sum = sum + lenght

next number

   print sum

There are 9 digits for the numbers 1-9.

The number of digits in the numbers 10-1999 will be a multiple of 10, which in turn means that the sum of these numbers will have a zero in the ones place.

The number of digits in the numbers 2000-2017 is 18x4 or 72. 72 +9 =81.

This means that the correct solution must have a 1 in the ones place.

The only answer ending in 1 is 6971.

This would not work unless this was a multiple choice question, but since it is, this solution works.

1
2
3
4

NUM = ''
for i in range(2017):
    NUM+=str(i+1)
len(NUM)

6961

• from 1-9 = 1*9 = 9
• from 10 - 99 = 2*90 = 180
• from 100 - 999 = 3*900 = 2700
• from 1000 - 2017 = 4*1018 = 4072

adding all these we get 6961.

(First of all in the options the unit digits of every number is different so we just need to bother about the unit digit instead of big calculations) Let's consider the numbers from 10 to 2014....... From 10 to 99 there are 90 numbers so number of digits 90x2 (unit digit=0), from 100 to 999 there are 900 numbers so number of digits 900x3 (unit digit=0), from 1000 to 2014 there are 1015 numbers so number of digits 1015x4 (here also unit digit will be zero as 5x4=20 whose unit digit is zero)...... Now sum of the above mentioned values will yield a zero as their unit digits.....Now the numbers left are from 1 to 9 and from 2015 to 2017 hence number of digits in these numbers will be= 9 + 3x4=21...... hence the unit digit will be 1 (as 1+0=1)........Hence the answer will be the first option. Thanks and enjoy using brilliant.

Step back for a moment, because only two calculations are needed to solve this.

9 numbers are single digit. 90 numbers (10-99) are double digit. 900 (100-999) have three digits. 1018 (1000-2017) have four digits.

The four options end in a different number, so we only need to consider the last digit, and in this case, the non-zero last digits:

9x1 = 9

(201)8 x 4 = 32

The total number of digits must end in 1, and the solution must therefore be 6961.

There are 9 numbers of 1 digit, and 900 numbers of 3 digits. If x is the number of numbers with two digits and y is the number of numbers with 4 digits then $9+2x+900*3+4y \equiv 2$ must be odd, and the unique possible solution is 6961

9 + 10x9x2 + 100x9x3 + 1018x1x4 = 6961

If one wished to use the order of the last number; namely 2017, in the calculation, then the solution may accurately be written as [(2017-1000)+1] x4 + [(999-100)+1] x3 + [(99-10)+1] x2 + [(9-1)+1] x1= 6961

9+(99-9)2+(999-99)3+(2017-999)4