2016 Digit Strings

Algebra Level 3

Following are the ten 2016-digit numbers with all their digits repeating in ascending (or increasing) order. What is the sum of the digits of the number obtained as a sum of these huge numbers?

0123456789012345678901234567890...123456789012345 123456789012345678901234567890...1234567890123456 23456789012345678901234567890...12345678901234567 3456789012345678901234567890...123456789012345678 456789012345678901234567890...1234567890123456789 56789012345678901234567890...12345678901234567890 6789012345678901234567890...123456789012345678901 789012345678901234567890...1234567890123456789012 89012345678901234567890...12345678901234567890123 9012345678901234567890...123456789012345678901234

Clarifications:

Despite leading with a 0, the first number has 2016 digits as well
Each of the numbers has a repeating part $\dots 0123456789 \dots$

Try Part 1 of this problem as well.

The answer is 18144.

2 solutions

Chung Kevin
Mar 3, 2016

This is simply the addition of ten 2016-digit numbers (yes, the first number on the top of the list is considered a 2016-digit as well).

By columnar addition,

The sum of digits in the $1^\text{st}$ column (from the right) is $0 + 1 + 2 + \cdots + 9 = 45$ , so the carry over is $c_1 = 4$ and the final digit at the end of this columnar sum is 5.

The sum of digits in the $2^\text{nd}$ column (from the right) is $0 + 1 + 2 + \cdots + 9 = 45$ , with the addition of the carry over, $c_1=4$ , the final digit at the end of this columnar sum is 9, and its carry over is $c_2 =4$ .

The sum of digits in the $3^\text{rd}$ column (from the right) is $0 + 1 + 2 + \cdots + 9 = 45$ , with the addition of the carry over, $c_2=4$ , the final digit at the end of this columnar sum is 9, and its carry over is $c_3 =4$ .

The sum of digits in the $2015^\text{th}$ column (from the right) is $0 + 1 + 2 + \cdots + 9 = 45$ , with the addition of the carry over, $c_{2014}=4$ , the final digit at the end of this columnar sum is 9, and its carry over is $c_{2015}=4$ .

The sum of digits in the $2016^\text{th}$ column (from the right) is $0 + 1 + 2 + \cdots + 9 = 45$ , with the addition of the carry over, $c_{2015}=4$ , the final digit at the end of this columnar sum is 9, and its carry over is $c_{2016}=4$ .

The final value can be written as

$\large 4\underbrace{99999\ldots9}_{2015 \text{ times}}5 .$

And so the sum of digits of this number is $4 + 9(2015) +5 = 9(2016) = \boxed{18144}$ .

Yeah! That's great.. it's exactly the same way that I expect to be followed to solve the problem.

Zeeshan Ali - 5 years, 3 months ago

Poca Poca
Jun 3, 2018

Let $S = 0+1+2+...+9=\frac{9\cdot 10}{2}=45$ and let $X$ be the total sum.

$\begin{aligned} X &= \sum_{k=0}^{2016}{10^kS} \\ &= 45\cdot \sum_{k=0}^{2016}{10^k} = 45\cdot \frac{1-10^{2017}}{1-10} &= -5 \cdot (1-10^{2017}) \\ &= 5\cdot \underbrace{9...9}_{2016} \\ &= 4 \underbrace{9...9}_{2015} 5 \end{aligned}$

Hence, the digital sum is $Q(X)=4+2015\cdot 9 + 5 = \boxed{18144}$ .

The proof of the fact that $5\cdot \underbrace{9...9}_{n} = 4 \underbrace{9...9}_{n-1} 5$ is left as an exercise to the reader.

For convenience:

for i in range(0,10):
    string = ''
    for i in range(i, 2016+i):
        string += str(i % 10)
    numbers.append(string)
sum = 0
for i in range(0,10):
    sum += int(numbers[i])
    print(numbers[i])
digitalsum = 0
sumstr = str(sum)
for i in range(0,len(sumstr)):
    digitalsum += int(sumstr[i])
print(digitalsum)

It yields the same.

Poca Poca - 3 years ago

2016 Digit Strings

The answer is 18144.

2 solutions

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