How Big is a Trillion

Part 1. First determine the sum of all the digits needed to write each counting number from 0 through 1,000,000,000,000. For example, the sum of all the digits needed to write each counting number from 18 through 23 would be 1+8+1+9+2+0+2+1+2+2+2+3=33.
Part 2. Next, find the sum of the digits of the sum found in Part 1. In other words, for 18 through 23, the sum of all of the digits was 33, and the sum of the digits of that sum is 3+3=6.
Part 3. Finally, only submit your sum of the digits for the sum of the digits or simply the answer you determined for Part 2.
Source: mathcontest.olemiss.edu 11/4/2013

SOLUTION
We first try a few smaller problems to get a feel for what’s happening. Let’s find the sum of the numbers from 0\to 99. The following table lists the numbers from 0\to 99 by columns. The sum of the numbers of each row is listed at the end of the row. For example, the sum of the numbers in the first row equals 450; the sum of the numbers in the second row equals 460; etc. Likewise, the sum of the numbers in each column is listed at the bottom of the column. For example, the sum of the numbers in the first column equals 45; the sum of the numbers in the second column equals 145; etc.


Sum of numbers 0\to 99
Rows: 450+460+470+\cdots +540=4950
Columns: 45+145+245+\cdots +945=4950


Sum of numbers 100\to 199
Rows: 1450+1460+1470+\cdots +1540=14950
Columns: 1045+1145+1245+\cdots +1945=14950


Sum of numbers 200\to 299
Rows: 2450+2460+2470+\cdots +2540=24950
Columns: 2045+2145+2245+\cdots +2945=24950

If we continue this method, we have the following results


Sum of numbers 0\to 1999
4950+14950+24950+\cdots +194950=4950\times 20+\left (1+2+3+\cdots +19\right )10^4
=99000+\left (190\right )10000
=1999000
As the numbers increase by 100 from row to row, the sums increase by 10000.  The 20 rows correspond to the 20 intervals of 100 into which the distance of 2000 from 0\to 1999 is divided. The sum \left (1+2+3+\cdots\right ) stops at 19 which is 1 less than 20.

We are now ready to calculate the sum of numbers from 0\to 999,999,999,999
Number of 100-intervals
\dfrac{10^{12}}{100}=10^{10}
4950\times 10^{10}+\left (1+2+3+\cdots +9,999,999,999\right )10^4=4950\times 10^{10}+\left (5\times 10^9\times 9,999,999,999\right )10^4
=495\times 10^{11}+49,999,999,995\times 10^{13}
=\left (495+49,999,999,995\times 10^2\right )10^{11}
=\left (4,999,999,999,995\right )10^{11}

Finally, we add 1,000,000,000,000 to the above sum
\left (4,999,999,999,995\right )10^{11}+10^{12}=\left (4,999,999,999,995+10\right )10^{11}
=\left (5,000,000,000,005\right )10^{11}

Sum of the digits
5+5=10

Answer: 10

Alternative solution
Imagine a sheet of paper with the digits 00,01,02,\cdots,99 written in a table of 100 rows by 2 columns. If you fold the paper in half by folding the bottom half to the top half, you end up with a table of 50 rows by 4 columns a portion of which is shown below
0\;9\;0\;9
0\;9\;1\;8
0\;9\;2\;7
0\;9\;3\;6
0\;9\;4\;5
0\;9\;5\;4
0\;9\;6\;3
0\;9\;7\;2
0\;9\;8\;1
\underline{0\;9\;9\;0}
1\;8\;0\;9
1\;8\;1\;8
1\;8\;2\;7
1\;8\;3\;6
1\;8\;4\;5
1\;8\;5\;4
1\;8\;6\;3
1\;8\;7\;2
1\;8\;8\;1
\underline{1\;8\;9\;0}
2\;7\;0\;9
2\;7\;1\;8
2\;7\;2\;7
2\;7\;3\;6
2\;7\;4\;5
2\;7\;5\;4
2\;7\;6\;3
2\;7\;7\;2
2\;7\;8\;1
\underline{2\;7\;9\;0}
\cdots
If you add the digits horizontally across the rows, each row sums to 2\times 9=18 for a total of 50\times 18=900.
The sum of the digits of 0\to 99 equals 900.

Imagine a sheet of paper with digits 000,000,000,000 through 999,999,999,999 written in a table of 10^{12} rows by 12 columns. Fold the paper in half and you end up with a smaller table of 5\times 10^{11} rows by 24 columns.
If you add the digits horizontally across the rows, each row sums to 12\times 9=108 for a total of 5\times 10^{11}\times 108=54\times 10^{12}.
The sum of the digits from 000,000,000,000\to 999,999,999,999 equals 54\times 10^{12}
Finally 5+4+1=10 where 1 comes from 1,000,000,000,000.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

3 Responses to How Big is a Trillion

  1. Allow me to shine a blinding light of clarity upon this for you and your readers.

    Visualize a table of 1e12 rows and 12 columns, append leading zeros to all numbers less than 1e12. Fill the rows of the table (right justified) with the numbers from 0 to ((1e12)-1). One digit per cell. Thinking about the variation in the columns reveals that each column has the exact same number of each digit, but in a different sort order or variation pattern. By inspecting the 10^0 (see note 1) column’s first 10 entries, we may determine that the average of all the cells (digits) in the table is 4.5 and there are 12e12 of them for a total sum of 54e12. To account for the contribution of 1e12 add 1 to the total sum. 5+4+(0*12)+1 = 10.

    Note 1: The right most column of the table. 10^0 is also known as the units position or the Least Significant Digit before the decimal point.

  2. As a further exercise: Apply your preferred method to a 32 bit binary number range. One may also do it in hexadecimal: 0000 0000 to FFFF FFFF. It is quite easy and fun; if you choose your method wisely. 3Ch8 = 3Cx16^8 = 257,698,037,760 in base 10.

  3. You may be asking: “Where did the 3C come from?” Okay, well, it comes from ((F+0)/2) * 8 . Or in simple English, the average value of each digit times the number of columns. 7.5 * 8 = 60 = 3C in base 16.

    Happy Trails.

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