Word problems on unit digit, tenth digit or digit sum.

**#1: How many digits are there in the positive integers 1 to 99 inclusive? **

Solution I: From 1 to 9, there are 9 digits.

From 10 to 99, there are 99 - 10 + 1 or 99 - 9 = 90 two digit numbers. 90 x 2 = 180

Add them up and the answer is

**189**.

Solution II: ___ There are 9 one digit numbers (from 1 to 9).

___ ___ There are 9 * 10 = 90 two digit numbers (You can't use "0" on the tenth digit but you

can use "0" on the unit digit.) 90 * 2 + 9 =

**189**
**# 2: A book has 145 pages. How many digits are there if you start counting from page 1?**

There are 189 digits from page 1 to 99. (See #1, solution I)

From 100 to 145, there are 145 - 100 + 1 or 145 - 99 = 46 three digit numbers.

189 + 46*3 =

**327 digits**.

**#3: "A book has N pages, number the usual way, from 1 to N. The total number of digits in the page number is 930. How many pages does the book have"? Similar to one Google interview question.**

Read the

questions and others here from the Wall Street Journal.

Solution I:

930 - 189 (digits of the first 99 pages) =741

741 divided by 3 = 247. Careful since you are counting the three digit numbers from 100 if the book has N

pages N - 100 + 1 or N - 99 = 247. N =

**346 pages**.

Solution II:

930 - 189 (total digits needed for the first 99 pages) = 741

741/3 = 247 (how far the three digit page numbers go).

247 + 99 =

**346 pages**

**#4: If you write consecutive numbers starting with 1, what is the 50th digit you write? **

Solution I:
50 - 9 = 41, 9 being the first 9 digits you need to use for the first 9 pages.

Now it's 2 digit. 41/2 = 20.1 , which means you will be able to write 20 two digit numbers + the first digit of the next two digit numbers.

10 to 29 is the first 20 two digit numbers so the next digit 3 is the answer. (first digit of the two digit number 30.)

Solution II: (50 - 9 ) / 2 = 20. 5 ; 20.5 + 9 = 29.5, so 29 pages + the first digit of the next two digit numbers, which is 3, the answer.

**#5: What is the sum if you add up all the digits from 1 to 100 inclusive?**
00 10 20 30 40 50 60 70 80 90

01 11 21 31 41 51 61 71 81 91

02 12 22 32 42 52 62 72 82 92

03 13 23 33 43 53 63 73 83 93

04 14 24 34 44 54 64 74 84 94

05 15 25 35 45 55 65 75 85 95

06 16 26 36 46 56 66 76 86 96

07 17 27 37 47 57 67 77 87 97

08 18 28 38 48 58 68 78 88 98

09 19 29 39 49 59 69 79 89 99

Solution I:

Do you see the pattern? From 00 to 99 if you just look at the unit digits.

There are 10 sets of ( 1+ 2 + 3 ... + 9) , which gives you the sum of 10 * 45 = 450

How about the tenth digits? There are another 10 sets of (1 + 2 + 3 + ...9) so another 450

Add them up and you have 450 * 2 = 900 digits from 1 to 99 inclusive.

900 + 1 ( for the "1" in the extra number 100) =

**901**
Solution II:

If you add the digits on each column, you have an arithmetic sequence, which is

45 + 55 + 65 ... + 135 To find the sum, you use average * the terms (how many numbers)

\(\dfrac {45+135} {2} * \left( \dfrac {135-45} {10}+1\right)\) =900

900 + 1 =

**901 **

Solution III :
2*45*10^{1} + 1 = **901**

**Problems to practice: Answers below.**
**#1: A book has 213 pages, how many digits are there?**

**#2: A book has 1012 pages, how many digits are there?**

**#3: If you write down all the digits starting with 1 and in the end there are a: 100, b: 501 and c: 1196 digits, what is the last digit you write down for each question?**

**#4: What is the sum of all the digits counting from 1 to 123? **

Answers:

#1: **531 digits. **

#2: **2941 digits.**

#3: a. **5**, b.** 3**, c.** 3**

#4: **1038 **

** **
** **