Unit digit, Tenth digit and Digit Sum

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*10¹ + 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 

Similar triangles, Trapezoids and Triangles that Share the Same Vertices

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

This is an interesting question that requires understanding of dimensional changes. (They are everywhere.)

Question: If D and E are midpoints of AC and AB respectively and the area of ΔBFC = 20, what is the 
a. area of Δ DFB? 
b. area of Δ EFC?
c. area of Δ DFE? 
d. area of Δ ADE? 
e. area of trapezoid DECB? 
f. area of Δ ABC?








Solution:

DE is half the length of BC (D and E are midpoints so DE : BC =  AD  : AB = 1 : 2

Δ DFE and ΔCFB are similar and their area ratio is 1² : 2²  = 1 : 4  (If you are not sure about this part, read this link on similar triangles.)

so the area of Δ DFE = (1/4) of ΔBFC = 20 = 5 square units. 

The area of Δ DFB = the area of Δ EFC = 5 x 2 = 10 square units because Δ DFE and Δ DFB,   
Δ DFE and ΔEFC share the same vertexs D and E respectively, so the heights are the same. 
Thus the area ratio is still 1 to 2. 

Δ ADE and ΔDEF share the same base and their height ratio is 3 to 1, so the area of
Δ ADE is 5 x 3 = 15 square units.

[DE break the height into two equal length and the height ratio of Δ DFE and ΔCFB is 1 to 2 (due to similar triangles) so the height ratio of Δ ADE and ΔDEF is 3 to 1.]

The area of trapezoid DEBC is 45 square units.

The area of Δ ABC is 60 square units. 


Extra problems to practice (answer below): 

The ratio of   AD and AB is 2 to 3,  DE//BC and the area of Δ BFC is 126, what is the area of

a. Δ DFE ? 

b. Δ DFB ?

c.  Δ EFC ? 

d.  Δ ADE? 

e. How many multiples is it of Δ ABC to ΔBFC?










Answer key: 
a. 56 square units
b. 84 square units
c. 84 square units
d. 280 square units
e. 5 times multiples.