Problem Solving

In tens - we have 7x10 = 70 (i.e. from 90-99 for every 100 and so on..)
In units - we have 9x7 = 63 (i.e. from 1-100, 101-200 and so on..)

Sum = 133.

1-100 = 19, 9's (9,19,29,39,49,59,69,79,89,99 & 90,91,92,93,94,95,96,97,98)
Hence,
1-705 = 19*7=133.

Counting problems are always tricky
Single digit no : 1
2 Digit No : 18
3 digit no upto 699:-
100~199 : 19
Hence fr 100~699 :19*6 = 114
Total = 114+18+1 = 133

How many numbers with 9's are there from 1-100?
9, 19, 29... =10
90, 91, 92... =9 (we already counted 99)

This gives me a total of 19. 19*7= 133 (since the hundreds digit will never contain a 9

Answer is D

(D) 133

705/10=70

705/100=7

70+7*9=70+63=133

Answer should be 126 as it asks how many pages.
There are only 18 pages on every hundred, thus it should be 18x7=126.

Page 9 , 19 , 29 , 39 , 49 , 59, 69, 79, 89, then 91, 92, 93, 94, 95, 96, 97, 98, 99. (total 18 pages).

There is a nine per each 10 pages (9,19,29 etc) --> Every ten pages there is one that contains the digit 9.
There are ten nines per each 100 pages (90,91,91-99/ 190,191,192/ 290,291,292) --> Every 100 pages there are 10 pages that contains the digit 9
So
705:10 =70 (Seventy times 9 is the third digit)
705:100= 7 7*10= 70 (Seventy times 9 is the second digit)
70+70=140

Now we need to discount the 7 times that nine is a the third and the second digit in the same page (99,199,299,399..) so 140-7 = 133 (D)

And what happen with the 90??? There are 19 pages on every 100 so 19*7=133

PiyushKashyap wrote:Answer should be 126 as it asks how many pages.
There are only 18 pages on every hundred, thus it should be 18x7=126.

Page 9 , 19 , 29 , 39 , 49 , 59, 69, 79, 89, then 91, 92, 93, 94, 95, 96, 97, 98, 99. (total 18 pages).

I still have a disturbance
"How many pages does digit 9 appear in the numeration?"
That means we only count one digit 9 in the number 99, right?

All pages of the book are numbered. If the first page is numbered 1 and the last page is numbered 705, on how many pages does digit 9 appear in the numeration?

(A) 70
(B) 77
(C) 126
(D) 133
(E) 140

How I did?
Pages 1 - 700

001 - 100 = 1*9 + 10 =19
101 - 200 = 1*9 + 10 =19
so on and add them all.
or 19*7 = 133.

A) 0 to 99 has second digit = 9, 10 times
B) 90 - 99 has first digit = 9, 10 times; but only 9 times if we exclude numbers from statement (A)
Therefore each set of 100 consecutive pages contains 10+9 = 19 pages (bearing at least one 9)
From 0 to 699 (as 700 to 705 contain no nines) = 7 x 100 pages
7 x 19 = 133

Using P&C approach

Let's form three digit numbers without using 9 such that

leftmost place is occupied any digit from 0 till 6 i.e. 7 ways
Middle place is occupied any digit from 0 till 8 i.e. 9 ways
Rightmost place is occupied any digit from 0 till 8 i.e. 9 ways

Total Such numbers = 7x9x8= 567

(These 567 numbers include all single digit number, two digit number and three digit numbers from 0 till 699 without using 9 in any number

rest numbers from 0 till 699 will have digit 9 in them so the numbers with digit 9 = 700-567 = 133

133 is the right answer.