M09-17

Question

In a book with numbered pages, the first page is numbered 1 and the last page is numbered 705. On how many pages does the digit 9 appear in the page numbering?

A. 70
B. 77
C. 126
D. 133
E. 140

Bunuel · Accepted Answer

Official Solution:

In a book with numbered pages, the first page is numbered 1 and the last page is numbered 705. On how many pages does the digit 9 appear in the page numbering?

A. 70
B. 77
C. 126
D. 133
E. 140

To calculate how many pages feature the digit 9 in the page numbering, first determine how many times 9 appears in the first 100 pages, and then multiply that result by 7, as there are 7 blocks of 100 numbers (1-100, 101-200, ..., 601-700). In the first 100 pages, there are 10 pages where 9 appears as a units digit (9, 19, 29, ..., 99) and 10 pages where 9 appears as a tens digit (90, 91, 92, ..., 99). Page 99 is counted twice, so the total number of pages on which 9 appears in the first 100 is 19. Therefore, there are  19 * 7 = 133  pages with the digit 9 in the page numbering.

Answer: D

deepthit · Answer

Here are the list of numbers 
9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, and 99
Total - 20 9's in 1 hundred. Total = 20*7 = 140

chetan2u · Answer

the Q ask us about the PAGES on which 9 occurs, so 99 will be ONE and not two separately
so 19*7

chickxxchococat · Answer

Another approach:

Pages from 1 to 705.

9 as units digit: 9- first number in set, 699-last number in set, set is inclusive. Distance between each occurrence of '9' in units digits = 10.
>> Number of occurrences =   + 1 = 70

9 as tenths digit: if hundredth digit is '0', we have 90-first number in set, 98-last number in set (omit 99 as already counted above), set is inclusive. 
>> Number of occurrences: (98-90) + 1 = 9
Possible hundredth digits: 0 1 2 3 4 5 6 >> total number of occurrences = 9*7=63

Finally: total pages with '9' = 63 +70 = 133

manojach87 · Answer

For those who want to solve this using combinatorics

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sayan640 · Answer

MartyMurray   KarishmaB  Can you please provide your solution for this question ?
I tried to solve in this way during the exam but could not do so under time pressure.
1---- 9   ====>  only one '9'

From 10 to 99 ======>  total numbers = 9*10 = 90 ; Numbers in which '9' is not there = 8*9 = 72
Hence numbers in which '9' is there = 90 - 72 =18

Hence , total numbers from 1 to 99 = 18 + 1 = 19

For three digit numbers,

numbers  in which  9 is absent  = 7 * 9 * 9 = 567  
Total numbers = 705 - 100 + 1 = 606

Numbers in which '9' is there = 606 - 567 = 39

Where am I making mistake here ?  MartyMurray   KarishmaB   gmatophobia   ..Till the first part , I think I was correct ...where did I make mistakes for three digit numbers ?

MartyMurray · Answer

7 * 9 * 9 = 567 is where you went wrong.

You already accounted for the first 99. So, 7 * 9 * 9  is too high.

Takeaway: Remember what you've already done so that you don't double count.

Also, what about 700 -  705?

Takeaway: Be careful of leaving out details because you used a formula that doesn't quite apply in a particular situation.

You can add the 6 for 700 to 705 to the total after you've done the multiplication or ignore them completely since none have 9s in them.

So, now, what do you need to do instead of 7 * 9 * 9?

MartyMurray · Answer

In a book with numbered pages, the first page is numbered 1 and the last page is numbered 705. On how many pages does the digit 9 appear in the page numbering?

To make the problem simpler, we can ignore 700 - 705 since none of those numbers have 9 in them.

So, we're dealing with 1 - 699, or 699 numbers in total.

To find the number of numbers that have 9s, we can simply subtract the number that don't have 9s in them from 699.

If we use 0s as placeholders for the empty spaces in the hundreds and tens places of the 1 and 2 digit numbers, we can calculate the number of numbers without 9 as follows.

For the hundreds place we have 0 - 6, or 7 possible digits.

For the tens and ones places, we have 0 - 8, or 9 possible digits.

7 × 9 × 9 - 1 = 566 (We subract 1 for 000 since the lowest number is 1.)

699 - 566 = 133

A. 70
B. 77
C. 126
D. 133
E. 140

Correct answer:  D

sayan640 · Answer

Thank you  marty  for the clarification . Much appreciated!!

Posted from my mobile device

suya1sh · Answer

Answer is wrong 9 comes 20 times in 100 number series 9, 19, 29, 39, 49, 59, 69, 79, 89 these are having 9 in 9 times now from 90 to 99 there are 11 9's so that is total of 20 and answer for this 20×7 that is 140

Bunuel · Answer

In a book with numbered pages, the first page is numbered 1 and the last page is numbered 705.  On how many pages  does the digit 9 appear in the page numbering?

The question is correct. You are not reading the question and solution carefully. We are NOT asked how many times the digit 9 appears; we are asked how many pages contain the digit 9. Those are not the same. Please re-read the question, the solution, and the discussion more carefully, and invest some more time in it.

sb995 · Answer

I did not quite understand the solution. Hi, could you please explain where I went wrong in my approach? I broke the digits down into two separate categories:

1) units digit is 7 (7*1*10) = 70

2) tens digit is 7 (7*10*1) = 70

70+70=140 however I see that the correct answer is 133. Could you please help me pinpoint where I might have gone wrong in the above approach? Thank you

Bunuel · Answer

Your doubt is addressed in this thread. Please review.

sb995 · Answer

I did not quite understand the solution. Hi - you used a similar methodology for another problem here and I can't get to the right answer using that method here. Could you please tell me where I'm going wrong?

3 digits each for numbers 0 to 700. Means there are 2100 digits. Ignore 700-705 since 9 doesn't appear.
There's no reason for any digit to appear more often than the others so divide 2100 by 10 should give 210 times 9 will appear.

If it can appear in any of the 3 digits then divide 210 by 3 gives 70 but this is not the correct answer. Thank you.

Bunuel · Answer

This method does not apply for two reasons. First, the page numbers run from 1 to 705, not up to 999, so the assumption of equal digit distribution does not hold. Second, the question is not asking how many times the digit 9 appears, but rather how many distinct pages contain the digit 9.

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