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Question 1150802: Patty uses stickers with digits printed from 0 to 9 to number the pages of her scrapbook. She has lots of all digits except the digit 3, and she has only 37 of these. How many pages can she number in her scrapbook with this limitation?
Found 2 solutions by jim_thompson5910, math_helper:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Answer: 172
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How I got that answer:
List out numbers that have at least one '3' in them and count how many '3's are used up so far.
Below is a handy table to show what I mean.
Note that the second column steadily increases by 1 each time except in the cases of 33 and 133 which have two copies of '3' (as opposed to one copy of '3').
| Value | Number of '3's used so far |
| 3 | 1 |
| 13 | 2 |
| 23 | 3 |
| 30 | 4 |
| 31 | 5 |
| 32 | 6 |
| 33 | 8 (note the jump by +2) |
| 34 | 9 |
| 35 | 10 |
| 36 | 11 |
| 37 | 12 |
| 38 | 13 |
| 39 | 14 |
| 43 | 15 |
| 53 | 16 |
| 63 | 17 |
| 73 | 18 |
| 83 | 19 |
| 93 | 20 |
| 103 | 21 |
| 113 | 22 |
| 123 | 23 |
| 130 | 24 |
| 131 | 25 |
| 132 | 26 |
| 133 | 28 (note the jump by +2) |
| 134 | 29 |
| 135 | 30 |
| 136 | 31 |
| 137 | 32 |
| 138 | 33 |
| 139 | 34 |
| 143 | 35 |
| 153 | 36 |
| 163 | 37 |
| 173 | 38 |
The second column stops at 38 and not 37. This will help us determine the ceiling, so to speak, that Patty can reach.
If she had 38 copies of '3' available, then Patty could form 173. But she doesn't have 38 copies, and instead has 37, so the highest number she can reach is 172.
edit: the tutor @math_helper has a much more elegant and simple solution, so it's best to go with that.
Answer by math_helper(2461)  (Show Source):
You can put this solution on YOUR website!
If you look at the digit placeholders
[] [] []
100's 10's 1's
For pages 1-100, she will need 10+10 = 20 "3"'s
For pages 101-200, she will need an additional 20 "3"'s
If we take away pages 193, 183, and 173 she uses 37 "3"'s.
Thus, the maximum page number with 37 "3"'s as the limiting factor is  .
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