r/learnmath • u/Gaudupada ie • 21h ago
Please help me!
Okay, so I have this sum:
"A printer numbers the pages of a book starting with 1 and uses 3189 digits in all. How many pages the book have?
Options:
a) 1074
b) 1075
c) 1000
d) None of the above"
My solution:
"The total number of pages have 3189 digits. (given)
⇒ the number of combinations= 9*10³¹⁸⁸.
∴ The total number of numbers= 9*10³¹⁸⁸*3189!
∴d) None of the above"
The Book's Solution:
"1 digit numbers have 9 digits (9 numbers)
2 digit numbers have 180 digits (90 numbers)
3 digit numbers have 2700 digits (900 numbers)
4 digit numbers have 300 digits (75 numbers)
[since total number of digits is 3189, hence
Number of digits for 4 digit numbers
=3189-(9+180+27000)= 300]
Total number of pages
= 9 +90 +900 +75= 1074"
So, according to them, the correct answer is a) 1074, but for me its d) None of the above.
Please help
Note: The solution of the book is almost verbatim, with marginal changes in the format.
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u/Southlander24 A friendly Redditor!👋 21h ago
You are grossly misinterpreting the question. Why are you assuming there's a 3189 digit number?
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u/themonstrousprotein New User 21h ago
that book solution is the standard way to solve these. you just break it down by how many digits each page number uses and subtract from the total. your approach went off the rails right at the start by treating the digit count like an exponent
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u/Gaudupada ie 18h ago
So, should I even be multiplyïng them to-gether, because I took '9 for single digit numbers, and 10 for the rest, and there 3189 pages, in total.'
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u/Gaudupada ie 21h ago
Because it says 'there are 3189 digits in all', I thought I should account for 3189 possible places.
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u/Muphrid15 New User 21h ago
You took the question to mean there are 3189 digits on the final page
That is not what the question means
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u/Southlander24 A friendly Redditor!👋 21h ago
So you're saying that there's a page number with 111....111 (3189 digits), another page number with 111....112, 111...113, all the way up to 999...999? Have you realised how many pages that would include in total?
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u/Gaudupada ie 18h ago
Yup... It would be huge... That's why I didn't bother multiplying all those numbers.
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u/Dazzling_Music_2411 New User 21h ago
The supplied solution is correct.
Your (main) error is that you view all the page numbers as joined up into one string of 3189 digits. Then you permute all the possibilities of this string! Obviously the number is huge.
You have imposed no constraints for order. So, to illustrate, one of the gazillion possibilities is that the first page is number 7028, followed by page 1113, followed by page 5, followed by 999, then 2658, etc, etc... Get what I'm driving at?
Once you impose an order of pages increasing by 1 every time, the problem becomes easy.
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u/Gaudupada ie 18h ago
So, how dœs that look like in Maths?
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u/Dazzling_Music_2411 New User 17h ago
The book explanation you have given in your OP covers it totally adequately.
For the first 9 pages you count 1 digit per page. The next 90 pages you have 2 digits per page. Total 189. For the next 900 pages you have 3 digits per page. Total 2889 digits in 999 pages, yes?
The next 9000 pages (if they existed) would have 4 digits per page). How many of them do you actually need? Well, you have to make up 3189 digits. You have 2889 so far, so you need 300 digits more. Each page has 4 digits, so you need 300/4 = 75 more pages. Plus the 999 you already have makes 1074 pages for 3189 digits.
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u/jeffcgroves New User 21h ago
Random comment: are they counting the two sides of a page as 2 pages or 1 page?
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u/EdgyMathWhiz New User 21h ago
Not a totally unreasonable question, but if they were counting 2-sides as one page the correct answer would be less than 1000.
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u/jeffcgroves New User 21h ago
Oh, I misread that as "the correct answer is none of the above". Really need to start reading posts before commenting on them ;)
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u/Calm_Relationship_91 New User 21h ago
It doesn't matter, each page corresponds to one number, so the answer is the same.
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u/Gaudupada ie 18h ago
Erm... Two sides, I believe? 'cause that's how I've thought to number my pages in my exams, and all the books I have read are numbered like that...
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u/hallerz87 New User 20h ago
Your solution misinterprets the question. I’m not even sure why you are considering all possible combinations of digits… have you even attempted to understand the solution? It’s completely different to what you’ve done, and for very good reason
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u/ZerinoWhellie New User 20h ago
I will point out, in addition to other comments explaining the correct understanding of the problem, that your answer doesn’t even include the number of pages, which is what the question asks.
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u/Gaudupada ie 18h ago
Yeah, I was like "This itself is this huge, so why bother?"
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u/ZerinoWhellie New User 18h ago
While I am definitely aware that not all problems are rooted in reality (“Bob went to the supermarket and bought 642 bananas”), I can safely teach you that, if you ever calculate an astronomically large number and it’s not an astronomy problem, you might want to take a step back and re-evaluate what you did.
This reminds me of the Matt Rose video where he reads absurd answers people gave on tests and one is “I calculated that a coffee was 300,000,000°C in temperature”.
Also, by your thought process, the question couldn’t even be completed, how do you get to figure out the number of pages? There is no additional data to get you there.
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u/Gaudupada ie 18h ago
It was because I took it like one of those "Find the number of words, without or without meaning, which use the letters in the word 'ROSE'" type questions.
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u/ZerinoWhellie New User 18h ago
Yes, anagrams work like that, but they have a unique solution. There are a lot of solutions for “how many pages can you number with x digits”, where x is your answer on the post.
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u/Gaudupada ie 18h ago
You know, sometimes you don't think of a certain way, or look over a detail, thinking they are not important, but those are what lead to your correct solution?
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u/Bounded_sequencE New User 21h ago
The assignment does not say every page number has 3189 digits, but that the total number of digits printed on all pages combined is 3189.