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  1. #1
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    number theory problem

    I have 2 very challenging secondary 1 questions here.

    1. A leaf was torn out from a book. The page numbers on the remaining pages of the book summed up to 1000. How many pages were there in this book at first? Which pages were torn out?[ans: 45 pages, 17 & 18 torn out]

    2. Given that a, b, and c are non-zero digits and satisfy acb+bac+bca+cab+cba=3194
    [ans: 358, hint given: consider S=abc+acb+bac+bca+cab+cba=222(a+b+c)]

    Anyone can help? Many Thanks!
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  2. #2
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    Hello acc100jt
    Quote Originally Posted by acc100jt View Post
    I have 2 very challenging secondary 1 questions here.

    1. A leaf was torn out from a book. The page numbers on the remaining pages of the book summed up to 1000. How many pages were there in this book at first? Which pages were torn out?[ans: 45 pages, 17 & 18 torn out]
    We like nice neat solutions to maths problems, but sometimes we have to use an intelligent 'trial and error' approach. So, if there were n pages to start with, the page numbers totalled \sum_{i=1}^ni=\tfrac{1}{2}n(n+1). So we're looking for a value of n for which \tfrac{1}{2}n(n+1) exceeds 1000, but not by so much that it is impossible to remove two consecutive integers \le n, in order to bring the total down to 1000. \sqrt{2000} = 44.7, so try n = 44.

    \tfrac{1}{2}\cdot 44\cdot 45 = 990, which is not big enough. So try n = 45.

    \tfrac{1}{2}\cdot 45\cdot 46 = 1035.

    Now two consecutive integers are removed from the sequence of page numbers, and, noting that in a book the right-hand pages are always odd numbers, the first of these will be an odd number. So if the pages removed were i and i+1, the total removed from this sum is 2i+1, and i is an odd number. So 2i+1 = 35 \Rightarrow i =17. So there's a solution: 45 pages, with pages 17 and 18 removed.

    To confirm that this is the only solution, try n = 46. Then \tfrac{1}{2}\cdot 46 \cdot 47 = 1081 \Rightarrow i = 40. This is not possible since i must be odd.

    Now try n = 47. Then \tfrac{1}{2}\cdot 47 \cdot 48 = 1128 \Rightarrow i is not an integer.

    n = 48 \Rightarrow i is again not an integer.

    n = 49 \Rightarrow \tfrac{1}{2}\cdot 49\cdot 50 = 1225 \Rightarrow i = 112, which is too large, since i \le 49. And similarly for any other n > 49.
    2. Given that a, b, and c are non-zero digits and satisfy acb+bac+bca+cab+cba=3194
    [ans: 358, hint given: consider S=abc+acb+bac+bca+cab+cba=222(a+b+c)]
    First, let's establish that abc+acb+bac+bca+cab+cba=222(a+b+c), as follows:

    If all six arrangements of the digits a, b and c are written down, one underneath the other to form an addition sum, there are 2 a's, 2 b's and 2 c's in each of the hundreds, tens and units columns. The total of each column is therefore 2(a + b + c), and their total, taking their place value into account is:

    100 x 2(a + b + c) + 10 x 2(a + b + c) + 2(a + b + c) = 222(a + b + c).

    Now we observe that when one of these arrangements, abc, is omitted the total is 3194. So 222(a + b + c) = 3194 + abc, where 0 < abc < 1000, and the sum of abc's digits, s, is, of course, a + b + c.

    So, we try some values for (a + b + c) bearing the above requirements in mind, working out abc each time, by subtracting 3194 from 222(a + b + c).

    If (a + b + c) < 15, 222(a + b + c) < 3194.

    If (a + b + c) = 15, 222(a + b + c) = 3330; abc = 136; s = 10

    If (a + b + c) = 16, 222(a + b + c) = 3552; abc = 358; s = 16

    If (a + b + c) = 17, 222(a + b + c) = 3774; abc = 580; s = 13

    If (a + b + c) = 18, 222(a + b + c) = 3996; abc = 802; s = 10

    If (a + b + c) > 19, abc > 1000, which is clearly impossible.

    So the only solution is (a + b + c) = 16; abc = 358.

    Grandad
    Last edited by Grandad; March 11th 2009 at 08:26 AM. Reason: Solution to qu 2
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