number theory problem

**acc100jt** · March 10th 2009, 06:59 PM

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!

**Grandad** · March 11th 2009, 07:43 AM

Hello acc100jt

Originally Posted by acc100jt

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

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