Hello acc100jtWe like nice neat solutions to maths problems, but sometimes we have to use an intelligent 'trial and error' approach. So, if there were pages to start with, the page numbers totalled . So we're looking for a value of for which exceeds 1000, but not by so much that it is impossible to remove two consecutive integers , in order to bring the total down to 1000. , so try .

, which is not big enough. So try .

.

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 and , the total removed from this sum is , and is an odd number. So . So there's a solution: 45 pages, with pages 17 and 18 removed.

To confirm that this is the only solution, try . Then . This is not possible since must be odd.

Now try . Then is not an integer.

is again not an integer.

, which is too large, since . And similarly for any other .

First, let's establish that abc+acb+bac+bca+cab+cba=222(a+b+c), as follows: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)]

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