, 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.