Bunuel
A book has 100 pages numbered from 1 to 100. Some of its pages are torn. If the sum of the numbers of the untorn pages equals 4949, how many pages are torn?
A. 1
B. 2
C. 3
D. 4
E. 5
The sum of the numbers on the full book is:
N(N+1)/2 = 50*101 = 5050
Since the total of untorn pages is 4949, the sum of the torn pages is:
5050-4949 = 101
Now, each page has two numbers, so for every page torn, two numbers are removed from the total, an odd and an even number.
If 1 page is torn, the two page numbers can be represented by:
2K+1 and 2K+2, their sum being:
4k+3 = 101 or 4k=98. Since 98 isn't divisible by 4, more than one page must be torn.
If two pages are torn their numbers can be represented by:
2k+1,2k+2,2L+1 and 2L+2 and their sum:
4(k+L)+6 = 101 or
4(k+L) = 95. Since 95 isn't divisible by 4, more than two pages must be torn.
How about 3 pages ? As above:
4(k+L+m) + 9 = 101 or:
4(k+L+m) = 92 or
K+L+M = 23
So, any combination of three different numbers adding to 23 will work.
For example:
K=12, l=8, m=3, which yield the following page numbers:
K:25 and 26 = 51
L:17 and 18 = 35
M: 7 and 8 = 15
Totaling 101
So three pages is the correct answer
Posted from my mobile device