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ANSWERS TO
LITTLE GAUSS AND A BOOK VANDAL
by Bill Graham
As you might have imagined, this has something to do with Gauss,
whom we discussed earlier.
You could guess and check with little Gauss’s method to find the length of
a series that adds up to 10,000.
But, luckily for me, there’s a formula to find the sum.
If n equals the number of terms in the series, then
[n (n + 1)] ÷ 2 = sum
Gauss had to find the sum of the numbers from one to 100.
[100 (101)] ÷ 2 = 5,050
For the book puzzle,
[n (n + 1)] ÷ 2 > 10,000
I used the quadratic formula to solve this and found out that the number of pages
had to be greater than 140.5. So I tried 141 pages.
[141(142)] ÷ 2 = 10,011
Aha! If I tear out page five (page six is on the other side of the leaf),
then I have subtracted 11 and the sum of the remaining pages is exactly 10,000.
Is this the only solution?
[142(143)] ÷ 2 = 10,153
You would have to tear out a page numbered 76 and 77. In a book, the odd page
number is always smaller. There is no such page.
[143(144)] ÷ 2 = 10,296
Too bad! The number in excess of 10,000 must be odd to be the sum of two
consecutive page numbers. Besides, the number would be larger than the
length of the book (144 page numbers).
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Vol 2, No 11 Answers
