Recall that the sum of the consecutive integers from 1 to n is given by n(n + 1)/2. Let’s analyze each Roman numeral:
I: 76 and 77
If the page numbers on the torn page was 76 and 77, then the sum of the pages of the novel was 10,000 + 76 + 77 = 10,153. To determine whether 10,153 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,153:
n(n + 1)/2 = 10153
n^2 + n = 20306
n^2 + n - 20306 = 0
(n + 143)(n - 142) = 0
n = -143 or n = 142
We see that 10,153 is the sum of the integers from 1 to 142; therefore it is possible that the page numbers on the torn page were 76 and 77.
II: 33 and 34
If the page numbers on the torn page was 33 and 34, then the sum of the pages of the novel was 10,000 + 33 + 34 = 10,067. To determine whether 10,067 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,067:
n(n+1)/2 = 10067
n^2 + n = 20134
n^2 + n - 20134 = 0
This equation has no integer roots. Thus, the page numbers on the torn page cannot be 33 and 34.
III: 5 and 6
If the page numbers on the torn page was 5 and 6, then the sum of the pages of the novel was 10,000 + 5 + 6 = 10,011. To determine whether 10,011 is the sum of consecutive integers, let’s solve n(n + 1)/2 = 10,011:
n(n+1)/2 = 10011
n^2 + n = 20022
n^2 + n - 20022 = 0
(n + 142)(n - 141) = 0
We see that 10,011 is the sum of the integers from 1 to 141; therefore it is possible that the page numbers on the torn page were 5 and 6.
Answer: E
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Hi
ScottTargetTestPrep,
Could you please tell how to split the middle term for such equations (n^2 + n - 20306 = 0
) in a quicker manner as I am finding difficulty in solving them.
Thanks!