Question 3461
To do this problem it is helpful to know that the sum of the numbers from 1 to x is {{{(n(n + 1))/2}}}. This means the sum of the page numbers from 1 to x is:<br>
{{{(x(x + 1))/2}}}.<br>
To find the remaining sum, think about it this way {{{5 + 4 = 1 + 2 + 3 + 4 + 5 - (1 + 2 + 3)}}}. In other words the sum from (x+1) to 10 is the sum of 1 to 10 minus the sum of 1 to x. This allows us to write a relatively simple expression for the sum from (x+1) to 10<br>
{{{(10(10 + 1))/2 - (x(x + 1))/2}}}<br>
Since the sum from one to x is 1 plus the sum of x+1 to 10, we get the following equation<br>
{{{(x(x + 1))/2 = 1 + (10(10 + 1))/2 - (x(x + 1))/2 = 1 + 55  - (x(x + 1))/2}}} or<br>
{{{(x(x + 1))/2 = 56  - (x(x + 1))/2}}}<br>
If we add {{{(x(x + 1))/2}}} to both sides we get<br>
{{{x(x + 1) = 56}}}<br>
The equation becomes {{{x^2 + x - 56 = 0}}} which factors nicely to {{{(x - 7)(x + 8) = 0}}}. So the solution to the equation is x = 7 and x = -8. Well, you cannot have negative pages, so the answer to the question must be 7. Let's check it.
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the sum from 1 to 7 is {{{(7(7+1))/2 = 28}}} and {{{8 + 9 + 10 = 27}}}, which is one more than the sum from 1 to 7. So the answer is indeed 7.