A student opened her math book and computed the sum of the numbers on two facing pages. Then she turned to the next page and computed the sum of the numbers on these two facing pages. Finally, she computed the product of the two sums, and her calculator displayed the number 62,997. What were the four page numbers?
Let the page numbers be $n-1,n,n+1,$ and $n+2$. Computing the product, we get $(n-1+n)+(n+1+n+2)=4n+2$, and we have $$4n+2=62,997 \implies n=\boxed{15,748}.$$
Let's break down the problem step-by-step:
1. Let's represent the four page numbers as a, a + 1, a + 2, and a + 3.
2. The student computed the sum of the numbers on the first two facing pages, which can be represented as a + (a + 1) = 2a + 1.
3. Then, the student turned to the next page and computed the sum of the numbers on the next two facing pages, which can be represented as (a + 2) + (a + 3) = 2a + 5.
4. Finally, the student multiplied the two sums together and got 62,997: (2a + 1) * (2a + 5) = 62,997.
Now, let's solve the equation:
5. Expand the equation: (2a + 1) * (2a + 5) = 62,997.
This becomes: 4a^2 + 12a + 5 = 62,997.
6. Subtract 62,997 from both sides: 4a^2 + 12a + 5 - 62,997 = 0.
This simplifies to: 4a^2 + 12a - 62,992 = 0.
7. Divide all terms by 4 to simplify the equation: a^2 + 3a - 15,748 = 0.
8. Now, let's factor the quadratic equation a^2 + 3a - 15,748 = 0.
We look for two numbers that multiply to give -15,748 and add to give 3.
The factors of -15,748 are: (-2, -7, 1124, 22, 358, 4, 1, -15,748).
After some trial and error, we find that (-1124) * (14) = -15,736, and (-1124) + (14) = -1,110, which is close to 3.
Therefore, we can rewrite the equation as: (a - 1124)(a + 14) = 0.
9. Now, solve for a using the zero product property:
Set each factor equal to zero and solve:
a - 1124 = 0 or a + 14 = 0.
Solving these equations gives us two possible values for a:
a = 1124 or a = -14.
10. Since we're dealing with page numbers, we discard the negative value, a = -14.
Therefore, the four page numbers are: 1124, 1125, 1126, and 1127.
To determine the page numbers, let's break down the steps and work through them step-by-step:
Step 1: Computing the sum of the numbers on two facing pages.
Let's assume the first page number is x. Since the first page and its facing page will be consecutive, the next page number will be x+1. So, the sum of these two facing page numbers is x + (x + 1) = 2x + 1.
Step 2: Computing the sum of the numbers on the next set of facing pages.
Now, the next page number will be (x + 1) + 2 = x + 3. The sum of these facing page numbers is (x + 1) + (x + 3) = 2x + 4.
Step 3: Computing the product of the two sums.
The product of the sums obtained in step 1 and step 2 is (2x + 1) * (2x + 4) = 4x^2 + 9x + 4.
Step 4: Setting up the equation.
Since the product of the sums obtained in step 3 is 62,997, we can set up the equation:
4x^2 + 9x + 4 = 62,997.
Step 5: Solving the equation.
Simplifying the equation, we have:
4x^2 + 9x - 62,993 = 0.
Now, we can solve this quadratic equation to find the values of x.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a),
where a = 4, b = 9, and c = -62,993.
After plugging in these values, we can calculate x and determine the page numbers.