Answer:
79
Explanation:
Let's start by writing out what we know:
A = 647 in base-n
A = 513 in base-(n+2)
To find A in base-10, we need to use the place value system and the definition of each base. Let's first convert A to base-10 using the given information.
In base-n, A is equal to:
A = 6n^2 + 4n + 7
In base-(n+2), A is equal to:
A = 5(n+2)^2 + 1(n+2) + 3
A = 5(n^2 + 4n + 4) + (n + 2) + 3
A = 5n^2 + 21n + 20
Now we have two expressions for A, so we can set them equal to each other and solve for n:
6n^2 + 4n + 7 = 5n^2 + 21n + 20
Simplifying and rearranging, we get:
n^2 + 8n + 13 = 0
Using the quadratic formula, we can solve for n:
n = (-8 ± sqrt(8^2 - 4(1)(13))) / (2(1))
n = (-8 ± 2) / 2
n = -3 or n = -5
Since n is a positive integer, we can disregard the negative solution and conclude that n = 3.
Now that we know n, we can substitute it back into the expression for A in base-n and solve for A in base-10:
A = 6n^2 + 4n + 7
A = 6(3^2) + 4(3) + 7
A = 79
Therefore, the number A written in base-10 is 79.
Hope this helps!