Question

An integer A written in an n-based number system is 647, while written in an (n+2)-based number system is 513. Write the number A in the 10 number system! (n is of course a positive integer)​

Answer 1

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!