Question

Consider \((xxx)_b = x^3\), where \(b\) is the base, and \(x\) is any digit in base \(b\). Find \(b\).

• A. 5
• B. 6
• C. 7
• D. 8
• E. None of the above

Hint:

When an equation demands a number be a perfect square, bound it between two consecutive squares to rule out possibilities quickly.

Answer 1

Answer: (E)

Step 1: The value of \((xxx)_b\) is \[ x b^2 + x b + x = x(b^2+b+1). \] Given \((xxx)_b = x^3\) and \(x\ne0\), divide by \(x\): \[ b^2+b+1 = x^2. \] Step 2: Note that for any integer \(b\ge2\), \[ b^2<b^2+b+1<(b+1)^2=b^2+2b+1. \] Hence \(b^2+b+1\) lies strictly between consecutive perfect squares and therefore cannot itself be a perfect square. Thus there is no integer \(x\) with \(x^2=b^2+b+1\).
So no base \(b\in\{5,6,7,8\}\) (nor any \(b\ge2\)) satisfies the condition; answer is \(\boxed{\text{None of the above}}\).