Question

Let \(b\) be a positive integer and \(a = b^2 - b\). If \(b \ge 4\), then \(a^2 - 2a\) is divisible by

• A. 15
• B. 20
• C. All of these
• D. 24

Hint:

Factorise \(a^2-2a\) in terms of \(b\); it turns out to be a product of four consecutive integers.

Answer 1

The Correct Option is D

Step 1: Factorise.
\(a=b^2-b=b(b-1)\). Then
\[a^2-2a=a(a-2)=b(b-1)\,(b^2-b-2)=b(b-1)(b-2)(b+1)\]since \(b^2-b-2=(b-2)(b+1)\).

Step 2: Interpret.
This is the product \((b-2)(b-1)\,b\,(b+1)\) of four consecutive integers. Among any four consecutive integers there is a multiple of 3 and two even numbers, one of which is a multiple of 4. So the product is always divisible by \(3\times 8=24\).

Step 3: Check 15 and 20.
Divisibility by 15 or 20 needs a factor 5, which is not guaranteed. For \(b=8\): \(6\cdot 7\cdot 8\cdot 9=3024\), and \(3024\) is not divisible by 15 or by 20 (\(3024=2^4\cdot 3^3\cdot 7\)). So only 24 always divides it. (For \(b=4\) to \(7\) the values 120, 360, 840, 1680 happen to be divisible by all three, which is why a single example can mislead.)

Final Answer:
The correct option is (4): 24.