Question

\(p, q\) and \(r\) are three non-negative integers such that \(p+q+r=10\). The maximum value of \(pq+qr+pr+pqr\) is

• A. \(\ge 40 \text{ and }<50\)
• B. \(\ge 50 \text{ and }<60\)
• C. \(\ge 60 \text{ and }<70\)
• D. \(\ge 70 \text{ and }<80\)
• E. \(\ge 80 \text{ and }<90\)

Hint:

Convert expressions like \(pq+pr+qr+pqr\) using \((p+1)(q+1)(r+1)\) to separate the symmetric sum from the fixed total. For integers with fixed sum, the product is maximized when the numbers are as equal as possible.

Answer 1

The Correct Option is C

Step 1: Use the identity \[(p+1)(q+1)(r+1)=pqr+pq+pr+qr+p+q+r+1.\] Hence the target \[ T=pq+pr+qr+pqr=(p+1)(q+1)(r+1)-(p+q+r)-1=(p+1)(q+1)(r+1)-11, \] since \(p+q+r=10\). Therefore maximizing \(T\) is equivalent to maximizing \((p+1)(q+1)(r+1)\).
Step 2: Let \(a=p+1,\ b=q+1,\ c=r+1\). Then \(a,b,c\) are positive integers with fixed sum \[ a+b+c=(p+q+r)+3=13. \] By AM–GM (or by the fact that for fixed sum the product is maximized when numbers are as equal as possible), the product \(abc\) is maximized at the near-equal split \(4,4,5\) (since \(13/3\approx 4.33\)). Thus take \((a,b,c)=(4,4,5)⇒ (p,q,r)=(3,3,4)\) in some order.
Step 3: Compute the maximum value: \[ T_{\max}=abc-11=4\cdot4\cdot5-11=80-11=69. \] Therefore, the maximum lies in the interval \(\boxed{[60,70)}\).