Question

Let $p, q$ and $r$ be three natural numbers such that their sum is 900, and $r$ is a perfect square whose value lies between 150 and 500. If $p$ is not less than $0.3q$ and not more than $0.7q$, then the sum of the maximum and minimum possible values of $p$ is

Hint:

When you have constraints like $ap \le q \le bp$ along with $p + q + r = \text{constant}$, try expressing $q$ in terms of $p$ and $r$, then convert the inequalities into bounds for $p$ in terms of $r$. After that, use monotonicity (increasing/decreasing behavior) to decide which extreme values of $r$ give the extreme values of $p$.

Answer 1

Correct Answer: 397

Approach: Fix \(r\) (a perfect square in range); then \(p+q = 900-r\) is fixed, and the constraint \(0.3q\le p\le 0.7q\) bounds \(p\). To make \(p\) as large as possible you want a small \(r\); to make \(p\) as small as possible you want a large \(r\). So check the smallest and largest valid squares.

Step 1: List the perfect squares with \(150 < r < 500\).
\(13^2=169,\ 14^2=196,\ \ldots,\ 22^2=484.\) Don't miss the endpoints \(169\) and \(484\).

Step 2: Turn the bound into a range for \(p\).
With \(S=p+q=900-r\), we have \(q=S-p\). The condition \(0.3q\le p\le 0.7q\) becomes:
Lower: \(p\ge 0.3(S-p)\Rightarrow 1.3p\ge 0.3S\Rightarrow p\ge \dfrac{0.3}{1.3}S=\dfrac{3}{13}S.\)
Upper: \(p\le 0.7(S-p)\Rightarrow 1.7p\le 0.7S\Rightarrow p\le \dfrac{0.7}{1.7}S=\dfrac{7}{17}S.\)
So for each \(r\): \(\dfrac{3}{13}S \le p \le \dfrac{7}{17}S,\ \ S=900-r.\)

Step 3: Maximum \(p\) — use the smallest \(r=169\).
\(S=900-169=731.\) Upper bound \(=\dfrac{7}{17}\times 731 = 301.\) This is an integer, and at \(p=301,\ q=430\) we get \(p=0.7q\) exactly (natural numbers). So \(p_{\max}=301.\)

Step 4: Minimum \(p\) — use the largest \(r=484\).
\(S=900-484=416.\) Lower bound \(=\dfrac{3}{13}\times 416 = 96.\) Integer, and at \(p=96,\ q=320\) we get \(p=0.3q\) exactly. So \(p_{\min}=96.\)

Step 5: Add the extremes.
\(p_{\max}+p_{\min} = 301+96 = 397.\)

\[\boxed{p_{\max}+p_{\min} = 397}\]

Answer 2

Alternate approach — testing the extreme cases directly:
Since \( p+q+r=900 \) and \( 0.3q \le p \le 0.7q \), for a fixed sum \( S=p+q=900-r \), the extreme values of \( p \) occur at the boundary ratios.
At \( p=0.7q \), \( p=\frac{7}{17}S \), which is largest when \( S \) is largest, i.e. \( r \) is smallest. The smallest perfect square strictly between 150 and 500 is \( 169 \), giving \( S=731 \) and \( p_{max}=\frac{7}{17}(731)=301 \).
At \( p=0.3q \), \( p=\frac{3}{13}S \), which is smallest when \( S \) is smallest, i.e. \( r \) is largest. The largest such perfect square is \( 484 \), giving \( S=416 \) and \( p_{min}=\frac{3}{13}(416)=96 \).
Sum of max and min \( = 301+96 = \) 397.