CAT 2025 Slot 3 QA Question 14

We have three natural numbers p, q, and r. Their sum is 900. So, p + q + r = 900. That's straightforward. Now, r is a perfect square, and its value is between 150 and 500. So, r is a perfect square such that 150 < r < 500. Also, p is not less than 0.3q and not more than 0.7q. So, mathematically, that translates to 0.3q ≤ p ≤ 0.7q. The question is asking for the sum of the maximum and minimum possible values of p.

Alright, so We find the maximum and minimum possible values of p given these constraints and then add them together. The answer is supposed to be 397, but I need to figure out how to get there.

First, let's break down the constraints:

Since r is a perfect square between 150 and 500, let me list all the perfect squares in that range. Perfect squares are numbers like 12²=144, 13²=169, 14²=196, and so on, up to 22²=484 because 22² is 484, which is less than 500, and 23² is 529, which is over 500. So, the possible values of r are 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.

So, r can be any of these values: 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.

Given that p + q + r = 900, we can express p + q = 900 - r. Let's denote S = p + q = 900 - r. So, S depends on the value of r.

Also, from the constraints on p, we have 0.3q ≤ p ≤ 0.7q. Let's express p in terms of q. So, p is between 0.3q and 0.7q. Let's write that as:

0.3q ≤ p ≤ 0.7q

But since p + q = S, we can write p = S - q. So, substituting into the inequality:

0.3q ≤ S - q ≤ 0.7q

Let me solve these inequalities for q.

First, the left part: 0.3q ≤ S - q

Adding q to both sides: 1.3q ≤ S

So, q ≤ S / 1.3

Similarly, the right part: S - q ≤ 0.7q

Adding q to both sides: S ≤ 1.7q

So, q ≥ S / 1.7

Therefore, combining both inequalities:

S / 1.7 ≤ q ≤ S / 1.3

So, q must be between S / 1.7 and S / 1.3.

But since q is a natural number, it has to be an integer. So, q must satisfy:

ceil(S / 1.7) ≤ q ≤ floor(S / 1.3)

Where ceil is the ceiling function (rounding up to the nearest integer) and floor is the floor function (rounding down to the nearest integer).

Now, since p = S - q, and p must also be a natural number, q must be less than S because p has to be positive.

So, for each possible value of r, we can compute S = 900 - r, then compute the range for q, and then find the possible values of p.

But since we need the maximum and minimum possible values of p, we need to find the extremes of p across all possible r.

Wait, but p is dependent on q and r. So, for each r, we can find the range of p, and then among all these ranges, find the overall maximum and minimum p.

Alternatively, maybe we can find the maximum and minimum p by considering the extreme values of r.

Since r is between 150 and 500, the smallest r is 169, and the largest is 484.

So, when r is smallest (169), S = 900 - 169 = 731.

When r is largest (484), S = 900 - 484 = 416.

So, S ranges from 416 to 731.

Now, for each S, we can find the range of q, and then p = S - q.

But we need to find the maximum and minimum p across all possible S.

Wait, but p is bounded by 0.3q ≤ p ≤ 0.7q.

So, for each S, the maximum p is 0.7q, and the minimum p is 0.3q.

But q is bounded by S / 1.7 ≤ q ≤ S / 1.3.

So, for each S, the maximum p would be when q is as small as possible, because p = S - q. So, if q is small, p is large.

Similarly, the minimum p would be when q is as large as possible, because p = S - q, so q large means p small.

But we also have the constraints on p in terms of q: p ≤ 0.7q and p ≥ 0.3q.

So, let's formalize this.

Given S, q must satisfy:

S / 1.7 ≤ q ≤ S / 1.3

And p = S - q must satisfy:

0.3q ≤ S - q ≤ 0.7q

Which we already used to get the bounds on q.

So, for each S, the maximum p is when q is minimum, which is q_min = ceil(S / 1.7). Then p_max = S - q_min.

Similarly, the minimum p is when q is maximum, which is q_max = floor(S / 1.3). Then p_min = S - q_max.

But we need to ensure that p_max and p_min satisfy the original inequalities.

Wait, but p_max is S - q_min, and p_min