CAT 2024 Slot 1 QA Question 16

The question says: If (a + b√n) is the positive square root of (29 - 12√5), where a and b are integers, and n is a natural number, then the maximum possible value of (a + b + n) is... and there are four options: A) 4, B) 18, C) 6, D) 22.

First, We analyze what's being asked. We have an expression under a square root, 29 - 12√5, and we're told that its positive square root can be written in the form a + b√n, where a and b are integers, and n is a natural number. Then, we have to find the maximum possible value of the sum a + b + n.

Okay, so my goal is to find integers a, b, and a natural number n such that when I square (a + b√n), I get 29 - 12√5. Then, among all possible such triples (a, b, n), We find the one where a + b + n is as large as possible.

Let me recall the formula for squaring a binomial with a radical. If I have (x + y√z)^2, it expands to x² + 2xy√z + y²z. So, applying that here, squaring (a + b√n) should give me a² + 2ab√n + b²n.

So, setting that equal to 29 - 12√5, we have:

a² + 2ab√n + b²*n = 29 - 12√5.

Now, since 29 and -12√5 are both terms, I can equate the rational and irrational parts separately. That is, the non-radical parts must be equal, and the coefficients of the radicals must be equal.

So, breaking it down:

Since we're dealing with square roots, and the original expression is a positive square root, both a and b must be positive integers. Wait, actually, the square root is positive, so (a + b√n) must be positive. So, a and b can't both be negative, but since n is a natural number, √n is positive, so if b is negative, a would have to be positive enough to make the whole expression positive. Hmm, but the problem says a and b are integers, so they could be positive or negative. But since we're looking for the maximum value of a + b + n, I think we can assume that a and b are positive because if they were negative, their sum would be smaller. So, I'll proceed under the assumption that a and b are positive integers.

So, from the irrational part: 2ab√n = -12√5. Wait, that's a problem because the left side is positive (since a, b, n are positive) and the right side is negative. That can't be. So, maybe I made a mistake in assuming both a and b are positive. Alternatively, perhaps one of them is negative.

Wait, but the square root is positive, so (a + b√n) must be positive. So, if b is negative, then a must be positive enough to offset it. Let's suppose that b is negative. Then, 2ab√n would be negative, which would match the right side being -12√5. So, maybe b is negative. So, let's write the irrational part as 2ab√n = -12√5. So, 2ab√n = -12√5. Since √n is positive, and √5 is positive, then 2ab must be negative. So, 2ab = -12. Therefore, ab = -6.

So, ab = -6. Since a and b are integers, possible pairs (a, b) could be (1, -6), (-1, 6), (2, -3), (-2, 3), (3, -2), (-3, 2), (6, -1), (-6, 1). But since we're looking for the maximum a + b + n, and a and b can be positive or negative, but n is a natural number, so positive. .

But wait, let's also consider the rational part: a² + b²*n = 29.

So, Taking each possible pair of (a, b) where ab = -6 and see if they satisfy a² + b²*n = 29, with n being a natural number.

Starting with (a, b) = (1, -6). Then, a² + b²*n = 1 + 36n = 29. So, 36n = 28. Then, n = 28/36 = 7/9, which is not a natural number. So, this pair doesn't work.

Next, (a, b) = (-1, 6). Then, a² + b²*n = 1 + 36n = 29. Same as above, n = 7/9, not natural.

Next, (a, b) = (2, -3). Then, a² + b²*n = 4 + 9n = 29. So, 9n = 25, n = 25/9, which is not natural.

Next, (a, b) = (-2, 3). Then, a² + b²*n = 4 + 9n = 29. Same as above, n = 25/9, not natural.

Next, (a, b) = (3, -2). Then, a² + b²*n = 9 + 4n = 29. So, 4n = 20, n = 5. That's a natural number. So, this pair works. So, a = 3, b = -2, n = 5. Then, a + b + n = 3 + (-2) + 5 = 6.

Next, (a, b) = (-3, 2). Then, a² + b²*n = 9 + 4n = 29. Same as above, n = 5. So, a = -3, b = 2, n = 5. Then, a + b + n = -3 + 2 +