Correct Answer: 42

N = 2⁶ × 3⁵ × 5³ × 7²

We want the number of divisors of N that are of the form (3r + 1), i.e.

divisors ≡ 1 (mod 3).

Step 1: Kill the 3-factor

Any divisor containing 3¹ or higher is divisible by 3 → it cannot be 3r + 1.

So exponent of 3 must be 0.

We only use: 2ᵃ × 5ᶜ × 7ᵈ

a: 0 to 6 → 7 values; c: 0 to 3 → 4 values; d: 0 to 2 → 3 values

Step 2: Work mod 3

2 ≡ -1 (mod 3); 5 ≡ -1 (mod 3); 7 ≡ 1 (mod 3)

So divisor ≡ (-1)^(a + c) (mod 3).

For divisor ≡ 1 (mod 3):

(-1)^(a + c) = 1 → a + c must be even.

Step 3: Count (a, c) with a + c even

a = 0,1,2,3,4,5,6; even: 0,2,4,6 → 4 values; odd: 1,3,5 → 3 values

c = 0,1,2,3; even: 0,2 → 2 values; odd: 1,3 → 2 values

a + c even when: a even, c even → 4 × 2 = 8; a odd, c odd → 3 × 2 = 6

Total (a, c) good pairs = 8 + 6 = 14. d has 3 values → 0,1,2.

Total divisors = 14 × 3 = 42.

Final Answer: 12

Given: 3ac = 8(a + b) ⇒ 8b = 3ac − 8a ⇒ b = (3ac/8) − a

Since a, b, c are natural numbers, 3ac must be divisible by 8.

Now try small values of a (since we want the minimum value of 3a + 2b + c).

Try a = 1

Then 3c must be divisible by 8 ⇒ c = 8

b = (3×1×8)/8 − 1 = 3 − 1 = 2

Value = 3a + 2b + c = 3 + 4 + 8 = 15

Try a = 2

Then 6c divisible by 8 ⇒ c = 4 is the smallest possible

b = (3×2×4)/8 − 2 = 3 − 2 = 1

Here a, b, c = 2, 1, 4 are distinct natural numbers

Value = 3a + 2b + c = 6 + 2 + 4 = 12

Try a = 3

Then 9c divisible by 8 ⇒ c = 8

b = (3×3×8)/8 − 3 = 9 − 3 = 6

Value = 9 + 12 + 8 = 29 (larger)

Hence the smallest possible value occurs at a = 2, b = 1, c = 4