CAT — Divisibility Rules

Step 1: Use the property of divisibility
If k divides two numbers, it also divides any integer linear combination of those numbers.
Given: k | (m + 2n) and k | (3m + 4n)

Step 2: Eliminate n
Multiply the first expression by 2: 2(m + 2n) = 2m + 4n
Subtract from the second: (3m + 4n) − (2m + 4n) = m
Hence: k | m

Step 3: Find another quantity divisible by k
Since k | (m + 2n) and k | m:
Subtract: (m + 2n) − m = 2n
Hence: k | 2n

Step 4: Conclusion
k is a common divisor of m and 2n.

Step 1: Write the given equation. Given, 3ac = 8(a + b). Rearranging: 8b = 3ac − 8a → b = a(3c − 8)/8. Since b is a natural number, a(3c − 8) must be divisible by 8.

Step 2: Find the smallest possible values. We need to minimize 3a + 2b + c. Try the smallest natural values of c.

Case 1: c = 1. b = −5a/8, which is not a natural number. Not possible.
Case 2: c = 2. b = −a/4, which is not a natural number. Not possible.
Case 3: c = 3. b = a/8. For b to be a natural number, a must be a multiple of 8. Smallest such value is a = 8. Then b = 1. Numbers are distinct: 8, 1, 3. 3a + 2b + c = 3×8 + 2×1 + 3 = 24 + 2 + 3 = 29.
Case 4: c = 4. b = a/2. For b to be a natural number, a must be even. Take a = 2 (smallest even keeping numbers distinct). Then b = 1. Numbers are 2, 1 and 4, all distinct. 3a + 2b + c = 3×2 + 2×1 + 4 = 6 + 2 + 4 = 12.

Step 5: Check whether a smaller value is possible. For c = 1 and c = 2, no natural number solution exists. For c = 3, the minimum value obtained is 29. For c ≥ 5, the value of c itself increases, and the corresponding values of a and b remain positive, making the expression larger than 12. Hence, the smallest possible value is 12.

Step 1: Write the general form of a divisor. The given number is 2⁶ × 3⁵ × 5³ × 7². A divisor is of the form 2ᵃ × 3ᵇ × 5ᶜ × 7ᵈ where 0 ≤ a ≤ 6, 0 ≤ b ≤ 5, 0 ≤ c ≤ 3, 0 ≤ d ≤ 2.

Step 2: Use the condition that the divisor is of the form 3r + 1. A number of the form 3r + 1 leaves remainder 1 when divided by 3. If b ≥ 1, then the divisor is divisible by 3. Hence, b = 0. Now the divisor becomes 2ᵃ × 5ᶜ × 7ᵈ.

Step 3: Find the remainder modulo 3. Modulo 3: 2 ≡ −1, 5 ≡ −1, 7 ≡ 1. Therefore, 2ᵃ × 5ᶜ × 7ᵈ ≡ (−1)ᵃ × (−1)ᶜ × 1ᵈ = (−1)^(a+c). For the remainder to be 1, a + c must be even.

Step 4: Count the valid values of a and c. Possible values of a: 0 to 6. Even values: 0, 2, 4, 6 → 4 choices. Odd values: 1, 3, 5 → 3 choices. Possible values of c: 0 to 3. Even values: 0, 2 → 2 choices. Odd values: 1, 3 → 2 choices. For a + c to be even: both even = 4 × 2 = 8 pairs; both odd = 3 × 2 = 6 pairs. Total valid pairs = 8 + 6 = 14.

Step 5: Choose the value of d. Since 7 ≡ 1 (mod 3), the value of d does not affect the remainder. Possible values of d are 0, 1 and 2, giving 3 choices. Total number of divisors = 14 × 3 = 42.