The number of divisors of (2⁶ × 3⁵ × 5³ × 7² ) , which are of the form (3r + 1) , where r is a non-negative integer, is

Correct Answer: 42 (Option 4)

Step-by-step Explanation:

We are given the prime factorization of a number N:

$N=2^6\times 3^5\times 5^3\times 7^2$

We need to find the number of divisors of N that are of the form (3r + 1), where r is a non-negative integer (i.e., r = 0, 1, 2, ...).

1. Analyzing the Condition (3r + 1):

A number of the form (3r + 1) gives a remainder of 1 when divided by 3. This means:

- The divisor cannot be divisible by 3. Therefore, the exponent of 3 in any such divisor must be 0 (since any positive power of 3 would make the divisor a multiple of 3, i.e., of the form 3k).

- Any divisor of N has the general form:

$d=2^a\times 3^b\times 5^c\times 7^d$

where 0 ≤ a ≤ 6, b = 0 (as established), 0 ≤ c ≤ 3, and 0 ≤ d ≤ 2.

2. Analyzing Modulo 3 Behavior of Prime Factors:

Let's check the values of the prime bases modulo 3:

- Base 2: 2 ≡ -1 (mod 3). Thus, 2^a ≡ (-1)^a (mod 3).

- If a is even, 2^a ≡ 1 (mod 3).

- If a is odd, 2^a ≡ -1 (mod 3).

- Base 5: 5 ≡ -1 (mod 3). Thus, 5^c ≡ (-1)^c (mod 3).

- If c is even, 5^c ≡ 1 (mod 3).

- If c is odd, 5^c ≡ -1 (mod 3).

- Base 7: 7 ≡ 1 (mod 3). Thus, 7^d ≡ 1^d ≡ 1 (mod 3) for all d (d = 0, 1, 2).

3. Determining Combinations for d ≡ 1 (mod 3):

Since 7^d is always 1 (mod 3), the modular condition for the entire divisor d becomes:

$2^a\times 5^c\equiv 1\text{(mod 3)}$

Using the powers of -1 modulo 3:

${(-1)}^a\times {(-1)}^c={(-1)}^{a+c}\equiv 1\text{(mod 3)}$

This holds true if and only if (a + c) is an even integer.

This occurs in two cases:

Case I: Both a and c are even

- Choices for a (even, from 0 to 6): {0, 2, 4, 6} → 4 choices

- Choices for c (even, from 0 to 3): {0, 2} → 2 choices

Number of ways for Case I = 4 × 2 = 8 combinations.

Case II: Both a and c are odd

- Choices for a (odd, from 0 to 6): {1, 3, 5} → 3 choices

- Choices for c (odd, from 0 to 3): {1, 3} → 2 choices

Number of ways for Case II = 3 × 2 = 6 combinations.

Total valid combinations for (a, c) = 8 + 6 = 14 combinations.

4. Including Choices for b and d:

- Choices for b: Only {0} → 1 choice

- Choices for d: {0, 1, 2} → 3 choices

5. Final Calculation:

Total number of divisors of the form (3r + 1) is:

$\text{Total Divisors}=14\times 1\times 3=42$