How many two digit numbers have exactly 4 positive factors? (Here 1 and the
number n are also considered as factors of n.)
  






A total fixed amount of N thousand rupees is given to three persons A, B, C, every
year, each being given an amount proportional to her age. In the first year, A got
half the total amount. When the sixth payment was made, A got six-seventh of the
amount that she had in the first year; B got Rs 1000 less than that she had in the
first year; and C got twice of that she had in the first year. Find N.

### Problem 1: Two-Digit Numbers with Exactly 4 Positive Factors

To solve this problem, we need to find two-digit numbers that have exactly 4 positive factors. Let's break down the concept:

1. **Prime Factorization and Number of Factors Formula:**
   - If a number \( n \) has a prime factorization \( n = p_1^{e_1} \times p_2^{e_2} \times \dots \times p_k^{e_k} \), then the total number of factors of \( n \) is given by \( (e_1 + 1)(e_2 + 1) \dots (e_k + 1) \).
   - We need to find two-digit numbers where this product equals 4.

2. **How Can a Number Have 4 Factors?**
   - There are two possible cases:
     1. The number is the product of exactly one prime raised to the power of 3. In this case, the number would have \( (3+1) = 4 \) factors. Example: \( p^3 \).
     2. The number is the product of two distinct primes, each raised to the power of 1. In this case, the number would have \( (1+1)(1+1) = 4 \) factors. Example: \( p_1 \times p_2 \).

Now, let's identify all two-digit numbers for both cases:

- **Case 1: \( n = p^3 \)**
  - Here, \( n \) must be a cube of a prime number.
  - Consider the following primes:
    - \( 2^3 = 8 \) (Not a two-digit number)
    - \( 3^3 = 27 \)
    - \( 5^3 = 125 \) (Not a two-digit number)
    - \( 7^3 = 343 \) (Not a two-digit number)
    
  Therefore, the only two-digit number of this form is 27.

- **Case 2: \( n = p_1 \times p_2 \)**
  - Here, \( n \) is the product of two distinct prime numbers.
  - Consider all pairs of primes where the product gives a two-digit number:
    - \( 2 \times 5 = 10 \)
    - \( 2 \times 7 = 14 \)
    - \( 2 \times 11 = 22 \)
    - \( 2 \times 13 = 26 \)
    - \( 2 \times 17 = 34 \)
    - \( 2 \times 19 = 38 \)
    - \( 2 \times 23 = 46 \)
    - \( 2 \times 29 = 58 \)
    - \( 2 \times 31 = 62 \)
    - \( 2 \times 37 = 74 \)
    - \( 2 \times 41 = 82 \)
    - \( 2 \times 43 = 86 \)
    - \( 2 \times 47 = 94 \)
    - \( 3 \times 5 = 15 \)
    - \( 3 \times 7 = 21 \)
    - \( 3 \times 11 = 33 \)
    - \( 3 \times 13 = 39 \)
    - \( 3 \times 17 = 51 \)
    - \( 3 \times 19 = 57 \)
    - \( 3 \times 23 = 69 \)
    - \( 3 \times 29 = 87 \)
    - \( 5 \times 7 = 35 \)
    - \( 5 \times 11 = 55 \)
    - \( 5 \times 13 = 65 \)
    - \( 5 \times 17 = 85 \)
    - \( 5 \times 19 = 95 \)
    - \( 7 \times 11 = 77 \)
    - \( 7 \times 13 = 91 \)

So, the two-digit numbers with exactly 4 factors are:  
\[ \{10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 94, 95\} \]

There are **29** such numbers.

---

### Problem 2: Finding \( N \) in the Distribution Problem

Given:
- A total fixed amount \( N \) thousand rupees is distributed among three persons A, B, and C.
- The distribution is proportional to their ages.
- In the first year, A got half of the total amount, so A got \( \frac{N}{2} \) thousand rupees.
- In the sixth payment:
  - A got six-sevenths of the amount she received in the first year, i.e., \( \frac{6}{7} \times \frac{N}{2} \).
  - B got Rs 1000 less than what she got in the first year.
  - C got twice the amount she got in the first year.

Let the amounts received by A, B, and C in the first year be \( A_1 = \frac{N}{2} \), \( B_1 \), and \( C_1 \), respectively. Since the distribution is proportional to their ages, we have:
\[
A_1 + B_1 + C_1 = N
\]

From the problem:
- \( A_6 = \frac{6}{7} \times A_1 \)
- \( B_6 = B_1 - 1 \) (since Rs 1000 is equivalent to 1 thousand rupees)
- \( C_6 = 2 \times C_1 \)

Now, using the fact that the total amount \( N \) remains the same, we can write:
\[
A_6 + B_6 + C_6 = N
\]

Substituting the values of \( A_6 \), \( B_6 \), and \( C_6 \):
\[
\frac{6}{7} \times \frac{N}{2} + (B_1 - 1) + 2 \times C_1 = N
\]

We already know that \( A_1 = \frac{N}{2} \), so the equation becomes:
\[
\frac{6}{7} \times \frac{N}{2} + B_1 - 1 + 2 \times C_1 = N
\]

Also, from the original distribution:
\[
\frac{N}{2} + B_1 + C_1 = N
\]

We can solve these two equations simultaneously to find \( N \).

Let's solve for \( N \).The solution shows that \( N \) and \( B_1 \) are expressed in terms of \( C_1 \):

- \( N = 14C_1 - 14 \)
- \( B_1 = 6C_1 - 7 \)

This implies that the total amount \( N \) depends on the amount \( C_1 \) that person C received in the first year. To find the exact value of \( N \), we need additional information about \( C_1 \). However, this equation shows that the total distribution and the amounts given to B are linearly related to C's first-year share.

If you have any further details or assumptions about \( C_1 \), we can determine \( N \) exactly.

Would you like further clarification on these equations, or is there another detail you'd like to explore?

Here are 5 related questions:
1. How do you derive the formula for the number of factors of a number?
2. What are the properties of numbers with exactly 4 factors?
3. How does the proportional distribution relate to the ages of A, B, and C?
4. Could we solve this problem using a different approach or additional assumptions?
5. What methods can we use to find the factors of larger numbers efficiently?

**Tip:** When dealing with problems involving factors, prime factorization is a key tool in determining the properties of the number.

How many two-digit numbers have exactly 4 positive factors? A total fixed amount of N thousand rupees is given to three persons A, B, C, every year, each being given an amount proportional to her age. In the first year, A got half the total amount. When the sixth payment was made, A got six-seventh of the amount that she had in the first year; B got Rs 1000 less than that she had in the first year; and C got twice of that she had in the first year. Find N.

Explore how to find two-digit numbers with exactly 4 positive factors using prime factorization, and solve a problem involving proportional distribution of a fixed amount of money among three people over time. This problem covers number theory concepts and is suitable for students in Grades 10-12.

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