Math Problem Statement

When three positive integers a , a, b , b, and c c are multiplied together, their product is 100. 100. Suppose a < b < c . a<b<c. In how many ways can the numbers be chosen?

0 0

1 1

2 2

3 3

4

Solution

We are tasked with finding the number of ways to choose three positive integers a,b,ca, b, c such that:

  1. a<b<ca < b < c,
  2. abc=100a \cdot b \cdot c = 100.

Step 1: Prime factorization of 100100

The prime factorization of 100100 is: 100=2252.100 = 2^2 \cdot 5^2.

Thus, any a,b,ca, b, c must be factors of 100100, and their product must still equal 100100.


Step 2: List the factors of 100100

The factors of 100100 are: 1,2,4,5,10,20,25,50,100.1, 2, 4, 5, 10, 20, 25, 50, 100.


Step 3: Find distinct triplets (a,b,c)(a, b, c)

We now determine all possible sets of three factors (a,b,c)(a, b, c) where a<b<ca < b < c, and their product equals 100100.

Case-by-case enumeration:

  1. Case 1: a=1a = 1 bc=100with b<c.b \cdot c = 100 \quad \text{with } b < c.

    • b=2,c=50(1250=100)b = 2, c = 50 \quad (1 \cdot 2 \cdot 50 = 100),
    • b=4,c=25(1425=100)b = 4, c = 25 \quad (1 \cdot 4 \cdot 25 = 100),
    • b=5,c=20(1520=100)b = 5, c = 20 \quad (1 \cdot 5 \cdot 20 = 100),
    • b=10,c=10(invalid since bc)b = 10, c = 10 \quad \text{(invalid since } b \neq c).

    Valid triplets: (1,2,50),(1,4,25),(1,5,20)(1, 2, 50), (1, 4, 25), (1, 5, 20).

  2. Case 2: a=2a = 2 bc=50with b<c.b \cdot c = 50 \quad \text{with } b < c.

    • b=5,c=10(2510=100)b = 5, c = 10 \quad (2 \cdot 5 \cdot 10 = 100).

    Valid triplet: (2,5,10)(2, 5, 10).

  3. Case 3: a=4a = 4 bc=25with b<c.b \cdot c = 25 \quad \text{with } b < c.

    • b=5,c=5(invalid since bc)b = 5, c = 5 \quad \text{(invalid since } b \neq c).

    No valid triplets.


Step 4: Count the number of triplets

The valid triplets are: (1,2,50),(1,4,25),(1,5,20),(2,5,10).(1, 2, 50), (1, 4, 25), (1, 5, 20), (2, 5, 10).

Thus, there are 4 ways to choose (a,b,c)(a, b, c) such that a<b<ca < b < c and abc=100a \cdot b \cdot c = 100.


Final Answer:

4


Would you like me to clarify any step or expand further?


Follow-up Questions:

  1. How do we handle cases where the integers are not distinct?
  2. What happens if one of the integers is fixed, e.g., a=2a = 2?
  3. How does the solution change for a different product, such as abc=200a \cdot b \cdot c = 200?
  4. Can you explain the reasoning behind using prime factorization in problems like this?
  5. How can this approach be extended to solve for four or more integers?

Tip:

When solving factorization problems with constraints, always start by listing all factors and work systematically to avoid missing cases.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Prime Factorization
Inequalities

Formulas

Product of integers: a * b * c = 100

Theorems

Fundamental Theorem of Arithmetic

Suitable Grade Level

Grades 9-11