A fruit vendor brings 1092 apples and 3432 oranges to a market. He arranges them in heaps of equal number of oranges as well as apples such that every heap consists of the maximum possible number of the fruits. What is this number?

Options

A. 78
B. 156
C. 312
D. 39

Answer

To solve the problem of arranging the apples and oranges in heaps of equal number, we need to find the highest common factor (HCF) of the two quantities: 1092 apples and 3432 oranges. Here’s the step-by-step solution:

### Step 1: Identify the quantities
We have:
- Number of apples = 1092
- Number of oranges = 3432

### Step 2: Find the prime factorization of each quantity
**Prime Factorization of 1092:**
1. Divide by 2: \( 1092 \div 2 = 546 \)
2. Divide by 2: \( 546 \div 2 = 273 \)
3. Divide by 3: \( 273 \div 3 = 91 \)
4. Divide by 7: \( 91 \div 7 = 13 \)
5. 13 is a prime number.

So, the prime factorization of 1092 is:
\[ 1092 = 2^2 \times 3^1 \times 7^1 \times 13^1 \]

**Prime Factorization of 3432:**
1. Divide by 2: \( 3432 \div 2 = 1716 \)
2. Divide by 2: \( 1716 \div 2 = 858 \)
3. Divide by 2: \( 858 \div 2 = 429 \)
4. Divide by 3: \( 429 \div 3 = 143 \)
5. Divide by 11: \( 143 \div 11 = 13 \)
6. 13 is a prime number.

So, the prime factorization of 3432 is:
\[ 3432 = 2^3 \times 3^1 \times 11^1 \times 13^2 \]

### Step 3: Identify the common prime factors
From the prime factorizations:
- For \( 1092 \): \( 2^2, 3^1, 7^1, 13^1 \)
- For \( 3432 \): \( 2^3, 3^1, 11^1, 13^2 \)

The common prime factors are:
- \( 2 \) (minimum power is \( 2^2 \))
- \( 3 \) (minimum power is \( 3^1 \))
- \( 13 \) (minimum power is \( 13^1 \))

### Step 4: Calculate the HCF
Now we multiply the common prime factors:
\[ \text{HCF} = 2^2 \times 3^1 \times 13^1 = 4 \times 3 \times 13 \]
Calculating this step-by-step:
1. \( 4 \times 3 = 12 \)
2. \( 12 \times 13 = 156 \)

Thus, the HCF (maximum number of fruits in each heap) is:
\[ \text{HCF} = 156 \]

### Final Answer
The maximum possible number of apples and oranges in each heap is **156**.

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