What is the least number that can be multiplied to 69120 to make it a perfect cube ?

Options

A. A) 10
B. B) 50
C. C) 25
D. D)5

Answer

To find the least number that can be multiplied to 69120 to make it a perfect cube, we need to follow these steps:

### Step 1: Prime Factorization of 69120
First, we will perform the prime factorization of 69120.

1. Divide by 2:
- 69120 ÷ 2 = 34560
- 34560 ÷ 2 = 17280
- 17280 ÷ 2 = 8640
- 8640 ÷ 2 = 4320
- 4320 ÷ 2 = 2160
- 2160 ÷ 2 = 1080
- 1080 ÷ 2 = 540
- 540 ÷ 2 = 270
- 270 ÷ 2 = 135 (stop here as 135 is not divisible by 2)

2. Now divide by 3:
- 135 ÷ 3 = 45
- 45 ÷ 3 = 15
- 15 ÷ 3 = 5 (stop here as 5 is a prime number)

So, the prime factorization of 69120 is:
\[ 69120 = 2^{8} \times 3^{3} \times 5^{1} \]

### Step 2: Determine the Exponents for Perfect Cube
For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3.

- The exponent of 2 is 8, which is \( 2 \mod 3 = 2 \). We need \( 3 - 2 = 1 \) more factor of 2.
- The exponent of 3 is 3, which is already a multiple of 3.
- The exponent of 5 is 1, which is \( 1 \mod 3 = 1 \). We need \( 3 - 1 = 2 \) more factors of 5.

### Step 3: Calculate the Least Number to Multiply
To make 69120 a perfect cube, we need to multiply it by:
\[ 2^{1} \times 5^{2} = 2 \times 25 = 50 \]

### Conclusion
Thus, the least number that can be multiplied to 69120 to make it a perfect cube is **50**.