Which is the smallest number that must be multiplied to 77175 to make it a perfect cube ?

Answer

To find the smallest number that must be multiplied to 77175 to make it a perfect cube, we will follow these steps:

### Step 1: Prime Factorization of 77175
We start by factorizing the number 77175 into its prime factors.

- Since 77175 is an odd number, it is not divisible by 2.
- Next, we check for divisibility by 3. The sum of the digits (7 + 7 + 1 + 7 + 5 = 27) is divisible by 3.
- Dividing 77175 by 3 gives us 25725.
- Dividing 25725 by 3 again gives us 8575.
- Now, we check 8575 for divisibility by 5 (since it ends in 5).
- Dividing 8575 by 5 gives us 1715.
- Dividing 1715 by 5 again gives us 343.
- Next, we check 343 for divisibility by 7.
- Dividing 343 by 7 gives us 49.
- Dividing 49 by 7 gives us 7.
- Finally, dividing 7 by 7 gives us 1.

Thus, the complete prime factorization of 77175 is:
\[ 77175 = 3^2 \times 5^2 \times 7^3 \]

### Step 2: Analyze the Exponents
To make a number a perfect cube, all the exponents in its prime factorization must be multiples of 3.

- For \(3^2\), the exponent is 2. To make it a multiple of 3, we need to add 1 (to make it \(3^3\)).
- For \(5^2\), the exponent is also 2. We need to add 1 here as well (to make it \(5^3\)).
- For \(7^3\), the exponent is already a multiple of 3.

### Step 3: Determine the Required Multiplication
To make \(3^2\) a perfect cube, we need to multiply by \(3^1\).
To make \(5^2\) a perfect cube, we need to multiply by \(5^1\).

Thus, the smallest number we need to multiply 77175 by is:
\[ 3^1 \times 5^1 = 3 \times 5 = 15 \]

### Conclusion
Therefore, the smallest number that must be multiplied to 77175 to make it a perfect cube is **15**.

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