Understanding and Calculating the Smallest Positive Integer ( n ) Such That ( n^3 ) is Divisible by 2020 Without Remainders

Understanding and Calculating the Smallest Positive Integer ( n ) Such That ( n^3 ) is Divisible by 2020 Without Remainders

Introduction to Prime Factorization

To understand the problem, we first need to break down the number 2020 into its prime factors. The prime factorization of 2020 is as follows:

Prime Factorization of 2020

The prime factors of 2020 are 22 5 101. This means that 2020 can be expressed as:

2020 22 5 101

Given this, we need to find the smallest positive integer ( n ) such that ( n^3 ) is divisible by 2020 without any remainders. This problem requires a deep understanding of divisibility and prime factorization.

Breaking Down the Factors for ( n^3 )

To ensure that ( n^3 ) is divisible by 2020, ( n ) itself must contain at least the prime factors 2, 5, and 101 in sufficient numbers to make ( n^3 ) divisible by 2020.

Let’s consider the prime factorization again:

Required Prime Factors for ( n )

The smallest ( n ) must include:

At least one 2 At least one 5 At least one 101

Hence, the smallest ( n ) would be the product of these prime factors,

Thus, the smallest ( n ) is:

2 5 101 1010

And this ( n ) will ensure that ( n^3 ) contains:

23 (since ( 1010^3 ) includes 101sup3; and 1010 contains 22) 53 1013

Therefore:

1010sup3; 23 53 1013 2020 510050

Conceptualizing the Smallest Cube Divisible by 2020

Alternatively, we can also deduce the smallest cube divisible by 2020. For this, we will round up the factors in the prime factorization to the nearest multiple of 3 to ensure divisibility by 3. Let's do this step by step:

2020 22 5 101

For ( n^3 ) to be divisible by 2020, the prime factorization of ( n ) must include:

23 (since 22 rounds up to 23) 53 1013

Thus, the smallest ( n ) is:

23 53 1013 10203 1030301000

Hence, the smallest ( n ) such that ( n^3 ) is divisible by 2020 is:

1030301000

Conclusion

The smallest positive integer ( n ) such that ( n^3 ) is divisible by 2020 without any remainders is:

1010

This solution is derived from understanding the prime factorization of 2020 and ensuring that the smallest ( n ) contains sufficient prime factors to make ( n^3 ) divisible by 2020. The process involves breaking down the problem using prime factorization, rounding up the factors to the nearest multiple of 3, and ensuring that ( n ) includes all necessary prime factors.