Step 1

Unique Factorization Theorem.

For every integer n>1, there exists a positive integer k and distinct prime numbers

\(p_{1},p_{2}, p_{3},....,p_{k}\) and positive integers such that,

\(n=p_{1}^{e_{1}}p_{2}^{e_{2}}...p_{k}^{e_{k}}\)

Step 2

Perform prime factorization of the number.

4,116 can be divided by 2.

\(4116=2\times 2058\)

2058 can also be divided by 2.

\(4116 = 2\times 2058\)

\(=2\times 2\times 1029\)

\(=2^{2}\times 1029\)

1029 can be divided by 3.

\(4116=2^{2}\times 1029\)

\(=2^{2}\times 3\times 343\)

343 can be divided by 7.

\(4116=2^{2}\times 1029\)

\(=2^{2}\times 3\times 7\times 49\)

Again divide 49 by 7.

\(4116=2^{2}\times 3\times 7\times 49\)

\(=2^{2}\times 3\times 7\times 7\times 7\)

\(=2^{2}\times 3\times 7^{3}\)

Thus, the factorization of 4116 is \(2^{2}\times 3\times 7^{3}\).

Unique Factorization Theorem.

For every integer n>1, there exists a positive integer k and distinct prime numbers

\(p_{1},p_{2}, p_{3},....,p_{k}\) and positive integers such that,

\(n=p_{1}^{e_{1}}p_{2}^{e_{2}}...p_{k}^{e_{k}}\)

Step 2

Perform prime factorization of the number.

4,116 can be divided by 2.

\(4116=2\times 2058\)

2058 can also be divided by 2.

\(4116 = 2\times 2058\)

\(=2\times 2\times 1029\)

\(=2^{2}\times 1029\)

1029 can be divided by 3.

\(4116=2^{2}\times 1029\)

\(=2^{2}\times 3\times 343\)

343 can be divided by 7.

\(4116=2^{2}\times 1029\)

\(=2^{2}\times 3\times 7\times 49\)

Again divide 49 by 7.

\(4116=2^{2}\times 3\times 7\times 49\)

\(=2^{2}\times 3\times 7\times 7\times 7\)

\(=2^{2}\times 3\times 7^{3}\)

Thus, the factorization of 4116 is \(2^{2}\times 3\times 7^{3}\).