Multiplication of Factorials

Algebra Level pending

Let $N!=1!\cdot 2!\cdot 3!\cdot .....\cdot 10!$ where $!$ is factorial symbol. Let $k$ be the greatest power of $2$ such that $N$ is divisible by ${ 2 }^{ k }$ . Compute $k$ .

The answer is 38.

1 solution

William Isoroku
Dec 5, 2014

Expanding $N$ gives us $N=1\cdot { 2 }^{ 9 }\cdot { 3 }^{ 8 }\cdot { 4 }^{ 7 }\cdot { 5 }^{ 6 }\cdot { 6 }^{ 5 }\cdot { 7 }^{ 4 }\cdot { 8 }^{ 3 }\cdot { 9 }^{ 2 }\cdot 10$

Prime factor it gives us $N={ 2 }^{ 38 }\cdot .......$

....... are basically other prime factors.

So the greatest power of $2$ , or the greatest value of $k$ is $\boxed {38}$

Multiplication of Factorials

The answer is 38.

1 solution

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