Factorial Product

Number Theory Level 3

Let $N = 1!\cdot 2! \cdot 3! \cdot 4! \ldots 9! \cdot 10!$ . Let $2^k$ be the largest power of $2$ that divides $N$ . What is the value of $k$ ?

Details and assumptions

$n! = n\times (n-1) \times (n-2) \ldots 2 \times 1$ .

The answer is 38.

1 solution

Arron Kau Staff
May 13, 2014

Clearly $2, 4, 6, 8, 10$ are the only numbers that contain a factor of $2$ . We break this up into the number of times each of $2, 4, 6, 8, 10$ appears in the expression: $N = 1\cdot (1 \times 2)\cdot (1 \times 2 \times 3) \ldots (1 \times 2 \times 3 \ldots \times 10)$ .

$2$ appears in $2!, 3!, \ldots, 9!, 10!$ , thus there are $9$ $2's$ . Since $2 = 2\cdot 1$ , thus it contributes $9$ factors of $2$ .

$4$ appears in $4!, 5!, \ldots, 9!, 10!$ , thus there are $7$ $4's$ . Since $4 = 2^2$ , thus it contributes $14$ factors of $2$ .

$6$ appears in $6!, 7!, \ldots, 9!, 10!$ , thus there are $5$ $6's$ . Since $6 = 2\cdot 3$ , thus it contributes $5$ factors of $2$ .

$8$ appears in $8!, 9!, 10!$ , thus there are $3$ $8's$ . Since $8 = 2^3$ , thus it contributes $9$ factors of $2$ .

$10$ appears in $10!$ , thus there is only one $10$ . Since $10 = 2\cdot 5$ , thus it contributes $1$ factor of $2$ .

Therefore there are $9 + 14 + 5 + 9 + 1 = 38$ factors of $2$ in $N$ . Hence $2 ^ { 38 }$ is the largest power of $2$ that divides $N$ .

You can also expand the factorial: $N=1!\cdot 2!\cdot 3!\cdot ....9!\cdot 10!$ which simplifies to $N={ 1 }^{ 10 }\cdot { 2 }^{ 9 }\cdot { 3 }^{ 8 }\cdot .....\cdot { 9 }^{ 2 }\cdot 10$

Prime factor the product and add up all the powers of $2$ will give us $2^{38}$ and some other numbers. Therefore $k=\boxed{38}$

William Isoroku - 6 years, 5 months ago

Factorial Product

The answer is 38.

1 solution

0 pending reports