Let N=1!⋅2!⋅3!⋅4!…9!⋅10!. Let 2^k be the largest power of 2 that divides N. What is the value of k?
Count how many 2's there are in each of the factorials,
1! = none
2! = 1
3! = 1
4! = 2*4 => 3
5! = 2*4 => 3
6! = 2*4*2 => 4
7! ...
10!=2*4*8*2 => 8
Add all the numbers on the right to get N.
38
To find the value of k, we need to determine the exponent of 2 in the prime factorization of N. Let's break down the expression N = 1!⋅2!⋅3!⋅4!…9!⋅10!.
First, let's calculate the prime factorization of each factorial term individually:
1! = 1
2! = 2
3! = 3⋅2
4! = 4⋅3⋅2
...
9! = 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2
10! = 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2
Now, let's count the number of factors of 2 in each factorial term:
1! has 0 factors of 2.
2! has 1 factor of 2.
3! has 1 factor of 2.
4! has 3 factors of 2.
...
9! has 7 factors of 2.
10! has 8 factors of 2.
Now, let's sum up the number of factors of 2 in each factorial term:
0 + 1 + 1 + 3 + ... + 7 + 8 = 1 + 1 + 3 + ... + 7 + 8 = 27.
Therefore, the largest power of 2 that divides N is 2^k = 2^27.
So the value of k is 27.