Consider the number 378, 000. (a) Count the number of positive integral factors. (b) Of these factors, how many are even? (c) How many are divisible by both 3 and 7? (d) How many zeros are at the end of 378, 000!

Step-by-step explanation:

A)   2^3 * 3^3 * 5 ^ 2 * 7 ^ 1  

      Total factor of 378,000 including  itself and 1  is

         (3+1)*(3+1)*(2+1)*(1+1) = 4*4*3*2 = 96

B) Knowing its prime factorization we can solve this problem

   378,000 = 378* 100

                  = 2 *27*7*10*10

                  = 2*3*3*3*7*5*2*5*2

                  All the above are indivisible  by any number except itself and one, or are now prime numbers,from here we regroup

                 = 2*2*2*3*3*3*5*5*7

  This can be given as 2^3 * 3^3 * 5 ^ 2 * 7 ^ 1

                making the x= 0 for 2 ^ x

                it becomes 2^0 * 3 ^ 3 * 5 ^ 2 * 7 ^ 1

                 We want the numbers of factors of the above

Therefore (0 + 1 )( 3+1)(2+1)(1+1) = 1 *4*3*2 = 24

C)  From answer (B)   2^0 * 3 ^ 3 * 5 ^ 2 * 7 ^ 1 Is a factor of 378,000

   And from it  three the power of three and 7 are been added to know the exact time 3 and 7 can successfully divide 378,000

3+1 = 4

D) Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow

378,000 Have three trailing zeros

Number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula

+

Solution:

378,000!

+ =  75,600 + 15,120 = 90,720 trailing zeros

22 and 22

step-by-step explanation:

10 plus 12 equals 22.

12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.