How many terminal zeroes will 1 * 2 *3... *50 have when it is written in standard form?
I would just write the list out if I could not guess :)
You could shorten the task as so:
1 ----- 10
11 -----20
21 -----30
31 .....40
41 .....50 :-)>
.
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The "*" are actually a multiplication sign. Please help! :(
Oh, 50 factorial
50 * 49 * 48 ..... 1
3.0414093e+64
so
64
Just GOOGLE
50 factorial
To find the number of terminal zeroes in the standard form of a factorial, we need to determine the number of times 10 divides the factorial.
The number 10 can be expressed as 2 * 5. So, essentially, we need to count the number of times 2 and 5 appear as factors in the factorial because the number of 2's will always be more than the number of 5's in the factorial.
To calculate the number of terminal zeroes, we need to find the power of 5 in the prime factorization of each number from 1 to 50.
Let's start by determining the number of multiples of 5 from 1 to 50:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50
There are 10 multiples of 5.
However, multiples of 25 contain two factors of 5:
25, 50
So, we add one more factor of 5 for each multiple of 25, totalling 2.
Finally, multiples of 125 contain three factors of 5:
50
Thus, we add one more factor of 5 for the multiple of 125, totalling 3.
Now, we can add up the count of factors of 5:
10 + 2 + 1 = 13
Therefore, the standard form of 1 * 2 * 3 * ... * 50 will have 13 terminal zeroes.