Can someone explain the steps to solve this problem?
If a number ends in zeros, the zeros are called terminal zeros. For example, 520000 has four terminal zeros, but 502,000 has just three terminal zeros. Let N equal the product of all natural numbers from 1 through 20: N=1x2x3x4x...x20.
How many terminal zeros will N have when it is written in standard form?
Thanks!
jennings
5 more than 2 times a number
-10=-1+3n
To find the number of terminal zeros in the product N when it is written in standard form, we need to determine the factors that contribute to these zeros. Terminal zeros occur because of the presence of factors of 10 in a number. Since 10 can be expressed as 2 x 5, we need to find the number of pairs of 2s and 5s in the prime factorization of N.
Here are the steps to solve this problem:
Step 1: Prime factorize the numbers from 1 to 20
Prime factorization of the numbers 1 to 20:
1 = 1
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 x 3
7 = 7
8 = 2^3
9 = 3^2
10 = 2 x 5
11 = 11
12 = 2^2 x 3
13 = 13
14 = 2 x 7
15 = 3 x 5
16 = 2^4
17 = 17
18 = 2 x 3^2
19 = 19
20 = 2^2 x 5
Step 2: Count the number of 2s and 5s in the prime factorization
We need to determine the number of 2s and 5s in the prime factorization of each number from 1 to 20.
Number of 2s:
- Each even number contributes at least one 2.
- Numbers like 4, 8, 16 have additional 2s.
Counting the number of 2s, we get: 1 + 1 + 1 + 2 + 1 + 1 + 1 + 3 + 2 + 1 + 1 + 2 + 1 + 1 + 1 + 4 + 1 + 1 + 1 + 2 = 38
Number of 5s:
- Each multiple of 5 contributes at least one 5.
- Numbers like 25 have an additional 5.
Counting the number of 5s, we get: 1 + 1 = 2
Step 3: Determine the number of pairs of 2s and 5s
Since each pair of 2s and 5s contributes a terminal zero, we need to determine the minimum number of pairs.
In this case, we have 2 5s and 38 2s. Since we only have 2 5s, we can only form 2 pairs. Therefore, the number of terminal zeros in N is 2.
So, when N is written in standard form, it will have 2 terminal zeros.