If a number ends in zeroes, the zeroes are called terminal zeroes.
For example, 520,000 has four terminal zeroes, but 502,000 has just 3. Let X equal the product of all natural numbers from 1 to 20.
N = 1 * 2 * 3... *20.
How many terminal zeroes will N have when it is written in standard form?
Scratch that. Never mind.
To determine the number of terminal zeroes in the product N, we need to find the number of factors of 10 that appear in the prime factorization of N.
10 can be represented as 2 * 5. Since 2 is more abundant in the set of natural numbers from 1 to 20, we only need to count the number of factors of 5.
To find this, we can divide each number from 1 to 20 by 5 and count the number of times the division is exact. However, this method will miss numbers like 25, where 5 appears more than once as a factor.
So, let's consider those numbers as well. We need to divide each number from 1 to 20 by 5^2 = 25 and count the number of times the division is exact.
Let's do the calculations step-by-step:
1. Count the number of times 5 divides the numbers 1 to 20:
- 5 ÷ 5 = 1 (exact division)
- 10 ÷ 5 = 2 (exact division)
- 15 ÷ 5 = 3 (exact division)
- 20 ÷ 5 = 4 (exact division)
2. Count the number of times 25 divides the numbers 1 to 20:
- 25 ÷ 25 = 1 (exact division)
Adding up the number of exact divisions, we find that there are 4 + 1 = 5 factors of 5 in the prime factorization of N.
Therefore, N will have 5 terminal zeroes when it is written in standard form.
To determine the number of terminal zeroes in N, we need to identify the number of times N is divisible by 10. Since 10 is a product of 2 and 5, we need to find the number of times N is divisible by both 2 and 5.
Let's analyze the powers of 2 and 5 in the prime factorization of all the numbers from 1 to 20:
1 = 2^0 * 5^0
2 = 2^1 * 5^0
3 = 2^0 * 5^0
4 = 2^2 * 5^0
5 = 2^0 * 5^1
6 = 2^1 * 5^0
7 = 2^0 * 5^0
8 = 2^3 * 5^0
9 = 2^0 * 5^0
10 = 2^1 * 5^1
11 = 2^0 * 5^0
12 = 2^2 * 5^0
13 = 2^0 * 5^0
14 = 2^1 * 5^0
15 = 2^0 * 5^1
16 = 2^4 * 5^0
17 = 2^0 * 5^0
18 = 2^1 * 5^0
19 = 2^0 * 5^0
20 = 2^2 * 5^1
We can observe that the exponent of 2 is greater than the exponent of 5 in almost all cases. Therefore, the number of times N is divisible by both 2 and 5 is determined by the exponent of 5 in the prime factorization of N.
To find the exponent of 5 in the prime factorization of N, we need to count the number of times 5 appears as a factor. In the numbers from 1 to 20, we have 5, 10, 15, and 20 which have a factor of 5. This means there are 4 factors of 5 in the prime factorization of N.
Hence, N will have 4 terminal zeroes when written in standard form.