Determine whether the statement is true or false. Explain or show why.
If a natural number is divisible by 7, then it must also be divisible by 21
Did you not like my answer I just gave you to the same question two postings ago ?
Sorry, didn't mean to re-post. Yes your answer helped me immensely! Thank you!
The statement is true. A natural number that is divisible by 7 will also be divisible by 21.
To understand why this is true, we need to understand the concept of divisibility. A number "a" is said to be divisible by another number "b" if dividing "a" by "b" leaves no remainder. In other words, "a" is a multiple of "b".
In this case, we are given that a natural number is divisible by 7. Let's call this number "n". Since "n" is divisible by 7, we can write it as 7k, where "k" is some other integer.
Now, we need to determine if "n" is also divisible by 21. Using the same logic, we can write 21 as 7 * 3. So, if "n" is divisible by 7, it can be written as 7k, and if it is divisible by 21, it can be written as 21m, where "m" is some other integer.
To show that "n" is divisible by 21, we can substitute the value of "n" we found earlier (n = 7k) into the expression for divisibility by 21:
21m = 7k
Now, we can simplify the equation:
3m = k
Since "k" is an integer, we can see that "3m" must also be an integer. Therefore, if "n" is divisible by 7 (meaning it can be written as 7k), it is also divisible by 21 (meaning it can be written as 21m).
Hence, the statement is true. A natural number that is divisible by 7 will also be divisible by 21.