If a natural number is divisible by 7, then it must also be divisible by 21.
not true
e.g. 35 is divisible by 7 but not by 21
But ....
"If a natural number is divisible by21 , then it must also be divisible by 7" is true.
Can you see the logic behind the last statement ?
To determine if a natural number is divisible by 7, we need to check if the sum of the odd-placed digits minus the sum of the even-placed digits is a multiple of 7 or the number itself is divisible by 7.
However, to determine if a natural number is divisible by 21, we need to check if the sum of the digits is a multiple of 3 and if the number itself is divisible by 7.
Therefore, the statement "If a natural number is divisible by 7, then it must also be divisible by 21" is not true. A number can be divisible by 7 but not by 21.
To understand why a natural number that is divisible by 7 must also be divisible by 21, we need to consider the properties of divisibility.
A number is said to be divisible by another number if it can be divided evenly without leaving a remainder. In this case, we are considering divisibility by 7 and 21.
Let's assume we have a natural number, n, that is divisible by 7. This means that n can be written as 7 multiplied by some integer k. We can express this as n = 7k.
Now, we want to show that if n is divisible by 7, it is also divisible by 21. To do this, we need to prove that n can be expressed as 21 multiplied by an integer.
Substituting the value of n from the previous equation into this equation, we get 7k = 21m, where m is some integer we want to find.
We can simplify this equation by dividing both sides by 7: k = 3m.
Now, we see that k is a multiple of 3, and we want to find a value of m such that 3m is equal to k.
Since k is a multiple of 3 (k = 3m), and we have already established that n = 7k, we can rewrite the equation as n = 7(3m).
Simplifying further, we get n = 21m, where m is an integer.
This equation tells us that if n is divisible by 7 (since n = 7k), it is also divisible by 21 (since n = 21m).
Therefore, we can conclude that if a natural number is divisible by 7, it must also be divisible by 21.