We will be learning more about how to check for divisibility.
Earlier, we learnt the divisibility rule by $2$ and $4$. Here we will learn the general divisibility rule by any number which is formed by multiplying any number of $2$'s.
Numbers like $2$, $4$ , $8$, $16$ can be formed by repeated multiplication of two.
$2 = 2$
$4 = 2 \times 2$
$8 = 2 \times 2 \times 2$
$16 = 2 \times 2 \times 2 \times 2$
(These are also called powers of two, which you will learn later).
Earlier we saw that if the last digit of a number is divisible by $2$ the number itself is divisible by $2$, likewise if the last two digits of a number is divisible by $4$, the number is divisible by $4$.
As a general rule, if a divisor is composed of only two's as factor:
If the divisor has $only\ 2's$ as factors, then the last as many digits of the dividend as there are $2$'s in the divisor, should be divisible by the divisor.
We take a few examples of this:
Is the number $198232$ divisible by $4$?
We know $4 = 2 \times 2$, so there are two $2$'s in the factor. We will check if the last two digits of the dividend is divisible by $4$.
The last two digits of $198232$ is $32$, which is divisible by $4$. Hence the number $198232$ is divisible by $4$.
Is the number $348914$ divisible by $4$? The last two digits of the given number is 14, which is not divisible by 4. Hence, the number $348914$ is not divisible by $4$.
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Is the number $2983128$ divisible by $8$?
We know that $8 = 2 \times 2 \times 2$, so there are three 2's in the divisor, and we need to check the divisibility of the last 3 digits of the given number, which is $128$.
Since $128$ is divisible by $8$, the number $2983128$ is divisible by $8$.
Is the number $3895218$ divisible by $8$?
The last three digits are $218$, which is not divisible by $8$. Therefore, $3895218$ is not divisible by $8$.
Let us look at the divisibility by any number that is formed by multiplying only $5$'s.
The numbers $5$, $25$, $125$, $625\ \ldots$ are formed by multiplying only $5$'s
$5 = 5$
$25 = 5 \times 5$
$125 = 5 \times 5 \times 5$
$625 = 5 \times 5 \times 5 \times 5$
The rule is exactly same as that of $5$. If the divisor has only $5$'s as factors, then check the last as many digits of the dividend as there are $5$'s in the dividend.
If the last digit is divisible by $5$, then the number is divisible by $5$.
If the last two digits of a number is divisible by $25$ then the number is divisible by $25$
If the last $3$ digits of a number is divisible by $125$ then the number is divisible by $125$
If the last $4$ digits of a number is divisible by $625$ then the number is divisible by $625$
Let us say a given number $N$ is divisible by $8$ and $12$ can we say for sure the number is divisible by $8 \times 12 = 96$? The answer is $\unicode{0x201C}No,\ we\ cannot! \unicode{0x201D}$. For example the number $120$ is divisible by both $8$ $(120 \div 8 = 5)$ and $12$ $(120 \div 12 = 10)$
What we can say for sure, is that the product is divisible by the $LCM$ of the two divisors. So if a number is divisible by $8$ and $24$ then the number $must\ be\ divisible$ by their $LCM$ which is $24$.
We also know that the $LCM$ of two co-prime numbers is their product.
Therefore, if a number $N$ is divisible by two numbers $A$ and $B$, where $A$ and $B$ are co-primes, then we can say for sure that $N$ is divisible by $A \times B$. But if $A$ and $B$ are not co-primes, then we cannot say for sure that $N$ is divisible by $A \times B$.
Let us take an example. The numbers $4$ and $9$ are co-prime numbers. Therefore, any number which is divisible by $4$ and $9$ is divisible by $36$