Some results on divisibility rule — lesson. Mathematics CBSE, Class 6.

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Result I: If a number can be divided by another number, then it can also be divided by each of its factor.

Example:

Lets take two number

$18$ and

$72$ .

$72\div18 = 4$ this show

$72$ is divisible by

$18$ ,

Factors of

$18 = 1, 2, 3, 6, 9, 18$ .

Now we divide the factors of

$18$ with

$72$ .

$72\div1=72$ ,

$72\div2=36$ ,

$72\div3=24$ ,

$72\div6=12$ ,

$72\div9=8$ and

$72\div18=4$ .

Thus,

$72$ is divisible by each of factors of

$18$ .

Result II: If a number is divided by two co-prime numbers, it is also divisible by its product.

Example:

Let's say

$90$ is divisible by

$5$ and

$9$ . As we know

$5$ and

$9$ are co-prime numbers.

Thus, the product of co-primes is (

$5\times9 = 45$ ).

$90\div45 = 2$ .

Therefore

$90$ is divisible by the product of co-primes

$45$ .

Result III: If two given numbers are divisible by a number, then, their sum is also divisible by that number.

Example:

Let us take

$21$ and

$18$ . Both the numbers are divisible by

$3$ .

$21\div3 = 7$ and

$18\div3 = 6$ .

Sum of the two numbers is

$21+18 = 39$ . Also

$39\div 3 = 13$ .

Therefore, if

$24$ and

$18$ are divisible by

$3$ , then their sum is

$39$ which is also divisible by

$3$ .

Result IV: If two given numbers are divisible by a number, then their difference is also divisible by that number.

Example:

Let us take

$58$ and

$54$ are divisible by

$2$ .

$958\div2 = 29$ and

$54\div2 = 27$ .

Difference of the two numbers that

$58-54 = 4$ and

$4\div2 = 2$ .

Therefore, if

$54$ and

$58$ are divisible by

$2$ , then their difference

$4$ is also divisible by

$2$ .

More examples:

1. A number is divisible by both

$7$ and

$15$ . By which other numbers will that number be always divisible?

Note that the number is divisible by

$7$ and

$15$ . Let us recall 'Result II'.

Result II: If a number is divided by two co-prime numbers, it is also divisible by its product.

Since

$7$ and

$15$ are co-prime numbers, the number must be divisible by the product

$7\times 15 = 105$ .

So, the given number will always be divisible by

$105$ .

2. A number is divisible by

$16$ . By what other numbers will that number be divisible?

As we need to find the more divisible for the given number, let us recall 'Result I'.

Result I: If a number can be divided by another number, then it can also be divided by each of its factor.

The number is divisible by

$16$ .

The factors of

$16$ are

$1, 2, 4, 8, 16$ .

Therefore, the number is also divisible by

$1, 2, 4, 8$ .

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4. Some results on divisibility rule

Theory: