Some other divisibility tests — lesson. Mathematics CBSE, Class 8.

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Before this, we discussed divisibility tests by

$10$ ,

$5$ , and

$2$ .

What do you think is common among all three divisibility tests?

These three divisibility tests solely depended only on one-digit, that is, the last digit.

Hence, these three divisibility tests followed a similar procedure.

Let us now look at a comparatively, unique divisibility test.

Divisibility by

$3$ or

$9$

Consider the number

$4239$

The expanded form is

$(4 \times 1000) + (2 \times 100) + (3 \times 10) + 9$ .

$= [4 \times (999 + 1)] + [2 \times (99 + 1) + [3 \times (9 + 1)] + 9$

$= (4 \times 999) + (4 \times 1) + (2 \times 99) + (2 \times 1) + (3 \times 9) + (3 \times 1) + (9 \times 1)$

$= (4 \times 999) + (2 \times 99) + (3 \times 9) + (4 + 2 + 3 + 9)$

Now,

$4 + 2 + 3 + 9 = 18$ .

$18$ is both divisible by

$3$ and by

$9$ . Therefore,

$4239$ is also divisible by

$3$ or

$9$ .

Similarly, consider the number

$2148$

The expanded form is

$(2 \times 1000) + (1 \times 100 + (4 \times 10) + 8$ .

$= [2 \times (999 + 1)] + [1 \times (99 + 1) + [4 \times (9 + 1)] + 8$

$= (2 \times 999) + (2 \times 1) + (1 \times 99) + (1 \times 1) + (4 \times 9) + (4 \times 1) + (8 \times 1)$

$= (2 \times 999) + (1 \times 99) + (4 \times 9) + (2 + 1 + 4 + 8)$

Now,

$2 + 1 + 4 + 8 = 15$ .

$15$ is divisible by

$3$ but not by

$9$ . Therefore,

$2148$ is also divisible by

$3$ but not by

$9$ .

Did you find an error?

6. Some other divisibility tests