Question

Write divisibility rule of 7 and 13.
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Answer 1

Answer:

Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.

Step-by-step explanation:

Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.

Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.

Answer 2

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➪Divisibility rule of 7⇒ RD(Remaining digits) - 2LD(last digit)

Example:- 343 (3 is the last digit and 34 will be the remaining digit)

=34-2×3

=34-6

=28 is divisible by 7 so

➪ 343 is also divisible by 7..

➪Divisibility rule of 13⇒RD+ 4(LD)

Example:- 169(9 is the last digit and 16 will be the remaining digit)

=16+9×4

=25×4

=120 (0 is the last digit and 12 will be the remaining digit)

=12×0

=0

➪ 169 is divisible by 13