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Divisibility test
Asked by prakharsrivastava | 08 May, 2010, 10:35: PM

2-the numer must be an even number

3-sum of digits must be divisible by 3

4-the number formed y the tens and units place must be divisible by 4

5- units place must be 0 or 5

11-take the sum of alternate digits. the difference of that sum is divisible by 1, then the num is divi by 11

e.g.

715

alt digits are 7,5 and 1

so 7+5=12

12-1=11 which is divi by 11 , hence 715 is divi by 11

for other nums take their factors if given num is divi by these factors, then, yes it'll be divi by the given number.e.g. for 6 you can chk if 2 and 3 are the factors of the given num, then the given num is divi by 6.

and so on.

for 7 you can subtract 2 times the last digit from the remaining part of the number.

e.g.

483

48-[2x 3]=42  which is divisible by 7 so, 483 is divi by 7

or

add 5 times the last digit two the remaining part of the num

48+[3 x 5]=63 which is divi y 7

if the hundreds digit is even examine the remaining part of the number whether it's divi by 7 or not

for 13-add 4 times the last digit to the rest and check

e.g.

533

53+[4 x3]=65 which is divi by 13, so given num is divi by 13

Answered by | 18 May, 2010, 12:49: PM
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