Consider a number , for example , 12345 . Then if someone ask you a question that , will this number divisible by 5. You will reply YES. Because its a small number , and the unit place is '5'. Hence you are sure that this number is divisible by 5.
Consider a 99 digit number created by writing side by side, the first fifty four natural numbers as follows
"12345678910111213141516171819.........5360" . Will this number divisible by 8?????
Is it possible to find even using calc?
Hence some concepts are important in solving this divisibility question.
If the number is divisible
by 2 - Unit place must be any of the digit 2, 4, 6, 8, 0
by 3 - Sum of the digits must be divisible by 3.
by 4 - Number formed by last two digits should be divisible by 4.
by 5 - Unit place must be either 5 or 0.
by 6 - should satisfy the divisibility test of 2 and 3.
by 7 - Take the last digit. double it . subtract from rest of the digit. ( e.g 427 -> last digit '7' double it , 14. (42 - 14 = 28 ) divisible by 4. Hence 427 is divisible by 7.
by 8 - Number formed by last 3 digits should be divisible by 8.
by 9 - Sum of its digits must be divisible by 9.
by 11 - Difference of sum of its digits at odd places and sum of its digits at even places is either '0' or divisible by '11' .
For testing , let us take the above question as example :
("123456789101112131415161718........5360" ) / 8 = ?
As per the above conditions , the number to be divisible by 8 has the last 3 digits should be divisible by 8. ( 360 / 8 ) = 45 . Thus the number is divisible by 8.
Take another example , a number ( 12345678 ) will this number divisible by 4 ?
Take last two digits - '78', (78 / 4) =19.5 , Hence it is not divisible by 4. It leaves reminder 2.
So 12345678+2 =12345680.
Now last two digits is '80' .It is divisible by 4. Hence 12345680 is divisible by 4 . (12345680/4=3086420) .
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Categories:
Aptitude
