how to find the divisibles of 3,4,5,7,9,11 out of lagre digit numbers - get me the shortcut?

Question

out of 5-7 digits numbers, what is the shortcut way to fine whether they are the divisibles of 3,4,5,7,11. i.e., that numbers should get remainder 0 if it is divided by the above said numbers(3,4,5,7,9,11)

Answer 1

The number is divisible by 3 if the sum of it's digits is divisible by 3

The number is divisible by 4 if the number acquired by it's last two digits is divisible by 4

The number is divisible by 5 if it ends with 5 or 0

The number is divisible by 7 if in its representation
an*10^n + ...+ a1*10 + a0 the number defined with
an*3^n + ... + a1*3 + a0 is divisible by 7

The number is divisible by 9 if the sum of it's digits is divisible by 9

The number is divisible by 11 if the alternating sum of it's digits is divisible by 11 (alternating sum means that for number 37846 the alternating sum is 3 - 7 + 8 - 4 + 6)

Answer 2

As far as I know:

3 - the last number is divisible by 3

5 - must end in a 5 or 0

I ran a test with the answer for 9, that it works by adding up all of the numbers and the result would be evenly divisible by 9. It worked with 754,569. This doesn't mean that it would work all of the time, but it's worth a try. Oh, and anything divisible by 9 will also be divisible by 3, but the reverse will not be true.

I am unaware of anything for 4,7,6,8, or 11. Sorry about that.

Hope the rest helps.

Answer 3

For 3:- sum of digits of number should be divisible by 3......
Suppose number is 123456789
Thus 1+2+3+4+5+6+7+8+9=45
Again 4+5=9 which is divisible by 3.....thus 123456789 is divisible by 3......................

For 4:-Last two digits taken together should be divisible by 4......
Suppose number is 1234567816.....Last 2 digits are '16' which is divisible by 4 thus number is also divisible..........

For 5:-If last digit is '5' or '0' then its divisible by 5..........

For 7:- there is no such rule for 7...............

For 9:-If the sum of the digits is divisible by 9 then number is divisible by 9......
Suppose number is 123456789
Thus 1+2+3+4+5+6+7+8+9=45
Again 4+5=9 which is divisible by 9.....thus 123456789 is divisible by 9......................

For 11:-If sum of the digits at odd plces=sum of digits at even places then the number is divisible by 11........
Suppose number is 1234567895.......
Sum of digits at odd places=1+3+5+7+9=25
Sum of digits at even places=2+4+6+8+5=25
Thus 1234567895 is divisible by 11............;-)

Answer 4

Forget the numbers and try polymath [in case you are aware]. You seem to be asking too much about integers.THAT RIDDLE [USED FOR DATA PROTECTION] HAS BEEN CRACKED.I would seriouslysuggst that you should first, understand `indivisibility paradox'[if you can access Pascal or Keppler]and then see whether you can expect answers about certain types of question.Incidently there is no single answer to either `shortcuts' or `longcuts'.

Answer 5

it will be 4620