Vasanth

• Aug 10th, 2005

 

381654729 
 
First of all, the fifth digit must be 5 for the number to be divisible by 5 
 
A) Digits in even numbered positions should be divisible by even numbers and hence should be one of 2,4,6,8 
B) As we have only four even numbers and four positions to fill, this leaves us with only 1,3,7,9 to fill the odd numbered positons 
C) If a number is divisible by 6, it is also divisible by 3 and as the number formed by first three digits is divisible by 3 and the number formed by first 6 digits is divisible by 6, the number formed with the second set of 3 digits ( i.e, with the 4th, 5th and 6th digit) should also be divisible by 3. 
D) If a number is divisble by 4 its last two digits are also divisible by 4. 
E) If a number is divisible by 8, it is also divisible by 4 
 
From A, B and C we can conclude that the number formed with the second set of 3 digits ( i.e, with the 4th, 5th and 6th digit) is one of 852, 654, 456 or 258. 
 
Using D, we can rule out 852 and 456 because the third digit being 1, 3, 7 or 9 ( from B ) and as the first four digits should be divisble by 4, any combination would not be divisible by 4 (possible combinations are: XX1852, XX3852, XX7852, XX9852, XX1456, XX3456, XX7456, XX9456) 
 
This should get one started and the number of possible combinations comes down to very manageable number. Applying the conditions it should not be tough to arrive at the final answer. 
 

rajani

• Jun 24th, 2006

 

hi,

  1st and 5th position numbers r fixed

   1_ _ _ 5_ _ _ _

    now in 2nd pos... 2,4,6,8 fit take 2try to fill the no.in rest of pos....

    it doesnt satisfy all the div rules

take 4 nby try   by trial n error the number is 147258369

dhanalakshmi02

 Profile Answers by dhanalakshmi02 

• Aug 16th, 2006

 

Hi Rajani

I think ur answer is wrong. check once again

Y Becoz  the no. upto 7 digits is not divisible by 7

The ans can be 147258963