A 10 digit number N has one 1, two 2's, three 3's and four 4's . Ia N a perfect square ?

1) Yes
2) No

• Yes
  in this case- _ _ _ _ _ _ _ _ 2 4
  it satisfied following condition :-
  ->N is even and divisible by 4.
  ->It not ending with 6,the ten's place is even.
• 11 years agoHelpfull: Yes(0) No(0)
• Two things one need to know to solve the above problem : (1) If any positive integer is divided by 9 the remainder is same when the its sum of its digits is divided by 9. (2) Any integer can be written in the following forms : 9k+r where k is some integer and the remainder r=0,1,2,3,4,5,6,7,8 which is same as saying that any integer can be written in the form : 9k + r where r = 0,1,2,3,4,-4,-3,-2,-1.
  Now if you square any integer it will be of the form : 81k^2+18kr+r^2 and if this number is divided by 9 the remainder will be same as when r^2 divided by 9 as the first two terms (81k^2 and 18 kr )are multiples of 9. Now the possible values of r^2 are 0,1,4,9,16 and if these values of r^2 are divided by 9 then the remainders are 0,1,4 and 7. So any perfect square when divided by 9 the possible remainders are 0,1,4,7. Now if we divided a number by 9 and get any remainders other than these four values we can clearly say it can not be a perfect square ( Note - the converse is not true ). Now in the problem it is given the no consists of One 1, Two 2s, Three 3s and Four 4s. The sum of the digits of the number, irrespective of the position of the digits is 1+2*2+3*3+4*4 = 30 . Now by Rule-(1) if the number is divided by 9 the remainder will be same as when 30 is divided by 9, which is 3. Therefore the number CAN NOT BE A PERFECT SQUARE by the Rule-(2) .
• 4 years agoHelpfull: Yes(0) No(0)