The number of positive integers which divide 2004 to leave a remainder of 24 ?

• Since 2004 - 36 = 1980, any number which divides 1980 but does not divide 24 will leave a remainder of 24. Since 1980 = 2^2*3^2*5*11, there are 36 divisors of 1980; of those, 1, 2, 3, 4, 6, and 12 also divide 24, so there are 36 - 6 = 30 numbers which leave a remainder of 24 when dividing 2004
• 12 years agoHelpfull: Yes(4) No(2)
• 22
  
  2004 -24 = 1980
  1980 = 2^2*3^2*5*11
  
  there are 36 divisors of 1980.
  out of which 1, 2, 3 ,4 ,5 ,6, 9, 10, 11, 12, 15, 18, 20, 22 are less than 24.
  so they can not leave remainder more than 24.
  
  so 36-14=22 integers are there which divide 2004 to leave a remainder of 24.
• 12 years agoHelpfull: Yes(3) No(0)
• the answer is 22. we know a number a=(x1^(k1))*(x2^(k2))*.....has (k1+1)*(k2+2)...factors. here 1980=2^2*3^2*11*5. therefore 3*3*2*2=36 factors. but we need to take into account those greater than 24. 1 2 3 4 5 6 9 10 11 12 15 18 20 22 are
• 12 years agoHelpfull: Yes(1) No(3)
• AS PER PRINCIPLE OF DIVISION,
  2004=Q*D+24
  =>Q*D=1980
  WE HAVE TO FIND NUMBER OF FACTORS FOR IT AND THAT IS ANSWER:
  36
  1980=2*2*3*3*5*11
  WE HAVE TO LOOK FOR DIFFERENT COMBINATIONS OUT OF IT........
  WE HAVE TWO 2'S,TWO 3'S,ONE 5 AND 11.
  NUMBER OF COMBINATIONS=(2+1)(2+1)(1+1)(1+1)-1=35
  1 IS ALSO FACTOR,SO ANSWER IS 36
• 12 years agoHelpfull: Yes(0) No(1)