you are given number N.give 2*N has 28 factor and 3*N has 30 factor.calculate how many factor will be in 6*N..?

• 35 factors.
  N= 2^5 *3^3
  2N= 2^6 * 3^3 .. so factors are 7*4=28
  3N= 2^5 *3^4... factors are 6*5=30
  6N = 2^6*3^4.. factors are 7*5=35
• 12 years agoHelpfull: Yes(40) No(20)
• HI Manav.. You are considering only one case as there can be many cases because the Maximum limit of N is not given in question.For such type of questions the Maximum Limit should be given.
  
  It is told in question that 2*N has 28 factors.
  28 = 7 * 4 or 14*2 or 7 * 2 * 2
  Manav is considering the case of (7 * 4) only.
  
  For this type of questions always start like
  N = 2^p * 3^q * 5^r * ......so on(beacuse 2,3,5,7,11,13,17.. are prime factors)
  
  since the maximum limit is not given in question so it is not possible to give a single answer...
  
  For the question how 7*4 = 28 is coming in MANAV's Solution---
  Lets for an example u need to represent 48 in prime factorization--
  u will write - 48 = 2^4 * 3^1
  Now if u want to know how many factors does 48 has.. we need to do [(4+1) * (1+1)]= 10 factors total
  
  IN General N = 2^p * 3^q * 5^r * 7^s.....so on
  Total no of factors are = (p+1) * (q+1) * (r+1) * (s+1)......
  
  Thanks....
• 12 years agoHelpfull: Yes(30) No(6)
• same as MANAV.this is right answer.
• 12 years agoHelpfull: Yes(4) No(1)
• please explain how it become factor 30 or 28,please help me out!
• 12 years agoHelpfull: Yes(3) No(1)
• i understand the solution
  
• 12 years agoHelpfull: Yes(1) No(5)
• Hey Manav can u plz help me to get ur solutions .. i ma understing ... plz help me yr
• 10 years agoHelpfull: Yes(0) No(0)
• Firstly we know that any no. N can be expressed as N=(p1^a).(p2^b).(p3^c)......
  Here p1,p2,p3....are prime numbers, and a, b, c and its respective powers.
  
  The no. of factors of N is given by (a+1)(b+1)(c+1).....
  ----------------------------------------------------------------------------------------------------------
  Given, 2N has 28 factors and 3N has 30 factors.
  
  Therefore for 2N----->(a+1)(b+1)(c+1)..... =28
  all possible ways to do that is
  
  1) 28=(1+1)(1+1)(6+1) ,and if so it would mean 2N=(p1^1).(p2^1).(p3^6)
  or
  2) 28=(1+1)(13+1) would correspond to 2N=(p1^1).(p2^13)
  or
  3) 28=(3+1)(6+1) would correspond to 2N=(p1^3).(p2^6)
  
  now since 2N is a multiple of 2, hence any one of p1, p2 or p3 has to be 2(prime no.).This is obviously true for all three cases.
  ---------------------------------------------------------------------------------------------------------------
  Knowing this we can guess 2N to be the following,
  
  form case 1) 2N=(2^1).(p2^1).(p3^6) or 2N=(p1^1).(p2^1).(2^6)
  =>N=(p2^1).(p3^6) =>N=(p1^1).(p2^1).(2^5)
  
  no.of factors=(1+1)(6+1)=14 (1+1)(1+1)(5+1)=24
  ---------------------------------------------------------------------------------------------------------------
  form case 2) 2N=(2^1).(p2^13) or 2N=(p1^1).(2^13)
  =>N=(p2^13) =>N=(p1^1).(2^12)
  
  no.of factors=(13+1)=14 (1+1)(12+1)=26
  --------------------------------------------------------------------------------------------------------------
  form case 3) 2N=(2^3).(p2^6) or 2N=(p1^3).(2^6)
  =>N=(2^2).(p2^6) =>N=(p1^3).(2^5)
  
  no.of factors=(2+1)(6+1)=21 (3+1)(5+1)=24
  --------------------------------------------------------------------------------------------------------------
  -----------------------------------------------------------------------------------------------------------
  
  similarly for 3N----->(a+1)(b+1)(c+1)..... =30
  all possible ways to do that is
  
  1) 30=(1+1)(2+1)(4+1) ,and if so it would mean 2N=(p1^1).(p2^2).(p3^4)
  or
  2) 30=(4+1)(5+1) would correspond to 2N=(p1^4).(p2^5)
  or
  3) 30=(2+1)(9+1) would correspond to 2N=(p1^2).(p2^9)
  or
  4) 30=(1+1)(14+1) would correspond to 2N=(p1^1).(p2^14)
  
  now since 3N is a multiple of 3, hence any one of p1, p2 or p3 has to be 3(prime no.).This is obviously true for all four cases.
  ---------------------------------------------------------------------------------------------------------------
  Knowing this we can guess 3N to be the following,
  
  form case 1) 2N=(3^1).(p2^2).(p3^4) or 2N=(p1^1).(3^2).(p3^4) or 2N=(p1^1).(p2^2).(3^4)
  =>N=(p2^2).(p3^4) =>N=(p1^1).(3^1).(p3^4) =>N=(p1^1).(p2^2).(3^3)
  
  no.of factors=(2+1)(4+1)=15 (1+1)(1+1)(4+1)=20 (1+1)(2+1)(3+1)=24
  ---------------------------------------------------------------------------------------------------------------
  form case 2) 2N=(3^4).(p2^5) or 2N=(p1^4).(3^5)
  =>N=(3^3).(p2^5) =>N=(p1^4).(3^4)
  
  no.of factors=(3+1)(5+1)=24 (4+1)(4+1)=25
  --------------------------------------------------------------------------------------------------------------
  form case 3) 2N=(3^2).(p2^9) or 2N=(p1^2).(3^9)
  =>N=(3^1).(p2^6) =>N=(p1^2).(3^8)
  
  no.of factors=(1+1)(9+1)=20 (2+1)(8+1)=27
  ----------------------------------------------------------------------------------------------------------------
  form case 4) 2N=(3^1).(p2^14) or 2N=(p1^1).(3^14)
  =>N=(p2^14) =>N=(p1^1).(3^13)
  
  no.of factors=(14+1)=15 (1+1)(13+1)=28
  ---------------------------------------------------------------------------------------------------------------
  from all the above cases we see, N=(p1^3).(2^5) whose sum of factors is 24 and,
  N=(3^3).(p2^5) whose sum of factors is also 24
  
  hence, N=(3^3).(2^5)
  6N=(3^4).(2^6)
  sum of factors of 6N is (4+1)(6+1)=5*7=35.
  
• 8 years agoHelpfull: Yes(0) No(1)