If N = 82pow3 - 62pow3 -
20pow3 then N is divisible by:
a) 31 and 41
b) 13 and 67
c) 17 and 7
d) none

• a) 31 and 41
  
  value of N is = 82^3 - 62^3 -20^3 = 3*82*62*20 [a+b+c=0 => a^3+b^3+c^3=3abc]
  
  = 3*(2*41)*(2*31)*20 divisible by both 41 & 31.
• 8 years agoHelpfull: Yes(36) No(1)
• L.C.M of ( 82, 62, 20) = 20*41*31
  now,
  from take the LCM of options one by one
  i.e, a) 31*41 b) 13*67 c) 17*7
  now , divide 20*41*31 by options
  so you will get option (A) that is dividing perfectly
• 8 years agoHelpfull: Yes(7) No(0)
• N = 2^3 ( 41^3 - 31^3 - 10^3 )
  N = 8 ( ( 41-31 ) ( 41^2 + 31^2 + 41*31 ) - 10^3 )
  N = 80 ( 1681 + 961 + 1271 - 100 )
  N = 80 ( 3813 )
  N = 80 ( 41 ) ( 93 )
  N = 80 "(41) (31)" 3
  Option a)
• 8 years agoHelpfull: Yes(5) No(0)
• STEP1
  Take 2^3 common from the given equation
  Hence, we'll get as:-
  2^3 {(41)^3 - (31)^3 - (10)^3}
  
  Step 2:-
  Now since from the basic arithmetic we know that : a^3 - b^3 = (a-b)*{(a)^2 + (b)^2 + a*b)}
  Hence just for the sake of ease of the following question I am considering: a=41 & b=31.
  So, we'll get 2^3* [(a-b)*{(a)^2 + (b)^2 + ab}]
  
  Step 3.
  Substitute the value accordingly and by solving it further we'll get the following:
  2^3*[{41-31}*{(41)^2 + (31)^2 + 41*31} - 10^3] which will in turn become (after taking 10 as common) as,
  8*10[{a^2 + b^2 + ab} - 10^2]
  
  Step 4.
  8*10[{a^2 + b^2 + 2ab - ab} - 10^2]
  80[{a + b}^2 - ab - 10^2 }]
  
  Step 5.
  Now finally substitute the values of a and b as 41 and 31 respectively .
  Finally you will land upon the equation as follows:
  2^3 *10 [ {(41)^2 + (31)^2 + 2*41*31} - {41*31} - {10}^2]
  8*10[{41 + 31}^2 - 10^2 - 41*31]
  80[ {62 * 82} - {41 * 31}]
  80[ 2{31 * 41} - {41*31}]
  80 * (41 * 31) [2 - 1]
  80*(41*31)
  this will be the outcome of the initial equation which is divisible by both 31 & 41.
  
  HOPE YOU GOT IT :)
  
• 8 years agoHelpfull: Yes(2) No(4)
• a). 31 and 41
  
  82^3-(62^3)-(20^3) => We can simplify this as
  
  (62=20)^3- (62^3)-(20^3)
  We know the formula that- (a + b)3 = a3 +b3 +3ab(a +b)
  =>(a + b)3 -( a3 +b3) = 3ab(a +b)
  that is 3*20*62*(20+62)
  =3*20*62*82
  
  
  by seeing we can say that is devisible by 31 and 41. As it has 62 and 82 in the multiplication.
  
  Hope this is simpler to understand.
• 8 years agoHelpfull: Yes(0) No(0)
• a) 31 and 41
  
  Value of N is = 82^3 - 62^3 -20^3 = 3*82*(-62)*(-20)
  because
  if a+b+c=0 then a^3+b^3+c^3=3abc
  
  = 3*(2*41)*(2*31)*20 divisible by both 41 & 31.
• 7 years agoHelpfull: Yes(0) No(0)