2^3-62^3-20^3 is divisible by

• lets first consider about unit digit of base....
  2^3=8
  62^3 :unit digit of 62 is 2,so 2^3=8
  20^3 :unit digit of 20 is 0,so 0^3=0
  now 8-8-0=0
  unit digit is 0
  so the number is divisible by 5 & 10
• 11 years agoHelpfull: Yes(36) No(2)
• options because it is divisible by multiple numbers..?
  
• 11 years agoHelpfull: Yes(12) No(0)
• 2^3(1-31^3-10^3)ie. = 8 * a no. having 0 at last
  therefore divisible by no. 2, 4, 5, 8, 10
• 11 years agoHelpfull: Yes(11) No(1)
• 2^3{1-31^3+10^3} so, div by 8
• 11 years agoHelpfull: Yes(5) No(4)
• 2^3=8
  62^3 :unit digit of 62 is 2,so 2^3=8
  20^3 :unit digit of 20 is 0,so 0^3=0
  now 8-8-0=0
  unit digit is 0
  so the o. isdivisible by 5 & 10
  
• 11 years agoHelpfull: Yes(1) No(1)
• 1/2*10*20*sinc=80(1/2absinc,area)
  sinc=4/5;cosc=3/5
  3/5=(10^2+20^2-c^2)/2*10*20;c=(260)^1/2;(cosine rule)
• 11 years agoHelpfull: Yes(0) No(3)