For all integral values of n, the expression ((7^2n)-(3^3n)) is a multiple of:
Options: a) 10 b) 31 c) 12 d)22

• As this equation should be divisible by a number for all Integral values, Assume n=1
  (7^2)-(3^3) = 22
  so, it's safe to assume that the equation is divisible by 22, you can test for n=2, but that's actually not necessary as the options does not contain ambiguities.
  
  Ans - 22
• 7 years agoHelpfull: Yes(14) No(7)
• yes,if n=1 then by option verification d holds true
  i.e ans:22
• 7 years agoHelpfull: Yes(5) No(0)
• take n=1 then verify options
  Ans: 10
• 7 years agoHelpfull: Yes(3) No(23)
• (7^2)=49
  (3^3)=27
  49-27=22
• 7 years agoHelpfull: Yes(3) No(0)
• n=1
  the solution after substituting n value, we got is 22
  answer:d
• 7 years agoHelpfull: Yes(0) No(2)
• PUT n=1,we get 49-27=22
• 7 years agoHelpfull: Yes(0) No(0)
• 22...put value of n=1 and check from options
• 6 years agoHelpfull: Yes(0) No(0)
• Option (a)
• 6 years agoHelpfull: Yes(0) No(1)
• substitute n=1
  (7^2)-(3^3)
  49-27=22
  Answer=22
• 6 years agoHelpfull: Yes(0) No(0)
• option : D
• 5 years agoHelpfull: Yes(0) No(0)
• Answer is 22
• 5 years agoHelpfull: Yes(0) No(0)
• It is 22.
  Check the sum for the value of n=1, 2
  You will find it is always divisible by 22.
• 5 years agoHelpfull: Yes(0) No(0)
• for n=1,the ans is 22
• 5 years agoHelpfull: Yes(0) No(0)
• Modulo 2:
  
  $49 equiv 1 pmod{2}$ (since 49 is odd),
  $27 equiv 1 pmod{2}$ (since 27 is odd).
  Thus:
  $ 49^n equiv 1^n equiv 1 pmod{2}, $
  $ 27^n equiv 1^n equiv 1 pmod{2}, $
  $ 49^n - 27^n equiv 1 - 1 equiv 0 pmod{2}. $
  The expression is divisible by 2 for all $n$.
  
  
  Modulo 11:
  
  $49 = 7^2$, and $7 equiv -4 pmod{11}$ (since $7 + 4 = 11$).
  $7^2 equiv (-4)^2 equiv 16 equiv 5 pmod{11}$ (since $16 - 11 = 5$).
  $27 equiv 5 pmod{11}$ (since $27 - 2 cdot 11 = 27 - 22 = 5$).
  Thus:
  $ 49^n equiv 5^n pmod{11}, $
  $ 27^n equiv 5^n pmod{11}, $
  $ 49^n - 27^n equiv 5^n - 5^n equiv 0 pmod{11}. $
  The expression is divisible by 11 for all $n$.
  
  
  
  Since the expression is divisible by both 2 and 11, and $gcd(2, 11) = 1$, it is divisible by $2 cdot 11 = 22$.
• 2 Months agoHelpfull: Yes(0) No(0)