Find numbers of n for which (100^n- (112^2)) divisible by 3?

• The min value of n have to be 3... So have to be => 3...
• 10 years agoHelpfull: Yes(15) No(2)
• least value be 3
  expl=when n=3
  x=1000000
  112^2=12544
  so y=1000000-12544 = 987456
  divisiblity rule for 3 is 9+8+7+4+5+6=39
  987456 is divisible be 3
• 10 years agoHelpfull: Yes(9) No(3)
• Least value will be n=6
  let x = ((100^n)-(112^2))
  As 1oo^6 = 1000000000000
  And 112^2 = 12544
  therefore
  x = 1000000000000 - 12544
  => x =999999987546
  now to check whether number is divisible by 3 , check whether summation of its digits is divisible by 3 or not
  as 9+9+9+9+9+9+9+8+7+5+4+6 is equal to 93
  so, 999999987546 is divisible by 3.
  
  like wise for each n>=6,
  number will be divisible by 3
  but least value will be 6.
  
  
  
  
  After subtraction
• 10 years agoHelpfull: Yes(4) No(2)
• 2,3,4,5,6,7,8,9,........................(from 2 all positive integer)
  
• 10 years agoHelpfull: Yes(3) No(4)
• all positive numbers...1,2,3.........
• 10 years agoHelpfull: Yes(3) No(8)
• n value should be 3 and above.......as we subtract 1000000-12544...only then we get positive answer....
  
• 10 years agoHelpfull: Yes(3) No(0)
• nd 1 more conditon n
• 10 years agoHelpfull: Yes(1) No(1)
• n is greater than 6 (n>6)
  options;-
  a. 6
  b.17
  c.16
  d.8
• 10 years agoHelpfull: Yes(1) No(0)
• If n=1 100^1/3=Reminder is 3
  if 3/3=1
  then 1-(112^2)/3=1-(1/3)
  =1-1=0
  IF n=2 then
  (100^2-(112^2)=(1^2)-(1^2)=1-1=0
  
  Then we can clearly say then N =Any Positive Integer number.....
  Ans:=1,2,3,..............................................+Infinity
• 10 years agoHelpfull: Yes(1) No(0)
• hey how to post a question here in e litmus page of m4 maths
• 10 years agoHelpfull: Yes(0) No(0)
• Given n
• 10 years agoHelpfull: Yes(0) No(0)
• Given n
• 10 years agoHelpfull: Yes(0) No(0)
• here question is 100^n-112^2 is positive and n
• 10 years agoHelpfull: Yes(0) No(0)
• this question is incomplete ....
• 10 years agoHelpfull: Yes(0) No(0)
• for all positive numbers from 1,2,3...............of n
• 10 years agoHelpfull: Yes(0) No(0)
• n can be anything....because
  consider it (100)^n/3-112^2/3 then
  (99+1)^n/3 gives modules 1
  112^2/3 gives modules 1
  so 1^n-1 wil gives 0...
  so n value can be any postive integer
• 10 years agoHelpfull: Yes(0) No(0)
• the condition will satisfy for all n>=3
• 10 years agoHelpfull: Yes(0) No(1)
• 112^2=12544 100^3=1000000-12544=987456 add all digit 9+8+7+4+5+6=39 and 39 is divisible by 3 so n=3
• 9 years agoHelpfull: Yes(0) No(0)
• any no greater or equal to 1 here it is not given in question that number should positive.
• 9 years agoHelpfull: Yes(0) No(0)
• It is not necessary for the result of subtraction to be +ve . Even -ve nos are divisible by 3
  So for n = 1
  100 - 12544 = 12444 is divisible by 3
  Hence for n = 1,2 3, 4,.....
• 9 years agoHelpfull: Yes(0) No(0)
• for all values of n=0,1,2,3,...
  because its not mentioned that the answer must be +ve
• 9 years agoHelpfull: Yes(0) No(0)
• given number = 3
  let n=3
  then 100 pow 3 = 1000000
  x=1000000
  112 pow 3 = 12544
  now
  x= 1000000-12544
  we get 987456
  now divisibility rule of 3
  9+8+7+4+5+6=39 and 39 vl be divisible by 3
  so 987456 will be divisible by 3
  
• 9 years agoHelpfull: Yes(0) No(0)
• n=3 becoz 112^2=12544 and 1000000-12544=987456 which is divisible to 3
• 8 years agoHelpfull: Yes(0) No(0)
• 112^2/3 will give remainder 1 ad 100/3 remainder 1/3 so 1^n/3 always gives remainder 1 so
  1-1=0 n=1,2,3... but we cant take 1 and 2 because i ll become negative thats y
  n>=3
• 7 years agoHelpfull: Yes(0) No(0)