How many 5-digit numbers are there such that the digits at hundred place, unit place and ten-thousand place form a gp? ans =3300 how?

• [Swarnabha Dey]
  if the digits of unit places,hundred places and ten places have to form a gp, Then it should consist of 111,124,421,139,931,248,842,222,333,444,555,666,777,888,999,000.
  now its given that the number is a five f=digit number so assuming 111, the 1st two digits can be filled in 9*10 = 90 ways,
  so 16 possible combinations are there ,total number of combinations will be = (16*90) =1440
• 10 years agoHelpfull: Yes(1) No(1)
• if the digits of unit places,hundred places and ten places have to form a gp, Then it should consist of 111,124,421,139,931,248,842 .
  now its given that the number is a five f=digit number so assuming 111, the 1st two digits can be filled in 9*10 = 90 ways,
  so 7 possible combinations are there ,total number of combinations will be = (7*90) =630
  
• 10 years agoHelpfull: Yes(0) No(2)
• 73300
  Let hundred unit and ten thousand no be a,ar,ar2 then
  If a=1 then r has 2 possible value 2,3 and no formed=400
  If r=1 then a cn hv 9 values so nos possible=72900
  Total =73300
• 10 years agoHelpfull: Yes(0) No(0)
• let the number be _ _ _ _ _
  
  if the number at ten-thousands place is:-
  
  1 then the other digits can be (1,1),(2,4),(3,9),(4,2),(9,3)----5
  
  2............(2,2),(4,8),(8,4)----------------------------------3
  
  3............(3,3)..............................................1
  
  4............(4,4),(2,1),(1,2)..................................3
  
  5............(5,5)..............................................1
  
  6............(6,6)..............................................1
  
  7............(7,7)..............................................1
  
  8............(8,8),(4,2),(2,4)..................................3
  
  9.............(9,9),(3,1),(1,3).................................3
  
  hence total no of distinct numbers are 21
  
  now for each of these 21 numbers there are 10 digits at tens place
  
  and 10 digits at thousands place so total answer is 21*10*10=2100(ans)
• 8 years agoHelpfull: Yes(0) No(0)
• abcde be the five digit number,
  for terms to be in gp ,middle term^2 = product of other two terms
  (1,2,4) will give 6 solutions
  (1,3,9)-6 solutions
  (2,4,8)-6 solutions
  (4,6,8)- solutions
  these 6 cases are for 0nly 3 terms that are in GP for rest of teh two places any of the 10 possible numbers can fit
  so total possible combinations = (6*4*100) = 2400
  now (1,1,1)(2,2,2)....(9,9,9) another 900 cases
  total combination = 2400+900 =3300
• 6 years agoHelpfull: Yes(0) No(0)