How many three digit no's are between 100-500,where sum of two digit is 3rd?
A)24 b)not rem. C) not rem d)84

• keeping in mind the range 100-500, we can form different sets.
  first set of nos: (101, 202, 303, 404). frm each set 2 3 digit nos can be formed. ....so total of 8 nos.
  2nd set of nos: { 112 ( 3 nos), 123 ( 6 nos), 134 (6 nos), 145(4 nos only as 541 and 514>500), 156(2 nos only, 156 nd 165), 167( 2 nos),178( 2 nos),189( 2 nos)}..... so total of 27 nos
  second set of nos: {224(3 nos),235(4 nos), 246(4 nos), 257(2 nos), 268( 2 nos),179( 2 nos)}.... total of 17 nos.
  third set of nos: {336( 2 nos),347( 4 nos),358( 2 nos),369( 2 nos)}... total of 10 nos.
  fourth set of nos:{448( 2 nos),459( 2 nos)}.... total of 4 nos.
  Adding all we find that there are total of 66 three digit nos between 100-500 are possible a/c to given conditions.
• 9 years agoHelpfull: Yes(40) No(7)
• For 100 - 500 following are the possible permutations
  
  1) 100-200
  101 - Keeping HUNDRED'S DIGIT FIXED this can be rewritten as 101 or 110 i.e = 2 permutations
  112 - Similarly 112 and 121 again i.e = 2 permutations
  123 - again 2 permutations
  134 - again 2 permutations
  145 - again 2 permutations
  156 - again 2 permutations
  167 - again 2 permutations
  178 - again 2 permutations
  189 - again 2 permutations
  TOTAL = 18 permutations between 100-200
  
  2) 200-300
  Similarly
  201 ---> 201 210 i.e 2 permutations
  213 - 2 permutations
  224 - 2 permutations
  235 - 2 permutations
  246 - 2 permutations
  257 - 2 permutations
  268 - 2 permutations
  279 - 2 permutations
  
  total of 16 permutations between 200-300
  
  3) 300 - 400
  
  303 - 2 permutations ( 303 and 330 )
  314 - 2 permutations
  325 - 2 permutations
  336 - 2 permutations
  347 - 2 permutations
  358 - 2 permutations
  369 - 2 permutations
  
  total = 14 between 300-400
  
  4) 400-500
  
  404 - 2 permutations
  415 - 2 permutations
  426 - 2 permutations
  437 - 2 permutations
  448 - 2 permutations
  459 - 2 permutations
  
  total = 12 between 400-500
  
  TOTAL PERMUTATIONS POSSIBLE = 18+16+14+12 = 60 Answer.
• 8 years agoHelpfull: Yes(11) No(9)
• @AAkash Chandnani....bro between 200-300, u forgot 211 (2 permutations), between 300-400, you forgot 312 (2 permutations) and between 400-500, you forgot 413 (2 permutations). So in all, you missed 6 permutations and thus answer will be 66. Otherwise, ,method of solution provided by you was awesome. :) :)
• 8 years agoHelpfull: Yes(7) No(0)
• total of 66 three digit nos between 100-500 are possible a/c to given conditions.
• 9 years agoHelpfull: Yes(4) No(4)
• 101 202 303 404
  112 213 314 415
  123 224 325 426
  134 235 336 437
  145 246 347 448
  156 257 358 459
  167 268 369 (6set*2permutation each=12)
  178 279 (7set*2permutation each=14)
  189 (8set*2permutation each=16)
  (9set *2 premutation each=18)
  
  (special cases):
  200-300: 211 also fit the condition but permutation will be 1
  300-400: 312 also fit the condition but permutation will be 2
  400-500: 413 also fit the condition but permutation will be 2
  &422 also fit the condition but permutation will be 1 =6
  so total ways will be(18+16+14+12+6=66)
  answer will be 66 ways
• 8 years agoHelpfull: Yes(4) No(0)
• Total 65 numbers are beteen 100 and 500.
  Solution :
  First fine the numbers having sum equal to 3rd number :
  Example : 101 => 1+0 =1
  112 =>1+1 =2
  123 =>1+2 = 3
  100 to 200 total 8 numbers && 200 to 300 total 8 numbers && 300 to 400 total 7 numbers && 400 to 500 total 6 numbers.
  So total 8+8+7+6 = 29
  
  Now ,
  Find the number having sum equal to middile number :
  Example
  121 => 1+1 =2,
  132 => 1+2=3,
  etc...
  
  100 to 200 total 8 numbers && 200 to 300 total 7 numbers && 300 to 400 total 6 numbers && 400 to 500 total 5 numbers.
  So total 8+7+6+5 = 26
  
  Now
  find the numbers having sum equal to first number...
  Example :
  
  110 => 1+0 =1,
  211=> 1=1 = 2,
  220=>2+0 =2,
  etc
  these are 10 numbers.
  
  So toal = 29+26+10 = 65
• 9 years agoHelpfull: Yes(2) No(6)
• given that the sum of two digits should be equal to the third digit,ie the sum of all the digits is equal to double the 3rd digit.
  let us consider the numbers from 100 to 199 : assume that the number is 1xy then as per the conditions I)1+x+y=2x ie x=y+1 or ii)1+x+y=2y ie y=x+1 or iii) 1+x+y =2 ie x+y =1
  in case 1 the possible combinations satisfying x=y+1 are(x,y) = (1,0),(2,1) and so on to (9,8) = 9 sets
  in case 2 the possible combinations satisfying y=x+1 are (x,y)=(0,1),(1,2) and so on to (8,9) = 9 sets
  in case 2 the possible combinations satisfying x+y=1 are(x,y) =(0,1),(1,0) = 2 sets
  
  let us consider the numbers from 200 to 299 : assume that the number is 2xy then as per the conditions I)2+x+y=2x ie x=y+2 or ii)2+x+y=2y ie y=x+2 or iii) 2+x+y =2(2)=4 ie x+y =2
  in case 1 the possible combinations satisfying x=y+2 are(x,y) = (2,0),(2,1) and so on to (9,7) = 8 sets
  in case 2 the possible combinations satisfying y=x+2 are (x,y)=(0,2),(1,3) and so on to (7,9) = 8 sets
  in case 2 the possible combinations satisfying x+y=2 are(x,y) =(1,1) = 1 sets since (2,0),(0,2) are alredy covered in above cases.
  
  we can follow the same procedure for numbers from 300 to 399 ,400 to 499 by taking 3xy,4xy as numbers respectively.solution may seem to be tedious but is very easy to work out
  
  
  
  
• 9 years agoHelpfull: Yes(2) No(0)
• in the range 100-200
  101,112,123,134,145,156,167,178,189 so total 9 sets
  200-300
  211,202,213,224,235,246,257,268,279 so total of 9 sets
  300-400
  303,312,314,325,336,347,358,369 8 sets
  400-500
  404,413,422,415,426,437,448,459 8 sets
  so 68 is the amswer
• 8 years agoHelpfull: Yes(2) No(4)
• explain how?
• 9 years agoHelpfull: Yes(0) No(0)
• Siddhartha bro how?
  Can u plz elaborate.
• 9 years agoHelpfull: Yes(0) No(0)
• with repetition
  case 1 : last two digit containg 0
  00,04,20,40(4 ways)
  5*6*4=120
  case 2: last two digit without 0
  12,24,32,44,52(5 ways)
  5*6*5=120
  answer:120+150=270.
  Without Repetition :
  case 1 : last two digits containg 0
  04,20,40(3 ways)
  4*3*3=36
  case 2: last two digits without 0
  12,24,32,52(4 ways)
  4*3*4=48
  answer:36+48=84-12( need to remove 4 digit numbers having leading zeroes)
  anser=72
• 8 years agoHelpfull: Yes(0) No(0)
• The ans is 29
• 8 years agoHelpfull: Yes(0) No(0)