5 digits number 3 pairs of sum is 11 each.last digit is 3 times first one,3rd digit is 3 less than 2nd,4rth digit is 4 more than the second one

• 1st Digit -> a, 2nd Digit -> b, 3rd Digit -> (b-3), 4th digit -> (b+4), 5th Digit -> 3a
  So the number is : (a)(b)(b-3)(b+4)(3a)
  Now, Let's analyze 1st and the 5th digit :
  Possible combinations -
  1 - 3
  2 - 6
  3 - 9
  (Since 4 will yield 12 which is obviously more than 2 digits)
  
  Now Let's analyze 2nd,3rd and 4th Digits :
  
  Possible Values of 2nd Digit i.e 'b' is :
  5,4,3
  As, (b-3)>0 i.e 3rd Digit and (b+4) 1+3+7 = 11
  Similarly, 24186 for 4-1-8 and 6+4+1=11
  3rd Combination 5-2-9 will get no possible match.
  Hence, 2 solutions : 13073 and 24186
  If Repetitions not allowed then Ans should be 24186 :)
• 10 years agoHelpfull: Yes(14) No(2)
• 25296
  
  let first digit be 'X'
  then 5th digit is '3X'
  let 2nd digit be 'Y'
  then 3rd digit is 'Y-3'
  and 4th digit is 'Y+4'
  then the no is '(X)(Y)(Y-3)(Y+4)(3X)'
  from the above we can say 3X
• 10 years agoHelpfull: Yes(10) No(10)
• Plz provide the whole ans...
  25296 is not satisfying the rule of 3 pairs sum is 11...which 3 pairs exactly??
• 10 years agoHelpfull: Yes(3) No(3)
• let first digit be 'X'
  then 5th digit is '3X'
  let 2nd digit be 'Y'
  then 3rd digit is 'Y-3'
  and 4th digit is 'Y+4'
  then the no is '(X)(Y)(Y-3)(Y+4)(3X)'
  from the above we can say 3X
• 10 years agoHelpfull: Yes(2) No(1)
• i think ans will be...26326
• 10 years agoHelpfull: Yes(2) No(3)
• ans:-65692
  if repetition of digits are allowed
• 10 years agoHelpfull: Yes(0) No(7)
• Using Shreevatsa's logic the answer should be 25936
  
• 10 years agoHelpfull: Yes(0) No(7)
• then 5th digit is '3X'
  let 2nd digit be 'Y'
  then 3rd digit is 'Y-3'
  and 4th digit is 'Y+4'
  then the no is '(X)(Y)(Y-3)(Y+4)(3X)'
  from the above we can say 3X
• 10 years agoHelpfull: Yes(0) No(1)
• "5 digits number 3 pairs of sum is 11 each" what does it mean?
• 10 years agoHelpfull: Yes(0) No(0)
• ashwin. please get a proper english tuition. then dare to post a question in here
• 10 years agoHelpfull: Yes(0) No(0)
• 65292 by trial and error
  
• 10 years agoHelpfull: Yes(0) No(0)
• RIP ENGLISH -_-
• 10 years agoHelpfull: Yes(0) No(0)
• In a 5-digit number, 3 pairs of sum is 11, each last digit is 3 times the first one, the 3rd digits are 3 less than the 2nd, and the 4th digit is 4 more than the second one. What is the number?
  Suppose, the five digit number be “abcde”.
  
  Since, it is a five digit number, so a ≠ 0.
  
  Now, three pairs should add up to 11. But, “three pairs” means six digits. But, only five digits can be accommodated. So, there has to be at least one repetition of digits.
  
  Now, c = (b - 3) and d = (b + 4)
  
  If we take, one pair as “c, d” adding up to 11, then: (b - 3) + (b + 4) = 11
  
  (2b + 1) = 11
  
  b = 5.
  
  c = 2 and d = 9.
  
  Also, e = 3a.
  
  Thus, probable considerations for (a, e) are (1, 3) or (2, 6) or (3, 9).
  
  If (1, 3) is considered; there is no repetition of digit in the five digit number. Hence, rejected.
  
  If (3, 9) is considered; the five digit number turns out to be 35299. The digit 9 been repeated here. But, only two pairs of digits, (third digit + fourth digit) and (third digit + fifth digit), add up to 11. Hence, this is also rejected.
  
  If (2, 6) is considered; the five digit number turns out to be 25296. The digit 2 been repeated here. Here, three pairs of digits, (first digit + fourth digit), (third digit + fourth digit) and (second digit + fifth digit), add up to 11. Thus, 25296 is the required five digit number.
• 2 years agoHelpfull: Yes(0) No(0)