in a 4 digit no,the sum of unit digit and tens digit is equal to sum of digits at thousands & hundreds place. The sum of the digits in the tens and hundrrds placeis twice the sum of othertwo digits. If the sum of all digits is more than 20,then the units place can be ?pls also tell the method.

1) 6
2) 7
3) 8
4) 5

• let a four digit No. = abcd
  given -
  a+b = c + d --------------------(1)
  b+c = 2*(a+d)------------------(2)
  a+b+c+d > 20-------------------(3)
  Now using equation 1 and 3
  2*(a+b) > 20 i.e. a+b+c+d is multiple of 2
  Now using equation 2 and 3
  3*(a+d) >20 i.e a+b+c+d is multiple of 3
  ....Conclusion------>>>>> a+b+c+d is multiple of 6
  
  As we know any single digit no. can have maximum value equal to 9
  so max of (b+c) = 18 i.e. b+c
• 8 years agoHelpfull: Yes(0) No(2)
• a+b+c+d >20 by using given data we have 3*(a+d)>20 and 2*(a+b)>20 so a+b+c+d is multiple of 6
  and max of b+c 20 (Now we know upper limit)
  20< 3*(a+d) < = 27 and they are multiple of 6 so only 24 is multiple of 6 in between 20 and 27
  So a+b+c+d = 24 , a+b=c+d = 12 , a+d = 8 ,b+c = 16
  and if we check each possible no. then only - 5793,4884,3975 satisfies it hence at unit place 3,4,5 is present
• 8 years agoHelpfull: Yes(0) No(0)
• Let the number be abcd
  We have a+b = c+d and b+c = 2(a+d) and a+b+c+d
  a+b+c+d = 2(a+b).......1
  a+b+c+d = 3(a+d).....2
  From 1 and 2 we can say that digit sum is a multiple of 6
  so the digit sum can be 24,30,36
  36 is not possible as it implies all digits 9 which contradicts b+c = 2(a+d)
  if digit sum is 30 then,
  a+b=c+d=15
  and a+d = 10 and b+c = 20 No solution as one digit can be 9 maximum.
  Hence 24 is the digit sum
  Now,
  a+b = c+d = 12
  b+c = 16 , a + d = 8
  from a+d = 8 ;
  (a,d) can be (4,4) ,(5,3),(3,5)
  which implies
  (b,c) = (8,8),(7,9),(9,7)
  
  (6,2) pair of (a,d) is rejected as it will make b= 10 which is not possible
  Therefore there are 3 values for unit digit = 3,4,5
• 7 years agoHelpfull: Yes(0) No(0)