If ever+since = darwin then d+a+r+w+i+n is ?

• 1. As it’s a sum of 3 numbers hence the maximum value of D could be 2 or 1, but then we look at S which could have maximum value 9 and if it gets a carry of 1 then the value of A will be 0 and the D=1.
  2. Then we looks at E and I which results in R, as 9 is already occupied by S so E could be 8 and I could be 7 and R will be 5 with carry 1.
  3. Then we place the values of R=5 and E=8 resulting in N=3 with carry 1.
  4. In ten’s digits – E=8 and and value of C could be 3,4,6 (As all others are occupied) and To make I=7 (previously assigned in step 2) E must be added by 9 which is already occupied by S. So this hit ends here.
  Now start again with reduced value of E or I, repeat above steps until you get a correct answer or a dead end.
  5. Repeating above steps one time comes with E=5 and I=7 which results in R=3 carry=1.
  6. Placing R=3 and E=5 in unit digits and their sum gives 8.
  7. in ten’s digits place E=5 now you can add 2 to make I=7 hence C=2.
  8. in 100′s digits you have N=8 (from step 6) and only digits that are left now are 4 and
  6 hence. So placing V=4 you will get W=2 which is not true because it is already assigned to C. So we are left with only V=6 as the ultimate choice and this leads the result in to W=4.
  Hence this way we got the answer
  ….5 6 5 3
  ….E V E R
  ..9 7 8 2 5
  + S I N C E
  ———————-
  D A R W I N
  1 0 3 4 7 8
  Hence D+A+R+W+I+N = 1+0+3+4+7+8 =23
  
• 10 years agoHelpfull: Yes(3) No(3)