Explanation: S I N C E + E V E R --------------------- D A R W I N Here, different alphabets signify different digits. So, in order to find the value of the alphabets, we have to solve. How? Note: 2 single digit number makes a minimum as well as maximum carry of 1 (e.g. 9+9=18 carry 1). In the first right column, there is only single alphabet D. And we know that it cannot be zero. And maximum carry possible is 1. So, its value will be 1. In the second column from right, A is directly beneath S. S will be added to nothing. In order to make S and A different, carry of 1 should be added to S as no other carry is possible. Now, from where this carry will come ?It will come from I+E. Now, there is only one possibility. Carry of 1 should be added to 9 to generate a carry of 1 (as 9+1=10). So, S= 9 and A=0 as A will have a carry of 1. This will generate carry of 1 which will be equal to D... Now the numbers- 0,1, 9 are already out. If I+E are generating a carry, there sum should be greater than 11. Lets, take it as 12. So, the value of R will become 0. But, if solve the whole problem with keeping R=2, tge solution won't come. So, the sum should be 13. So, now R=3. Now, on splitting 13 into 2 or 3 part(carry 1), Possibility of 8+3+1, 8+4, 7+5+1,7+6 will come. On solving by hit and trial. I=7, E=5. Now on solving the values by hit and trial method. The values of alphabets are as follows: D=1,S=9, A=0, I=7, E=5, N=8, V=6, C=2, R=3, W=4.

Explanation: W O R L D + T R A D E --------------------- C E N T E R Here, different alphabets signify different digits. So, in order to find the value of the alphabets, we have to solve. How? Note: 2 single digit number makes a minimum as well as maximum carry of 1 (e.g. 9+9=18 carry 1). In the first right column, there is only one alphabet C and possible carry is 1. So, its vale should be 1. Now, from where this carry will come ?It will come from W+T.. Now, if we look towards left end, E and D can not be zero, otherwise the value of R would be same as either D or E. So, E can not be 1 or 0. So, consider it as 2 and solve rest. We can also take value of D as 3. But, E is coming many times in the question so we are taking E. We can take value of D as 3 as 3 is available. But, then value of L should be 9 and tge remaining numbers left would be 0,4,6,8,7. Which wont solve the question. So, we will take D=4 and see the remaining numbers. After trying hit and try method, we get the values as: C=1, W=5, T=7, O=3, R=6, E=2, A=0, N=9, D=4, L=8.