Cryptarithmetic 2

[jamesravid]

Solve this,

DONALD
GERALD+
--------
ROBERT

[Barbie]

1 4 2 5 2 1 3 0 4 5 2 1 + ----------- 4 4 7 0 4 2

[kalayama]

526485
197485
______
723970

This was a hard one! (For me )

[psuresh1982]

Barbie ur answer was wrong.
The condition was all letters have unique number.
Check it in the equation.

--------------------------
suresh

[kalayama]

(Well I tried many other ways too... This way yielded)
1)After a lot of permutation and combinations, I assumed D= 5, then T = 0
2)Now, O + E = O, then e Must be 9
3) Also, L + L + 1 (1 is carry from adding D+D) = R, then R must be an odd number
4)Now, R must be greater than 5 because 5 + G = R. Since E=9, R must be 7 as 7 is the only odd number between 5 and 9.
5)Now, 5 + 1 + G = 7. So G will have to be 1.
6)L+L+1=R. R=7. So, L is either 3 or 8.
Now, A+A=E. But,E=9 which means, there has to be a carry.
Which makes L=8
7)N+7=B (There are 3 unused numbers now. 2,3 and 6)
=> N=6 and B=3
8) O=2 (@ is the only left over number)

Boy! Wasn't that hard?

[kalayama]

Barbie, the rule says no two letters will have same digit assigned though not mentioned it is implicit. Guess you used same digit for many alphabets....

[Barbie]

Ya.. I used so. I am not aware of that.
Thank you.

[kalayama]

That's no problem. Now you know ;-)