Originally Posted by
nixontkl
My answer K = 3
using logic K + K = H, H <=9 thus K can only be 1-4
so if K = 1, then H = 2 11A2 + 2JED = 2F1G
since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
now left with 4,3,8,0
3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out
there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah
when u get to K = 3 33A6 + 6JED = 6F3G
Numbers left = 1,2,4,5,6,7,8,9,0
same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
so far numbers left are ( 2,7,8,0 )
last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
numbers left ( 7,0 )
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