I agree with the magic man. His answer and what a fun challenge this one is. Deserves more than one twenty couple click thrus in my opinion hence the bump. Gonna post my methodology but strongly suggest giving it a go. Know KS usually doesnt post answers and in no way would I want to suggest he should or do anything that might otherwise distract him from posting more fabulous teasers so here goes (besides, I'm a little proud I actually got one)
11
MALTHU5
THOMA5
+ 3600
=======
RRRRR00
M+1=R (A+T<18 and if A+T<10 then M=R)
H+M+7= R or R+10 or R+20
if H+M+7=R then M+1=H+M+7 or H=-6 so no
if H+M+7=R+20 then M+1=H+M-13 or H=12 so no
if H+M+7=R+10 then M+1=H+M-3 or H=4
1 111
MALT4U5
T4OMA5
+ 3600
=======
RRRRR00
Solve for R in the left most column of the equation and in the third column from the right and set them equal to one another:
M,O,S,T are unique
MO5T/23 is an integer
M=(1,2,6,7,8) (M cannot equal 0 (condition of the OP),3(if M=3 then R=4 but H=4),4(H=4),5(S=5)
or 9(the sum would have too many digits))
I used a spreadsheet and with the above restrictions, narrowed down MOST=(1058,1357,7153,7659,7958,8257)
Limit the possibilities of MOST:
1 111
MALT4U5
T4OMA5
+ 3600
=======
RRRRR00
A+T(+1 if L+4>9)=R+10
T+O+4= R or R+10 or R+20
if T+O+4=R then A+T-(10 or 9)=T+O+4 or A-(14 or 13)=O so no
if T+O+4=R+20 and R>1 (since M+1=R) then T+O>17 or T=O=9 so no
if T+O+4=R+10 then T+O+4=A+T(+1) or O+(3 or 4)=A so:
(O,A)=(0,3)(2,6)(3,6)(3,7)(6,9)
now:
if O=6 then the fourth column of the equation yields T=R so no
if O=3 then MOST=1357 (see STEP 3) and T=7 and plugging in O=3 and T=7 in the fourth column
yields R=4 but H=4 so no
if O=2 then MOST=8257 and M=8 R=9 so looking at the third column from the left L would have
to equal 5 but S=5 so no.
so O=0 and A=3 and MOST=1058 and ...plug and chug
Am sure there are other methods and undoubtedly a more elegant solution but that's what I gotsk. Much fun KS!
Solve the second column from the left and the fourth column simultaneously: