Spoiler for New Scientist Enigma 1707: (I skipped 1706.)
I was teaching my nephew about arithmetic progressions – sequences like 17, 23, 29,… 677,… in which the common difference between successive terms is a constant. In this one the difference is 6 and the 111th term is 677. Later, he devised his own progression, but consistently replaced digits with capital letters, with different letters for different digits. His sequence was ONE, TWO,… THREE,… with the 111th term being THREE. The common difference was odd and more than the number represented by SIX.
Tell me the number represented by SENT?
If d is the difference then we know: d is odd, TWO-ONE=d, and ONE+110d=THREE. The latter two statements can be combined to make:
TWO00 + TWO0 + ONE -ONE00 - ONE0 ------ THREE
Denote the carries into the first four columns (which, since there are subtractions, can be negative or positive) as c1, c2, c3, c4. So for instance we have c1+T-O=T. O must equal c1, and O cannot be 0 if leading zeroes are disallowed (and it can’t be negative!), and c1 cannot be more than 1, so O=c1=1. Then TWO is odd, so ONE must be even; E must be even.
In column 3, O+W+O-E-N=2+W-(E+N); E+N must be at least 2, so 2+W-(E+N)<10 and c2 < 1.
In column 2, c2+W+T-N-1=10+H and so W+T=11+H+N>13. So W>4.
Consider column 4; c4=0, so O+N-E=E or O+N-E=10+E or O+N+E=E-10. That is, N+1=2E or N-9=2E or N+11=2E. So if E = 0, 2, 4, 6, 8 then N = 9, 3, 7, 1 (not allowed), or 5.
Try E=2, N=3. Then c3=0 and column 3 is W-3=R. W=5 and W=6 are ruled out, since R cannot be 2 or 3. If W=7, R=4. Then W+T-N-1=3+T=10+H and H must be 1 or 2, neither of which is allowed. If W=8, R=5, and W+T-N-1=4+T=10+H and the only possibility is T=6, H=0.
Does that work? ONE=132, TWO=681, d=549, and THREE=60522=132+110*549 as required. The three remaining digits are 4, 7, and 9; if S=4 then SIX=479 or 497 either of which is less than d. All requirements are satisfied, and SENT=4236.
I haven’t checked for additional solutions, but presumably there are none.