Answer is:





The brackets contain the line (from the bottom) used.

$D=1$ as sum of two 2-digit numbers $\lt200$ (123). $C=2$ by the same argument (234), therefore, because $U+V$ doesn't carry, $V=3$ (345). We have $V+R=E$ next to $V+R=D=1$ (456), so $V+R$ must carry, and $D=E+1$, so $E=0$. Also $V+C+1=S$ (456), so $S=6$. $Q=R+D=7+1=8$ (567), and the second line up tells us $P=Q+1=9$ (123). $P+S=DT$ (678), so $T=5$, giving $U=4$ (123).

  • $\begingroup$ Got it......... $\endgroup$ – Uvc Jul 1 '19 at 11:15
  • 1
    $\begingroup$ Alternate approach: The first 3-digit numbers all end in a doubled digit? Surely OP wouldn't have used the actual Fibonacci sequence? $\endgroup$ – Bass Jul 1 '19 at 15:36

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