Fibonacci again..Distinct Digits..Detail all your Deductive Steps [closed]

$$ABB$$

$$CDE$$

$$GFB$$

$$DPGB$$

$$QPFR$$

$$RDFD$$

• Puzzled by downvotes for this..concise..purely logic based..no ambiguities..clear – Uvc Jul 3 '19 at 10:46
• You are asking the solver to make a lot of assumptions about what to solve. – Jay Jul 3 '19 at 12:22
• @Uvc The reason (I think) you are getting a lot of downvotes is that you post a lot of puzzles, particularly of the kind that don't take any effort at all to create. Case in point: this puzzle is a strict subset of the puzzle you posted two days ago. One of these puzzles shouldn't exist, and if you had tried to solve both variations before posting, you would have known which is the better one, and posted only that. – Bass Jul 3 '19 at 12:41
• @Uvc, since one can't really take a puzzle back after posting it, I'd like to recommend sleeping on every puzzle, and then trying to solve it again in the morning. This has saved many of my own puzzles from being a disaster. Also, it's not like the internet is going anywhere, and this isn't a "post as much as you can" competition, so waiting until the next day before posting increases puzzle quality without costing much anything. Cheers, and happy puzzling! – Bass Jul 3 '19 at 12:45
• Puzzles need to be self contained. Looking at just what is posted here, there is no explanation whatsoever of what this puzzle is asking for, making it literally unclear what you’re asking here. @Bass has given some good advice, and taken together with earlier comments you’ve received about avoiding genre fatigue, I hope this will encourage you to slow things down. Too much of even a good thing isn’t good, and it’s not a race here. – Rubio Jul 3 '19 at 12:56

Solution 1 - expected solution

$$ABB+CDE=GFB\implies E=0\\CDE+GFB<2000\implies D=1\\GFB+DPGB<3000\implies Q=2\\R=D+Q+1=4\,\text{carrying due to}\,P+P=D=1\\1PGB+2PF4=41F1\implies B=(1)1-4=7\\A77+C10=GF7\implies F=7+1=8\\C10+G87=1PG7\implies G=1+8=9\\1P97+2P84=4181\implies P+P=(1)1\implies P=5\\\text{remaining integers are}\,3,6\implies A=3,C=6\\\boxed{A=3,\quad B=7,\quad C=6,\quad D=1,\quad E=0,\quad F=8,\\G=9,\quad P=5,\quad Q=2,\quad R=4.}$$

Solution 2 - uses known Fibonacci list

Step 1:

List of three- and four-digit Fibonacci numbers: $$144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$$

Step 2:

Thus we see that $$B=7$$ since we have $$ABB$$ and $$GFB$$. Therefore $$A=3$$ and $$DPGB=1597$$.

Step 3:

Next, since $$G=9$$, we have that $$F=8$$. Since $$D=1$$, we have that $$RDFD=4181$$ and $$QPFR=2584$$.

Step 4:

Finally, $$CDE=610$$ since $$D=1$$.

Step 5:

So $$A=3,\quad B=7,\quad C=6,\quad D=1,\quad E=0,\quad F=8,\\G=9,\quad P=5,\quad Q=2,\quad R=4.$$ These are distinct from $$0-9$$.

• You cannot just see..every number has to be logically deduced right from step 1 – Uvc Jul 3 '19 at 9:42
• What do you mean? Do you want me to elaborate, i.e. B=7 since we have ABB and GFB both ending in B and 7 is the only number that does this (377 and 987)? – TheSimpliFire Jul 3 '19 at 9:43
• No..you cannot take from literature..you have to mathematically deduce every digit through logical reasoning – Uvc Jul 3 '19 at 9:46
• It's not literature: I can write out the list of Fibonacci numbers by hand from the definition. – TheSimpliFire Jul 3 '19 at 9:47
• Then it is not a puzzle..every digit can be mathematically deduced using fib properties – Uvc Jul 3 '19 at 9:50