A, B, C, D are distinct digits and can vary from 1 to 9. DB is a concatenated number.

From the following relationship, solve for A,B,C,D. Give your deductive reasoning.

Please read all the comments, I have dropped an important clue. This problem is pure logic at work. But you need to take the first right step.

If you are voting to close this puzzle, look at all the related puzzles you have accepted. This needs lot more deductive reasoning than many of others.

Otherwise, you need lot of brute force thinking!!

$$A! + C! + D! = (A+B)^2 + (2C+B)^2 + (DB)^2 - 3B$$

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    $\begingroup$ These questions are getting awfully repetitive. $\endgroup$ – greenturtle3141 Jun 4 '19 at 6:52
  • $\begingroup$ @greenturtle3141 Still great fun though! $\endgroup$ – Rand al'Thor Jun 4 '19 at 7:51
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    $\begingroup$ For those who cast votes to close, I urge you to think through. This is not designed to test your mathematical knowledge but your logical mettle!! You have to peer underneath. This is not designed to be brute force problem. $\endgroup$ – Uvc Jun 4 '19 at 9:00
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    $\begingroup$ I would have less of a complaint if each of these puzzles had a unique spin on them. Alas, there is nothing remarkable here. To add, I'd definitely look at your most up-voted puzzles. They are either 1) Aesthetically pleasing or 2) Have a very straight, pretty line to the solution. This puzzle is neither because it doesn't showcase some remarkable relationship, answer isn't pretty, and the solution involves some bashy casework. $\endgroup$ – greenturtle3141 Jun 4 '19 at 15:32
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    $\begingroup$ To add one more thought - too much of even a good thing isn't good. You've saturated the site with a large number of puzzles of a similar appearance and, mostly, theme. At some point, people get tired of too much of the same thing; I think you're seeing many more weak to negative votes of late due to genre fatigue. Please (strongly!) consider both slowing down, and mixing up a bit the type of puzzle you post. $\endgroup$ – Rubio Jun 4 '19 at 21:57

The solution is:

$A=4, B=1, C=5, D=7$

$4! + 5! + 7! = 5^2 + 11^2 + 71^2 - 3$

I found this by guessing that $D=7$, especially because $71^2-1=7!$. Then I noticed $B=1$,and that we can use the other (known) two solutions to Brocard's Problem, namely $5^2-1=4!$ and $11^2-1=5!$ to get the answer.

  • $\begingroup$ I will look at all the answers posted in the next 10 hrs... ..will select most well reasoned answers ...also post my own reasoning.. $\endgroup$ – Uvc Jun 4 '19 at 13:35

Partial answer (could be used as a springboard by somebody else)

Preliminary size considerations:

The right-hand side can't be larger than around $10,000$, so the left-hand side can't include $8!$ or $9!$, i.e. $A,C,D$ are all at most 7. But since $A,B,C,D$ are all positive and distinct, the right-hand side must be quite a bit bigger than $144$, so at least one of $A,C,D$ must be bigger than 5.

Case 1

If one of $A,C,D$ is 7, then

the left-hand side is between $5043$ and $5880$, so $DB$ must be seventy something, i.e. $D=7$.

[to be completed]

Case 2

If none of $A,C,D$ is 7, then

the left-hand side is at most $864$, so $DB$ is less than $30$, so $D$ is 1 or 2. We also know that one of $A,C$ must be 6.

Case 2a

If $A=6$, then the right-hand side modulo 4 is $B^2+B^2+B^2-3B=3B(B-1)$ which is even. All factorials after the first one are even, so we can't have $D=1$, which means $D=2$. So the possibilities for the left-hand side are $723,728,746,842$.

Considering each of these possibilities in turn:

  • $C=1,723=(6+B)^2+(2+B)^2+(20+B)^2-3B=3B^2+53B+440$, impossible.

  • $C=3,728=(6+B)^2+(6+B)^2+(20+B)^2-3B=3B^2+61B+472$, impossible.

  • $C=4,746=(6+B)^2+(8+B)^2+(20+B)^2-3B=3B^2+65B+500$, impossible.

  • $C=5,842=(6+B)^2+(10+B)^2+(20+B)^2-3B=3B^2+69B+536$, impossible.

So Case 2a is impossible.

Case 2b

If $C=6$, then ...

[to be completed]

  • $\begingroup$ Try rearranging some of the terms, job becomes so much easier. It is not as hard as you think.Restricted choice makos the digits pop out. $\endgroup$ – Uvc Jun 4 '19 at 8:50
  • $\begingroup$ @Uvc $3B^2 + (2A+4C+20D-3)B + (A^2+4C^2+100D^2)$ - does this help? $\endgroup$ – Rand al'Thor Jun 4 '19 at 9:06
  • $\begingroup$ Nope..but B plays an important part. $\endgroup$ – Uvc Jun 4 '19 at 9:10

!Key to solving this is rearranging terms, otherwise it becomes messy.

!$(A!+B) $- $(A+B)^2$

!+$ (C!+B)$ -$ (2C+B)^2$

!+$(D!+B)$ - $(DB)^2$ = $0$

!For 1st term, simple inspection yields A=4, B=1.

!For second term, C=3, B=1.

!For third term, based on Factorial values ! D has to be 7

!Everything falls into place neatly.


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