From the given Square - Factorial Relationship, deduce the unknowns

Question

$Given$:

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!!

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

Answer 1

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.

Answer 2

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

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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]

Answer 3

!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.