Puzzle for April 9, 2022  ( )

Scratchpad

Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 21 eq.2) B – C = D – B eq.3) E – D = F – E eq.4) A – D – E = B + C + D – A eq.5) D + F – A = A – B + C – E eq.6)* C = (E × F) mod B

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  "(E × F) mod B" equals the remainder of (E × F) divided by B.

Scratchpad

 

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Hint #1


Add the left and right sides of eq.5 to the left and right sides of eq.4, respectively: A – D – E + D + F – A = B + C + D – A + A – B + C – E which becomes –E + F = 2×C + D – E Add E to both sides of the equation above: –E + F + E = 2×C + D – E + E which becomes eq.4a) F = 2×C + D


  

Hint #2


In eq.3, replace F with 2×C + D (from eq.4a): E – D = 2×C + D – E Add D and E to both sides of the equation above: E – D + D + E = 2×C + D – E + D + E which becomes 2×E = 2×C + 2×D Divide both sides by 2: 2×E ÷ 2 = (2×C + 2×D) ÷ 2 which becomes eq.3a) E = C + D


  

Hint #3


Add C and B to both sides of eq.2: B – C + C + B = D – B + C + B which becomes 2×B = D + C which may be written as 2×B = C + D In the above equation, replace C + D with E (from eq.3a): 2×B = E


  

Hint #4


In eq.6, substitute 2×B for E: C = (2×B × F) mod B which means eq.6a) C = remainder of (2×B × F) ÷ B


  

Hint #5


In eq.6a, B is an integer factor of (2×B × F), which means: (2×B × F) ÷ B = 2×F and which means remainder (2×B × F) ÷ B = 0 Substitute 0 for remainder (2×B × F) ÷ B, in eq.6a: C = 0


  

Hint #6


Substitute 0 for C in eq.3a: E = 0 + D which makes E = D and also makes 2×B = E = D


  

Hint #7


Substitute 0 for C in eq.4a: F = 2×0 + D which makes F = D and also makes F = D = E = 2×B


  

Hint #8


Substitute 2×B for D and E, and 0 for C in eq.4: A – 2×B – 2×B = B + 0 + 2×B – A which becomes A – 4×B = 3×B – A Add 4×B and A to both sides of the above equation: A – 4×B + 4×B + A = 3×B – A + 4×B + A which becomes 2×A = 7×B Divide both sides by 2: 2×A ÷ 2 = 7×B ÷ 2 which makes A = 3½×B


  

Solution

Substitute 3½×B for A, 0 for C, and 2×B for D and E and F in eq.1: 3½×B + B + 0 + 2×B + 2×B + 2×B = 21 which simplifies to 10½×B = 21 Divide both sides of the above equation by 10½: 10½×B ÷ 10½ = 21 ÷ 10½ which means B = 2 making A = 3½×B = 3½ × 2 = 7 D = E = F = 2×B = 2 × 2 = 4 and ABCDEF = 720444