Puzzle for March 7, 2023  ( )

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

eq.1) A + B + C + D + E + F = 37 eq.2) A = B + C eq.3) D = B + F eq.4) A + B = E + F eq.5) A + C = B + E + F eq.6) C = F + (C ÷ B)

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


In eq.5, replace E + F with A + B (from eq.4): A + C = B + A + B which becomes A + C = 2×B + A Subtract A from both sides of the above equation: A + C - A = 2×B + A - A which makes C = 2×B


  

Hint #2


In eq.2, replace C with 2×B: A = B + 2×B which makes A = 3×B


  

Hint #3


In eq.6, replace C with 2×B: 2×B = F + (2×B ÷ B) which becomes 2×B = F + 2 Subtract 2 from each side of the equation above: 2×B - 2 = F + 2 - 2 which becomes eq.6a) 2×B - 2 = F


  

Hint #4


In eq.3, substitute 2×B - 2 for F (from eq.6a): D = B + 2×B - 2 which becomes eq.3a) D = 3×B - 2


  

Hint #5


Substitute 3×B for A, and 2×B - 2 for F (from eq.6a) in eq.4: 3×B + B = E + 2×B - 2 which becomes 4×B = E + 2×B - 2 In the above equation, subtract 2×B from both sides, and add 2 to both sides: 4×B - 2×B + 2 = E + 2×B - 2 - 2×B + 2 which becomes eq.4a) 2×B + 2 = E


  

Hint #6


Substitute 3×B for A, 2×B for C, 3×B - 2 for D (from eq.3a), 2×B + 2 for E (from eq.4a), and 2×B - 2 for F (from eq.6a) in eq.1: 3×B + B + 2×B + 3×B - 2 + 2×B + 2 + 2×B - 2 = 37 which simplifies to 13×B - 2 = 37 Add 2 to both sides of the above equation: 13×B - 2 + 2 = 37 + 2 which makes 13×B = 39 Divide both sides by 13: 13×B ÷ 13 = 39 ÷ 13 which means B = 3


  

Solution

Since B = 3, then: A = 3×B = 3×3 = 9 C = 2×B = 2×3 = 6 D = 3×B - 2 = 3×3 - 2 = 9 - 2 = 7 (from eq.3a) E = 2×B + 2 = 2×3 + 2 = 6 + 2 = 8 (from eq.4a) F = 2×B - 2 = 2×3 - 2 = 6 - 2 = 4 (from eq.6a) and ABCDEF = 936784