Puzzle for July 5, 2021  ( )

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

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

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

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Hint #1


In eq.6, replace B with A + F (from eq.3): A + F – D = A + D In the equation above, subtract A from both sides, and add D to both sides: A + F – D – A + D = A + D – A + D which makes F = 2×D


  

Hint #2


In eq.2, replace F with 2×D: C = D + 2×D which makes C = 3×D


  

Hint #3


In eq.4, substitute 3×D for C, and 2×D for F: 3×D + 2×D = A which makes 5×D = A


  

Hint #4


Substitute 5×D for A, and 2×D for F in eq.3: B = 5×D + 2×D which makes B = 7×D


  

Hint #5


Substitute 5×D for A, and 3×D for C in eq.5: 5×D + D = 3×D + E which becomes 6×D = 3×D + E Subtract 3×D from each side of the above equation: 6×D – 3×D = 3×D + E – 3×D which makes 3×D = E


  

Solution

Substitute 5×D for A, 7×D for B, 3×D for C and E, and 2×D for F in eq.1: 5×D + 7×D + 3×D + D + 3×D + 2×D = 21 which simplifies to 21×D = 21 Divide both sides of the above equation by 21: 21×D ÷ 21 = 21 ÷ 21 which means D = 1 making A = 5×D = 5×1 = 5 B = 7×D = 7×1 = 7 C = E = 3×D = 3×1 = 3 F = 2×D = 2×1 = 2 and ABCDEF = 573132