Puzzle for March 14, 2022  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) D = B + C eq.3) F = B + D eq.4) A = C + D eq.5) D + E = A + C eq.6) C + E = D – E

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 A with C + D (from eq.4): D + E = C + D + C which becomes D + E = 2×C + D Subtract D from each side of the above equation: D + E – D = 2×C + D – D which makes E = 2×C


  

Hint #2


In eq.6, replace E with 2×C: C + 2×C = D – 2×C which becomes 3×C = D – 2×C Add 2×C to both sides of the above equation: 3×C + 2×C = D – 2×C + 2×C which makes 5×C = D


  

Hint #3


In eq.4, substitute 5×C for D: A = C + 5×C which makes A = 6×C


  

Hint #4


Substitute 5×C for D in eq.2: 5×C = B + C Subtract C from each side of the equation above: 5×C – C = B + C – C which makes 4×C = B


  

Hint #5


Substitute 4×C for B, and 5×C for D in eq.3: F = 4×C + 5×C which makes F = 9×C


  

Solution

Substitute 6×C for A, 4×C for B, 5×C for D, 2×C for E, and 9×C for F in eq.1: 6×C + 4×C + C + 5×C + 2×C + 9×C = 27 which simplifies to 27×C = 27 Divide both sides of the above equation by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 6×C = 6 × 1 = 6 B = 4×C = 4 × 1 = 4 D = 5×C = 5 × 1 = 5 E = 2×C = 2 × 1 = 2 F = 9×C = 9 × 1 = 9 and ABCDEF = 641529