Puzzle for January 1, 2019  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) A + C = E eq.3) C = A + D eq.4) A = B + D eq.5) B + E = C + F eq.6) A + F = 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.6, substitute B + D for A (from eq.4): B + D + F = D + E Subtract D from both sides of the above equation: B + D + F - D = D + E - D which becomes B + F = E In eq.5, replace E with B + F: B + B + F = C + F Subtract F from both sides: B + B + F - F = C + F - F which means 2×B = C


  

Hint #2


Substitute 2×B for C in eq.3: 2×B = A + D Replace A with B + D (from eq.4): 2×B = B + D + D Subtract B from both sides: 2×B - B = B + D + D - B which makes B = 2×D Divide both sides by 2: B ÷ 2 = 2×D ÷ 2 which means ½×B = D


  

Hint #3


In eq.4, replace D with ½×B: A = B + ½×B which makes A = 1½×B


  

Hint #4


Substitute 1½×B for A, and 2×B for C in eq.2: 1½×B + 2×B = E which means 3½×B = E


  

Hint #5


Substitute 1½×B for A, ½×B for D, and 3½×B for E in eq.6: 1½×B + F = ½×B + 3½×B Subtract 1½×B from both sides: 1½×B + F - 1½×B = ½×B + 3½×B - 1½×B which makes F = 2½×B


  

Solution

Substitute 1½×B for A, 2×B for C, ½×B for D, 3½×B for E, and 2½×B for F in eq.1: 1½×B + B + 2×B + ½×B + 3½×B + 2½×B = 22 which simplifies to 11×B = 22 Divide both sides by 11: 11×B ÷ 11 = 22 ÷ 11 which means B = 2 making A = 1½×B = 1½ × 2 = 3 C = 2×B = 2 × 2 = 4 D = ½×B = ½ × 2 = 1 E = 3½×B = 3½ × 2 = 7 F = 2½×B = 2½ × 2 = 5 and ABCDEF = 324175