Puzzle for April 30, 2019  ( )

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

eq.1) A + B + C + D + E + F = 17 eq.2) D - A - B - F = A + B eq.3) A + D = C eq.4) C - B = B + D eq.5) D - A - B = A + B

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

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Hint #1


Add B to both sides of eq.4: C - B + B = B + D + B which becomes C = 2×B + D In the equation above, replace C with A + D (from eq.3): A + D = 2×B + D Subtract D from each side: A + D - D = 2×B + D - D which makes A = 2×B


  

Hint #2


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


  

Hint #3


Substitute 2×B for A, and 6×B for D in eq.3: 2×B + 6×B = C which means 8×B = C


  

Hint #4


Substitute 6×B for D, and 2×B for A in eq.2: 6×B - 2×B - B - F = 2×B + B which becomes 3×B - F = 3×B Subtract 3×B from each side: 3×B - F - 3×B = 3×B - 3×B which becomes -F = 0 which means F = 0


  

Hint #5


Substitute 2×B for A, 8×B for C, 6×B for D, and 0 for F in eq.1: 2×B + B + 8×B + 6×B + E + 0 = 17 which becomes 17×B + E = 17 Subtract 17×B from both sides of the equation above: 17×B + E - 17×B = 17 - 17×B which becomes eq.1a) E = 17 - 17×B


  

Solution

To make eq.1a true, check several possible values for B and E:   If B = 0, then E = 17 - 17×0 = 17 - 0 = 17 If B = 1, then E = 17 - 17×1 = 17 - 17 = 0 If B = 2, then E = 17 - 17×2 = 17 - 34 = -17 If B > 2, then E < -17   Since E must be a one-digit non-negative integer, then E = 0 and therefore B = 1 making A = 2×B = 2 × 1 = 2 C = 8×B = 8 × 1 = 8 D = 6×B = 6 × 1 = 6 and ABCDEF = 218600