Puzzle for January 1, 2022  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 41 eq.2) B + F = A + C eq.3) A + F = B + D eq.4) C + E = A + B + D eq.5) E + F – D = A + D eq.6) D + E = A + B + C – E

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

Scratchpad

 

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Hint #1


Subtract the left and right sides of eq.5 from the left and right sides of eq.4, respectively: C + E – (E + F – D) = A + B + D – (A + D) which is equivalent to C + E – E – F + D = A + B + D – A – D which becomes C – F + D = B Add F to both sides of the above equation: C – F + D + F = B + F which becomes eq.4a) C + D = B + F


  

Hint #2


In eq.2, replace B + F with C + D (from eq.4a): C + D = A + C Subtract C from each side of the above equation: C + D – C = A + C – C which makes D = A


  

Hint #3


In eq.3, replace D with A: A + F = B + A Subtract A from both sides of the equation above: A + F – A = B + A – A which makes F = B


  

Hint #4


Substitute B for F in eq.2: B + B = A + C which becomes eq.2a) 2×B = A + C


  

Hint #5


eq.6 may be written as: D + E = A + C + B – E Substitute 2×B for A + C (from eq.2a) into the above equation: D + E = 2×B + B – E which becomes D + E = 3×B – E Subtract E from both sides: D + E – E = 3×B – E – E which makes D = 3×B – 2×E and also makes eq.6a) A = D = 3×B – 2×E


  

Hint #6


Substitute 3×B – 2×E for A (from eq.6a) in eq.2a: 2×B = 3×B – 2×E + C In the above equation, subtract 3×B from both sides, and add 2×E to both sides: 2×B – 3×B + 2×E = 3×B – 2×E + C – 3×B + 2×E which becomes –B + 2×E = C which may be written as eq.2b) 2×E – B = C


  

Hint #7


Substitute 2×E – B for C (from eq.2b), and 3×B – 2×E for both A and D (from eq.6a) in eq.4: 2×E – B + E = 3×B – 2×E + B + 3×B – 2×E which becomes 3×E – B = 7×B – 4×E Add B and 4×E to both sides of the above equation: 3×E – B + B + 4×E = 7×B – 4×E + B + 4×E which makes 7×E = 8×B Divide both sides by 8: 7×E ÷ 8 = 8×B ÷ 8 which makes ⅞×E = B and also makes F = B = ⅞×E


  

Hint #8


Substitute ⅞×E for B in eq.2b: 2×E – ⅞×E = C which makes 1⅛×E = C


  

Hint #9


Substitute (⅞×E) for B in eq.6a: A = D = 3×(⅞×E) – 2×E which becomes A = D = 2⅝×E – 2×E which makes A = D = ⅝×E


  

Solution

Substitute ⅝×E for A and D, ⅞×E for B and F, and 1⅛×E for C in eq.1: ⅝×E + ⅞×E + 1⅛×E + ⅝×E + E + ⅞×E = 41 which simplifies to 5⅛×E = 41 Divide both sides of the above equation by 5⅛: 5⅛×E ÷ 5⅛ = 41 ÷ 5⅛ which means E = 8 making A = D = ⅝×E = ⅝ × 8 = 5 B = F = ⅞×E = ⅞ × 8 = 7 C = 1⅛×E = 1⅛ × 8 = 9 and ABCDEF = 579587