Puzzle for April 1, 2021  ( )

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

eq.1) A + B + C + D + E + F = 39 eq.2) B + E = A eq.3) A – B + C = B + F eq.4) D + F – C = C + E eq.5) B + F = A + C + E – F eq.6) A – D + E + F = B + D

A, B, C, D, E, and F each represent a one-digit non-negative integer.
 
We thank Abby S (age 10) for sending us this puzzle!  Thank you, Abby!

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Hint #1


In eq.5, replace B + F with A – B + C (from eq.3): A – B + C = A + C + E – F Add B and F to both sides of the above equation: A – B + C + B + F = A + C + E – F + B + F which becomes A + C + F = A + C + E + B Subtract A and C from both sides: A + C + F – A – C = A + C + E + B – A – C which becomes F = E + B which may be written as eq.5a) F = B + E


  

Hint #2


In eq.2, replace B + E with F (from eq.5a): F = A


  

Hint #3


In eq.3, substitute A for F: A – B + C = B + A In the above equation, subtract A from both sides, and add B to both sides: A – B + C – A + B = B + A – A + B which simplifies to C = 2×B


  

Hint #4


Substitute 2×B for C into eq.4: D + F – 2×B = 2×B + E Add 2×B to both sides of the above equation: D + F – 2×B + 2×B = 2×B + E + 2×B which becomes D + F = 4×B + E which may be written as eq.4a) D + F = 3×B + B + E


  

Hint #5


Substitute F for B + E (from eq.5a) in eq.4a: D + F = 3×B + F which makes D = 3×B


  

Hint #6


Substitute F for A, and 3×B for D in eq.6: F – 3×B + E + F = B + 3×B which becomes 2×F – 3×B + E = 4×B Add 3×B to both sides of the equation above: 2×F – 3×B + E + 3×B = 4×B + 3×B which becomes eq.6a) 2×F + E = 7×B


  

Hint #7


In eq.6a, substitute (B + E) for F (from eq.5a): 2×(B + E) + E = 7×B which is equivalent to 2×B + 2×E + E = 7×B which becomes 2×B + 3×E = 7×B Subtract 2×B from each side of the equation above: 2×B + 3×E – 2×B = 7×B – 2×B which makes 3×E = 5×B Divide both sides by 3: 3×E ÷ 3 = 5×B ÷ 3 which makes E = 1⅔×B


  

Hint #8


Substitute 1⅔×B for E in eq.5a: F = B + 1⅔×B which makes F = 2⅔×B and also makes A = F = 2⅔×B


  

Solution

Substitute 2⅔×B for A and F, 2×B for C, 3×B for D, and 1⅔×B for E in eq.1: 2⅔×B + B + 2×B + 3×B + 1⅔×B + 2⅔×B = 39 which simplifies to 13×B = 39 Divide both sides of the above equation by 13: 13×B ÷ 13 = 39 ÷ 13 which means B = 3 making A = F = 2⅔×B = 2⅔ × 3 = 8 C = 2×B = 2 × 3 = 6 D = 3×B = 3 × 3 = 9 E = 1⅔×B = 1⅔ × 3 = 5 and ABCDEF = 836958