Puzzle for May 11, 2021  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) D – F = B + F eq.3) F – B = B + C eq.4) B + C + F = A eq.5) A + F = D + E – A eq.6) C + F = D – B – C

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

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Hint #1


Add B and C to both sides of eq.6: C + F + B + C = D – B – C + B + C which becomes 2×C + F + B = D which is the same as eq.6a) 2×C + B + F = D


  

Hint #2


In eq.6a, replace B + F with D – F (from eq.2): 2×C + D – F = D In the equation above, subtract D from both sides, and add F to both sides: 2×C + D – F – D + F = D – D + F which makes eq.6b) 2×C = F


  

Hint #3


In eq.3, replace F with 2×C: 2×C – B = B + C In the equation above, add B to both sides, and subtract C from both sides: 2×C – B + B – C = B + C + B – C which makes C = 2×B


  

Hint #4


In eq.6b, substitute (2×B) for C: 2×(2×B) = F which makes 4×B = F


  

Hint #5


Substitute (2×B) for C, and 4×B for F in eq.6a: 2×(2×B) + B + 4×B = D which becomes 4×B + B + 4×B = D which makes 9×B = D


  

Hint #6


Substitute 2×B for C, and 4×B for F in eq.4: B + 2×B + 4×B = A which makes 7×B = A


  

Hint #7


Substitute 7×B for A, 4×B for F, and 9×B for D in eq.5: 7×B + 4×B = 9×B + E – 7×B which becomes 11×B = 2×B + E Subtract 2×B from both sides of the above equation: 11×B – 2×B = 2×B + E – 2×B which makes 9×B = E


  

Solution

Substitute 7×B for A, 2×B for C, 9×B for D and E, and 4×B for F in eq.5: 7×B + B + 2×B + 9×B + 9×B + 4×B = 32 which simplifies to 32×B = 32 Divide both sides of the above equation by 32: 32×B ÷ 32 = 32 ÷ 32 which means B = 1 making A = 7×B = 7 × 1 = 7 C = 2×B = 2 × 1 = 2 D = E = 9×B = 9 × 1 = 9 F = 4×B = 4 × 1 = 4 and ABCDEF = 712994