Puzzle for March 29, 2021  ( )

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

eq.1) A + B + C + D + E + F = 21 eq.2) B + F = A + D eq.3) F = D – A eq.4) D = B + C eq.5) E – C = B – E eq.6) C = D ÷ E

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

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Hint #1


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


  

Hint #2


Add C and E to both sides of eq.5: E – C + C + E = B – E + C + E which becomes 2×E = B + C In the above equation, replace B + C with D (from eq.4): 2×E = D


  

Hint #3


In eq.6, replace D with 2×E: C = 2×E ÷ E which makes C = 2


  

Hint #4


In eq.4, substitute 2×E for D, and 2 for C: 2×E = B + 2 Subtract 2 from both sides of the above equation: 2×E – 2 = B + 2 – 2 which makes eq.4a) 2×E – 2 = B


  

Hint #5


Substitute 2×A for B in eq.4a: 2×E – 2 = 2×A Divide both sides of the above equation by 2: (2×E – 2) ÷ 2 = 2×A ÷ 2 which makes eq.4b) E – 1 = A


  

Hint #6


Substitute 2×E for D, and (E – 1) for A (from eq.4b) in eq.3: F = 2×E – (E – 1) which becomes F = 2×E – E + 1 which makes eq.3a) F = E + 1


  

Solution

Substitute E – 1 for A (from eq.4b), 2×E – 2 for B (from eq.4a), 2 for C, 2×E for D, and E + 1 for F (from eq.3a) in eq.1: E – 1 + 2×E – 2 + 2 + 2×E + E + E + 1 = 21 which simplifies to 7×E = 21 Divide both sides of the above equation by 7: 7×E ÷ 7 = 21 ÷ 7 which means E = 3 making A = E – 1 = 3 – 1 = 2 (from eq.4b) B = 2×E – 2 = 2×3 – 2 = 6 – 2 = 4 (from eq.4a) D = 2×E = 2×3 = 6 F = E + 1 = 3 + 1 = 4 (from eq.3a) and ABCDEF = 242634