Puzzle for May 14, 2021  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 29 eq.2) B + C = A eq.3) F = A + E eq.4) D + E = A – E + F eq.5) A + B = D + F – A eq.6) E + F – D = B + D – E

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

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Hint #1


In eq.5, subtract F from both sides, and add A to both sides: A + B – F + A = D + F – A – F + A which becomes 2×A + B – F = D In eq.4, replace D with 2×A + B – F: 2×A + B – F + E = A – E + F In the above equation, subtract A from both sides, add E and F to both sides: 2×A + B – F + E – A + E + F = A – E + F – A + E + F which simplifies to eq.4a) A + B + 2×E = 2×F


  

Hint #2


In eq.4a, substitute (A + E) for F (from eq.3): A + B + 2×E = 2×(A + E) which becomes A + B + 2×E = 2×A + 2×E Subtract A and 2×E from each side of the equation above: A + B + 2×E – A – 2×E = 2×A + 2×E – A – 2×E which simplifies to B = A


  

Hint #3


In eq.2, replace A with B: B + C = B Subtract B from both sides of the equation above: B + C – B = B – B which makes C = 0


  

Hint #4


In eq.6, substitute A for B: E + F – D = A + D – E In the above equation, subtract D from both sides, and add E to both sides: E + F – D – D + E = A + D – E – D + E which becomes eq.6a) 2×E + F – 2×D = A


  

Hint #5


Substitute 2×E + F – 2×D for A (from eq.6a) into eq.4: D + E = 2×E + F – 2×D – E + F which becomes D + E = E + 2×F – 2×D In the equation above, subtract E from both sides, and add 2×D to both sides: D + E – E + 2×D = E + 2×F – 2×D – E + 2×D which becomes 3×D = 2×F Divide both sides by 2: 3×D ÷ 2 = 2×F ÷ 2 which makes eq.4b) 1½×D = F


  

Hint #6


Substitute 1½×D for F in eq.3: 1½×D = A + E Subtract E from each side of the above equation: 1½×D – E = A + E – E which becomes eq.3a) 1½×D – E = A


  

Hint #7


In eq.4, substitute 1½×D – E for A (from eq.3a), and 1½×D for F: D + E = 1½×D – E – E + 1½×D which becomes D + E = 3×D – 2×E In the above equation, subtract D from both sides, and add 2×E to both sides: D + E – D + 2×E = 3×D – 2×E – D + 2×E which becomes 3×E = 2×D Divide both sides by 2: 3×E ÷ 2 = 2×D ÷ 2 which makes 1½×E = D


  

Hint #8


Substitute (1½×E) for D in eq.3a: 1½×(1½×E) – E = A which becomes 2¼×E – E = A which makes 1¼×E = A and also makes B = A = 1¼×E


  

Hint #9


Substitute (1½×E) for D in eq.4b: 1½×(1½×E) = F which makes 2¼×E = F


  

Solution

Substitute 1¼×E for A and B, 0 for C, 1½×E for D, and 2¼×E for F in eq.1: 1¼×E + 1¼×E + 0 + 1½×E + E + 2¼×E = 29 which simplifies to 7¼×E = 29 Divide both sides of the above equation by 7¼: 7¼×E ÷ 7¼ = 29 ÷ 7¼ which means E = 4 making A = B = 1¼×E = 1¼ × 4 = 5 D = 1½×E = 1½ × 4 = 6 F = 2¼×E = 2¼ × 4 = 9 and ABCDEF = 550649