Puzzle for November 10, 2020  ( )

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

eq.1) A + B + C + D + E + F = 25 eq.2) A + D = B + C eq.3) B = C – D eq.4) C + F = B + E – F eq.5) E – B = A – E eq.6) E ÷ F = A ÷ B

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

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Hint #1


Add D to both sides of eq.3: B + D = C – D + D which becomes B + D = C In eq.2, replace C with B + D: A + D = B + B + D which becomes A + D = 2×B + D Subtract D from each side of the equation above: A + D – D = 2×B + D – D which makes A = 2×B


  

Hint #2


In eq.6, replace A with 2×B: E ÷ F = 2×B ÷ B which becomes E ÷ F = 2 Multiply both sides of the equation above by F: E ÷ F × F = 2 × F which makes eq.6a) E = 2×F


  

Hint #3


In eq.4, substitute 2×F for E: C + F = B + 2×F – F which becomes C + F = B + F Subtract F from each side of the above equation: C + F – F = B + F – F which makes C = B


  

Hint #4


Substitute B for C in eq.3: B = B – D Subtract B from each side of the above equation: B – B = B – D – B which makes 0 = –D which means 0 = D


  

Hint #5


Substitute 2×B for A in eq.5: E – B = 2×B – E Add B and E to both sides of the equation above: E – B + B + E = 2×B – E + B + E which makes 2×E = 3×B Divide both sides by 2: 2×E ÷ 2 = 3×B ÷ 2 which makes E = 1½×B


  

Hint #6


Substitute 1½×B for E in eq.6a: 1½×B = 2×F Divide both sides of the equation above by 2: 1½×B ÷ 2 = 2×F ÷ 2 which makes ¾×B = F


  

Solution

Substitute 2×B for A, B for C, 0 for D, 1½×B for E, and ¾×B for F in eq.1: 2×B + B + B + 0 + 1½×B + ¾×B = 25 which simplifies to 6¼×B = 25 Divide both sides of the above equation by 6¼: 6¼×B ÷ 6¼ = 25 ÷ 6¼ which means B = 4 making A = 2×B = 2×4 = 8 C = B = 4 E = 1½×B = 1½×4 = 6 F = ¾×B = ¾×4 = 3 and ABCDEF = 844063