Puzzle for June 1, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
eq.6 may be written as: A = (C + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (C + D + E) ÷ 3 which becomes eq.6a) 3×A = C + D + E
Hint #2
In eq.6a, replace D + E with A + B – D (from eq.4): 3×A = C + A + B – D Subtract A from both sides of the above equation: 3×A – A = C + A + B – D – A which becomes 2×A = C + B – D which may be written as eq.6b) 2×A = B + C – D
Hint #3
In eq.6b, substitute A + D + E for B + C (from eq.3): 2×A = A + D + E – D which becomes 2×A = A + E Subtract A from each side of the equation above: 2×A – A = A + E – A which makes A = E
Hint #4
Substitute A for E in eq.2: B + A = A + C Subtract A from each side of the above equation: B + A – A = A + C – A which makes B = C
Hint #5
Substitute A for E into eq.4: D + A = A + B – D In the equation above, subtract A from both sides, and add D to both sides: D + A – A + D = A + B – D – A + D which makes 2×D = B and also makes C = B = 2×D
Hint #6
In eq.6b, substitute 2×D for both B and C: 2×A = 2×D + 2×D – D which makes 2×A = 3×D Divide both sides of the above equation by 2: 2×A ÷ 2 = 3×D ÷ 2 which makes A = 1½×D and also makes A = E = 1½×D
Hint #7
In eq.5, substitute 2×D for both B and C, and 1½×D for E: 2×D + 2×D + 1½×D – F = D + F which becomes 5½×D – F = D + F In the equation above, add F to both sides, and subtract D from both sides: 5½×D – F + F – D = D + F + F – D which becomes 4½×D = 2×F Divide both sides by 2: 4½×D ÷ 2 = 2×F ÷ 2 which makes 2¼×D = F
Solution
Substitute 1½×D for A and E, 2×D for B and C, and 2¼×D for F in eq.1: 1½×D + 2×D + 2×D + D + 1½×D + 2¼×D = 41 which simplifies to 10¼×D = 41 Divide both sides of the above equation by 10¼: 10¼×D ÷ 10¼ = 41 ÷ 10¼ which means D = 4 making A = E = 1½×D = 1½ × 4 = 6 B = C = 2×D = 2 × 4 = 8 F = 2¼×D = 2¼ × 4 = 9 and ABCDEF = 688469