Puzzle for January 12, 2021 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
Add F and D to both sides of eq.5: C – F + D + F = A – D + D + F which becomes C + D = A + F In the above equation, replace A + F with C + E (from eq.3): C + D = C + E Subtract C from both sides: C + D – C = C + E – C which makes D = E
Hint #2
In eq.2, replace E with D: F = D + D which makes F = 2×D
Hint #3
In eq.4, substitute D for E: B + D = C + D Subtract D from both sides of the equation above: B + D – D = C + D – D which makes B = C
Hint #4
Substitute B for C, and 2×D for F in eq.5: B – 2×D = A – D Add 2×D to both sides of the above equation: B – 2×D + 2×D = A – D + 2×D which becomes eq.5a) B = A + D
Hint #5
Substitute D for E, 2×D for F, and A + D for B (from eq.5a) in eq.1: D + D + 2×D = A + A + D which becomes 4×D = 2×A + D Subtract D from each side of the equation above: 4×D – D = 2×A + D – D which makes 3×D = 2×A Divide both sides by 2: 3×D ÷ 2 = 2×A ÷ 2 which makes 1½×D = A
Hint #6
Substitute 1½×D for A in eq.5a: B = 1½×D + D which makes B = 2½×D and also makes C = B = 2½×D
Solution
Substitute (1½×D) for A, (2×D) for F, and 2½×D for both B and C in eq.6: (1½×D)×(2×D) = 2½×D + 2½×D + D which becomes 3×D² = 6×D Divide both sides of the equation above by 3×D: 3×D² ÷ 3×D = 6×D ÷ 3×D which makes D = 2 making A = 1½×D = 1½ × 2 = 3 B = C = 2½×D = 2½ × 2 = 5 D = E = 2 F = 2×D = 2 × 2 = 4 and ABCDEF = 355224