Puzzle for August 11, 2023 ( )
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
eq.5 may be written as: F = (A + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (A + C + E) ÷ 3 which becomes eq.5a) 3×F = A + C + E
Hint #2
In eq.5a, replace E with C + F (from eq.2): 3×F = A + C + C + F which becomes 3×F = A + 2×C + F Subtract F from each side of the above equation: 3×F - F = A + 2×C + F - F which becomes eq.5b) 2×F = A + 2×C
Hint #3
Add F and A to both sides of eq.4: D - F + F + A = F - A + F + A which becomes eq.4a) D + A = 2×F
Hint #4
In eq.5b, replace 2×F with D + A (from eq.4a): D + A = A + 2×C Subtract A from each side of the equation above: D + A - A = A + 2×C - A which makes D = 2×C
Hint #5
In eq.3, substitute 2×C for D: B + C = 2×C + F Subtract C from each side of the above equation: B + C - C = 2×C + F - C which becomes eq.3a) B = C + F
Hint #6
Substitute E for C + F (from eq.2) in eq.3a: B = E
Hint #7
Subtract 2×C from both sides of eq.5b: 2×F - 2×C = A + 2×C - 2×C which becomes eq.5c) 2×F - 2×C = A
Hint #8
Substitute 2×F - 2×C for A (from eq.5c), and C + F for B (from eq.3a) and for E (from eq.2) in eq.6: C × F = 2×F - 2×C + C + F + C + F which becomes C × F = 4×F Divide both sides of the above equation by F: (C × F) ÷ F = 4×F ÷ F which makes C = 4 and makes D = 2×C = 2×4 = 8
Hint #9
Substitute 4 for C in eq.5c: 2×F - 2×4 = A which makes eq.5d) 2×F - 8 = A
Hint #10
Substitute 4 for C in eq.2: E = 4 + F which also means eq.2a) B = E = 4 + F
Hint #11
Substitute 2×F - 8 for A (from eq.5d), 4 + F for B and E (from eq.2a), 4 for C, and 8 for D in eq.1: 2×F - 8 + 4 + F + 4 + 8 + 4 + F + F = 37 which simplifies to 5×F + 12 = 37 Subtract 12 from both sides of the equation above: 5×F + 12 - 12 = 37 - 12 which makes 5×F = 25 Divide both sides by 5: 5×F ÷ 5 = 25 ÷ 5 which means F = 5
Solution
Since F = 5, then: A = 2×F - 8 = 2×5 - 8 = 10 - 8 = 2 (from eq.5d) B = E = 4 + F = 4 + 5 = 9 (from eq.2a) and ABCDEF = 294895