Puzzle for October 10, 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 non-negative integer.
* "C^E" means "C raised to the power of E".
Scratchpad
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Hint #1
Add A, F, and C to both sides of eq.3: C – A – F + A + F + C = D – C + A + F + C which becomes eq.3a) 2×C = D + A + F In eq.4, add D to both sides, and subtract F from both sides: A + D – F + D – F = C – D + F + D – F which becomes eq.4a) A + 2×D – 2×F = C
Hint #2
In eq.3a, substitute (A + 2×D – 2×F) for C (from eq.4a): 2×(A + 2×D – 2×F) = D + A + F which becomes 2×A + 4×D – 4×F = D + A + F In the equation above, add 4×F to both sides, and subtract 4×D and A from both sides: 2×A + 4×D – 4×F + 4×F – 4×D – A = D + A + F + 4×F – 4×D – A which simplifies to eq.3b) A = 5×F – 3×D
Hint #3
In eq.3a, replace A with 5×F – 3×D (from eq.3b): 2×C = D + 5×F – 3×D + F which becomes 2×C = 6×F – 2×D Divide both sides of the above equation by 2: 2×C ÷ 2 = (6×F – 2×D) ÷ 2 which becomes eq.3c) C = 3×F – D
Hint #4
In eq.2, substitute 5×F – 3×D for A (from eq.3b): D + E = 5×F – 3×D – B In the equation above, subtract D from both sides, and add B to both sides: D + E – D + B = 5×F – 3×D – B – D + B which becomes eq.2a) E + B = 5×F – 4×D
Hint #5
eq.1 may be written as: A + C + D + E + B + F = 21 In the above equation, substitute 5×F – 3×D for A (from eq.3b), 3×F – D for C (from eq.3c), and 5×F – 4×D for E + B (from eq.2a): 5×F – 3×D + 3×F – D + D + 5×F – 4×D + F = 21 which becomes 14×F – 7×D = 21 Divide both sides by 7: (14×F – 7×D) ÷ 7 = 21 ÷ 7 which becomes 2×F – D = 3 Add D to both sides, and subtract 3 from both sides: 2×F – D + D – 3 = 3 + D – 3 which makes eq.1a) 2×F – 3 = D
Hint #6
Substitute 2×F – 3 for D (from eq.1a) into eq.3c: C = 3×F – (2×F – 3) which is equivalent to C = 3×F – 2×F + 3 which becomes eq.3d) C = F + 3
Hint #7
Substitute 2×F – 3 for D (from eq.1a) into eq.3b: A = 5×F – 3×(2×F – 3) which becomes A = 5×F – 6×F + 9 which becomes eq.3e) A = 9 – F
Hint #8
eq.6 may be written as: F = (C + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × ((C + D + E) ÷ 3) which becomes eq.6a) 3×F = C + D + E
Hint #9
Substitute F + 3 for C (from eq.3d), and 2×F – 3 for D (from eq.1a) in eq.6a: 3×F = F + 3 + 2×F – 3 + E which becomes 3×F = 3×F + E Subtract 3×F from each side of the equation above: 3×F – 3×F = 3×F + E – 3×F which makes 0 = E
Hint #10
Substitute 0 for E, and (2×F – 3) for D (from eq.1a) in eq.2a: 0 + B = 5×F – 4×(2×F – 3) which becomes B = 5×F – 8×F + 12 which becomes eq.2b) B = 12 – 3×F
Solution
Substitute 0 for E, 12 – 3×F for B (from eq.2b), and 2×F – 3 for D (from eq.1a) in eq.5: C^0 = 12 – 3×F + 2×F – 3 – F which becomes 1 = 9 – 2×F (implies C ≠ 0) In the above equation, add 2×F to both sides, and subtract 1 from both sides: 1 + 2×F – 1 = 9 – 2×F + 2×F – 1 which makes 2×F = 8 Divide both sides by 2: 2×F ÷ 2 = 8 ÷ 2 which makes F = 4 and makes A = 9 – F = 9 – 4 = 5 (from eq.3e) B = 12 – 3×F = 12 – 3×4 = 12 – 12 = 0 (from eq.2b) C = F + 3 = 4 + 3 = 7 (from eq.3d) D = 2×F – 3 = 2×4 – 3 = 8 – 3 = 5 (from eq.1a) and ABCDEF = 507504