Puzzle for December 11, 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.
* "B ^ E" means "B raised to the power of E".
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Hint #1
eq.4 may be written as: C = (A + B + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (A + B + E + F) ÷ 4 which becomes eq.4a) 4×C = A + B + E + F Add C, D, and E to both sides of eq.3: B – C – D + C + D + E = A + C – E + C + D + E which becomes eq.3a) B + E = A + 2×C + D
Hint #2
In eq.4a, replace B + E with A + 2×C + D (from eq.3a): 4×C = A + A + 2×C + D + F which becomes 4×C = 2×A + 2×C + D + F Subtract 2×C from each side of the equation above: 4×C – 2×C = 2×A + 2×C + D + F – 2×C which becomes eq.4b) 2×C = 2×A + D + F
Hint #3
eq.1 may be written as: A + B + E + F + C + D = 15 In the equation above, replace A + B + E + F with 4×C (from eq.4a): 4×C + C + D = 15 which becomes 5×C + D = 15 Subtract 5×C from both sides: 5×C + D – 5×C = 15 – 5×C which becomes eq.1a) D = 15 – 5×C
Hint #4
In eq.4b, substitute 15 – 5×C for D (from eq.1a): 2×C = 2×A + 15 – 5×C + F In the above equation, subtract 2×A and 15 from both sides, and add 5×C to both sides: 2×C – 2×A – 15 + 5×C = 2×A + 15 – 5×C + F – 2×A – 15 + 5×C which becomes eq.4c) 7×C – 2×A – 15 = F
Hint #5
Substitute A + 2×C + D for B + E (from eq.3a), and (7×C – 2×A – 15) for F (from eq.4c) into eq.5: A + 2×C + D – (7×C – 2×A – 15) = (A × C) + D which becomes A + 2×C + D – 7×C + 2×A + 15 = (A × C) + D which becomes 3×A – 5×C + D + 15 = (A × C) + D In the above equation, subtract D from both sides, and add 5×C to both sides: 3×A – 5×C + D + 15 – D + 5×C = (A × C) + D – D + 5×C which becomes 3×A + 15 = (A × C) + 5×C which may be written as 3×(A + 5) = C × (A + 5) Divide both sides by (A + 5): 3×(A + 5) ÷ (A + 5) = C × (A + 5) ÷ (A + 5) which makes 3 = C
Hint #6
Substitute 3 for C in eq.1a: D = 15 – 5×3 which becomes D = 15 – 15 which makes D = 0
Hint #7
Add A, B, and C to both sides of eq.2: D – A – C + A + B + C = C – B + A + B + C which becomes D + B = 2×C + A which may be written as eq.2a) D + B = A + 2×C
Hint #8
In eq.3a, substitute D + B for A + 2×C (from eq.2a): B + E = D + B + D which becomes B + E = 2×D + B Subtract B from each side of the equation above: B + E – B = 2×D + B – B which makes eq.3b) E = 2×D
Hint #9
Substitute 0 for D in eq.3b: E = 2×0 which means E = 0
Hint #10
Substitute 3 for C in eq.4c: 7×3 – 2×A – 15 = F which becomes 21 – 2×A – 15 = F which becomes eq.4d) 6 – 2×A = F
Hint #11
Substitute 0 for D, and 3 for C in eq.2a: 0 + B = A + 2×3 which makes eq.2b) B = A + 6
Solution
Substitute 3 for C, 0 for D and E, (6 – 2×A) for F (from eq.4d), and (A + 6) for B (from eq.2b) in eq.6: 3 + 0 – (6 – 2×A) = (A + 6) ^ 0 which becomes 3 – 6 + 2×A = 1 which becomes –3 + 2×A = 1 Add 3 to both sides of the equation above: –3 + 2×A + 3 = 1 + 3 which makes 2×A = 4 Divide both sides by 2: 2×A ÷ 2 = 4 ÷ 2 which means A = 2 making B = A + 6 = 2 + 6 = 8 (from eq.2b) F = 6 – 2×A = 6 – 2×2 = 6 – 4 = 2 (from eq.4d) and ABCDEF = 283002