Puzzle for July 26, 2020 ( )
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.
* "A ^ ((A ÷ E) ÷ B)" means "A raised to the power of ((A ÷ E) ÷ B)".
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Hint #1
eq.4 may be written as: A + E = C + F + D In the equation above, replace C + F with A + B (from eq.2): A + E = A + B + D Subtract A from each side: A + E – A = A + B + D – A which becomes eq.4a) E = B + D
Hint #2
In eq.1, replace B + D with E (from eq.4a): E = A
Hint #3
Add E and D to each side of eq.3: D – E + E + D = C – D + E + D which becomes 2×D = C + E In the equation above, substitute B + D for E (from eq.4a): 2×D = C + B + D Subtract D from each side of the above equation: 2×D – D = C + B + D – D which becomes eq.3a) D = C + B
Hint #4
Substitute C + B for D (from eq.3a) in eq.1: B + C + B = A which becomes eq.1a) 2×B + C = A
Hint #5
Substitute 2×B + C for A (from eq.1a) in eq.2: 2×B + C + B = C + F which becomes 3×B + C = C + F Subtract C from both sides of the above equation: 3×B + C – C = C + F – C which becomes 3×B = F
Hint #6
Multiply both sides of eq.5 by B: B × (B + (C ÷ B)) = B × (A ÷ B) which becomes B² + C = A Subtract B² from both sides of the equation above: B² + C – B² = A – B² which makes eq.5a) C = A – B²
Hint #7
Substitute A – B² for C (from eq.5a) in eq.1a: 2×B + A – B² = A In the equation above, subtract A from each side, and add B² to each side: 2×B + A – B² – A + B² = A – A + B² which simplifies to 2×B = B² Divide both sides by B: 2×B ÷ B = B² ÷ B which makes 2 = B making F = 3×B = 3 × 2 = 6
Hint #8
In eq.6, substitute 2 for B, and A for E: C – 2 = A ^ ((A ÷ A) ÷ 2) which becomes C – 2 = A ^ (1 ÷ 2) which may be written as C – 2 = A ^ ½ Square both sides of the above equation: (C – 2) × (C – 2) = (A ^ ½) × (A ^ ½) which becomes C² – 2×C – 2×C + 4 = A ^ (½ + ½) which becomes C² – 4×C + 4 = A ^ 1 which is equivalent to eq.6a) C² – 4×C + 4 = A
Hint #9
Substitute 2 for B in eq.1a: 2×2 + C = A which makes 4 + C = A and also makes eq.1b) E = A = 4 + C
Solution
Substitute 4 + C for A (from eq.1b) in eq.6a: C² – 4×C + 4 = 4 + C In the above equation, subtract 4 from each side, and add 4×C to each side: C² – 4×C + 4 – 4 + 4×C = 4 + C – 4 + 4×C which makes C² = 5×C Divide both sides by C: C² ÷ C = 5×C ÷ C which means C = 5 making A = E = 4 + C = 4 + 5 = 9 (from eq.1b) D = C + B = 5 + 2 = 7 (from eq.3a) and ABCDEF = 925796