Puzzle for July 10, 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 non-negative integer.
* "B ^ A" means "B raised to the power of A". "D ^ C" means "D raised to the power of C".
Scratchpad
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Hint #1
In eq.5, replace D with A + E (from eq.3): A + F = A + E + E which becomes A + F = A + 2×E Subtract A from each side of the above equation: A + F – A = A + 2×E – A which makes F = 2×E
Hint #2
In eq.2, replace F with 2×E: 2×E = A + B + E Subtract E from each side of the equation above: 2×E – E = A + B + E – E which becomes eq.2a) E = A + B
Hint #3
In eq.4, substitute E for A + B (from eq.2a): E = E + C Subtract E from each side of the equation above: E – E = E + C – E which becomes 0 = C
Hint #4
Substitute 0 for C in eq.6: B ^ A = D ^ 0 which makes eq.6a) B ^ A = 1 (implies D > 0, and B > 0) To make eq.6a true, then: B = 1 or A = 0
Hint #5
Begin checking: A = 0 ... Substituting 0 for A in eq.3 would yield: D = 0 + E which would become D = E Substituting 0 for A and C in eq.4 would yield: E = 0 + B + 0 which would become E = B
Hint #6
Finish checking: A = 0 ... Substituting 0 for A and C, E for B and D, and 2×E for F in eq.1 would yield: 0 + E + 0 + E + E + 2×E = 17 which would become 5×E = 17 Dividing both sides of the above equation by 5 would yield: 5×E ÷ 5 = 17 ÷ 5 which would make E = 3⅖ Since E must be an integer, then E ≠ 3⅖ which means A ≠ 0 and therefore means B = 1
Hint #7
Substitute 1 for B in eq.2a: E = A + 1 Subtract 1 from both sides of the equation above: E – 1 = A + 1 – 1 which becomes eq.2b) E – 1 = A
Hint #8
Substitute E – 1 for A (from eq.2b) in eq.3: D = E – 1 + E which becomes eq.3a) D = 2×E – 1
Solution
Substitute E – 1 for A (from eq.2b), 1 for B, 0 for C, 2×E – 1 for D (from eq.3a), and 2×E for F in eq.1: E – 1 + 1 + 0 + 2×E – 1 + E + 2×E = 17 which simplifies to 6×E – 1 = 17 Add 1 to both sides of the equation above: 6×E – 1 + 1 = 17 + 1 which makes 6×E = 18 Divide both sides by 6: 6×E ÷ 6 = 18 ÷ 6 which means E = 3 making A = E – 1 = 3 – 1 = 2 (from eq.2b) D = 2×E – 1 = 2×3 – 1 = 6 – 1 = 5 (from eq.3a) F = 2×E = 2×3 = 6 and ABCDEF = 210536