Puzzle for July 19, 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.
* "C ^ D" means "C raised to the power of D". "E ^ A" means "E raised to the power of A".
Scratchpad
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Hint #1
Add D to both sides of eq.4: B – D + F + D = A + C + E + D which becomes B + F = A + C + E + D which may be written as eq.4a) B + F = A + C + D + E Add C, E, and F to both sides of eq.1: B – C – E + C + E + F = A – F + C + E + F which becomes eq.1a) B + F = A + C + E
Hint #2
In eq.4a, replace A + C + E for B + F (from eq.1a): A + C + E = A + C + D + E Subtract A, C, and E from both sides of the equation above: A + C + E – A – C – E = A + C + D + E – A – C – E which simplifies to 0 = D
Hint #3
In eq.3, replace D with 0: A – 0 = C + 0 which makes A = C
Hint #4
In eq.5, substitute 0 for D: F = C ^ 0 which makes F = 1
Hint #5
Substitute A for C, and 1 for F in eq.2: A + E + 1 = B – 1 Add 1 to both sides of the equation above: A + E + 1 + 1 = B – 1 + 1 which becomes eq.2a) A + E + 2 = B
Hint #6
Substitute 1 for F, and A for C in eq.1a: B + 1 = A + A + E B + 1 = 2×A + E Subtract 1 from each side of the equation above: B + 1 – 1 = 2×A + E – 1 which becomes eq.1b) B = 2×A + E – 1
Hint #7
Substitute 2×A + E – 1 for B (from eq.1b) in eq.2a: A + E + 2 = 2×A + E – 1 Subtract A and E from each side of the above equation: A + E + 2 – A – E = 2×A + E – 1 – A – E which makes 2 = A – 1 Add 1 to both sides: 2 + 1 = A – 1 + 1 which makes 3 = A and also makes C = A = 3
Hint #8
Substitute 3 for A in eq.2a: 3 + E + 2 = B which makes eq.2b) E + 5 = B
Hint #9
Substitute 3 for A and C, and E + 5 for B (from eq.2b) in eq.6: E ^ 3 = ((E + 5) × 3) + E which is equivalent to E³ = 3×E + 3×5 + E which becomes E³ = 4×E + 15 Subtract 4×E and 15 from both sides of the equation above: E³ – 4×E – 15 = 4×E + 15 – 4×E – 15 which becomes E³ – 4×E – 15 = 0 which may be written as eq.6a) E³ + 0×E² – 4×E – 15 = 0
Solution
eq.6a is a cubic equation in standard form. Using the cubic equation solution formula to solve for E in eq.6a yields: E = cu.rt.[(0 + 0 – (–15÷(2×1))) + sq.rt.[(0 + 0 – (–15÷(2×1)))² + (–4÷(3×1))³]] + cu.rt.[(0 + 0 – (–15÷(2×1))) – sq.rt.[(0 + 0 – (–15÷(2×1)))² + (–4÷(3×1))³]] – 0 which becomes E = cu.rt.[(0 – (–7½)) + sq.rt.[(0 – (–7½))² + (–1⅓)³]] + cu.rt.[(0 – (–7½)) – sq.rt.[(0 – (–7½))² + (–1⅓)³]] which becomes E = cu.rt.[7½ + sq.rt.[56.25 + (–2.37037037037)]] + cu.rt.[7½ – sq.rt.[56.25 + (–2.37037037037)]] which becomes E = cu.rt.[7½ + sq.rt.[53.87962962963]] + cu.rt.[7½ – sq.rt.[53.87962962963]] which becomes E = cu.rt.[7.5 + 7.3402744928] + cu.rt.[7.5 – 7.3402744928] which becomes E = cu.rt.[14.8402744928] + cu.rt.[0.1597255072] which becomes E = 2.4574271078 + 0.5425728922 which makes E = 3 making B = E + 5 = 3 + 5 = 8 (from eq.2b) and ABCDEF = 383031