Puzzle for January 22, 2023 ( )
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.
* "E ^ D" means "E raised to the power of D".
** "sq.rt.(F)" means "square root of F".
Scratchpad
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Hint #1
Add C to both sides of eq.2: A + D + C = B - C + C which becomes A + D + C = B which may be written as eq.2a) A + C + D = B
Hint #2
In eq.2a, replace A + C with B + D (from eq.3): B + D + D = B which becomes B + 2×D = B Subtract B from each side of the equation above: B + 2×D - B = B - B which makes 2×D = 0 which means D = 0
Hint #3
eq.4 may be written as: E = (A + B + C + D + F) ÷ 5 Multiply both sides of the above equation by 5: 5 × E = 5 × (A + B + C + D + F) ÷ 5 which becomes eq.4a) 5×E = A + B + C + D + F
Hint #4
eq.1 may be written as: A + B + C + D + F + E = 30 In the above equation, replace A + B + C + D + F with 5×E (from eq.4a): 5×E + E = 30 which makes 6×E = 30 Divide both sides by 6: 6×E ÷ 6 = 30 ÷ 6 which makes E = 5
Hint #5
In eq.5, substitute 5 for E, and 0 for D: F - A = 5 ^ 0 which becomes F - A = 1 Add A to both sides of the equation above: F - A + A = 1 + A which makes eq.5a) F = 1 + A
Hint #6
Substitute 5 for E in eq.6: eq.6a) A - 5 = sq.rt.(F) Square both sides of eq.6a: (A - 5) × (A - 5) = sq.rt.(F) × sq.rt.(F) which becomes A² - 5×A - 5×A + 25 = F which becomes eq.6b) A² - 10×A + 25 = F
Hint #7
In eq.6b, substitute 1 + A for F (from eq.5a): A² - 10×A + 25 = 1 + A Subtract 1 and A from each side of the equation above: A² - 10×A + 25 - 1 - A = 1 + A - 1 - A which becomes eq.6c) A² - 11×A + 24 = 0
Hint #8
eq.6c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for A in eq.6c yields: A = { (-1)×(-11) ± sq.rt.[(-11)² - (4 × (1) × (24))] } ÷ (2 × (1)) which becomes A = {11 ± sq.rt.(121 - 96)} ÷ 2 which becomes A = {11 ± sq.rt.(25)} ÷ 2 which becomes A = (11 ± 5) ÷ 2 In the above equation, either A = (11 + 5) ÷ 2 = 16 ÷ 2 = 8 or A = (11 - 5) ÷ 2 = 6 ÷ 2 = 3
Hint #9
Check: A = 3 ... Substituting 3 for A in eq.6a would yield: 3 - 5 = sq.rt.(F) which would make -2 = sq.rt.(F) Since the square root of any real number cannot be negative, then: -2 ≠ sq.rt.(F) which means A ≠ 3 and therefore makes A = 8
Hint #10
Substitute 8 for A in eq.5a: F = 1 + 8 which makes F = 9
Hint #11
Substitute 8 for A, and 0 for D in eq.2: 8 + 0 = B - C which becomes eq.2b) 8 = B - C
Hint #12
Substitute 5 for E, 8 for A, 0 for D, and 9 for F in eq.4a: 5×5 = 8 + B + C + 0 + 9 which becomes 25 = 17 + B + C Subtract 17 from each side of the above equation: 25 - 17 = 17 + B + C - 17 which becomes eq.4b) 8 = B + C
Hint #13
Substitute B + C for 8 (from eq.4b) into eq.2b: B + C = B - C In the above equation, subtract B from both sides, and add C to both sides: B + C - B + C = B - C - B + C which makes 2×C = 0 which means C = 0
Solution
Substitute 0 for C in eq.4b: 8 = B + 0 which makes 8 = B and makes ABCDEF = 880059