Puzzle for February 18, 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.
Scratchpad
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Hint #1
In eq.5, replace E with A + F (from eq.2): C + A + F = B + D + F Subtract F from each side of the above equation: C + A + F - F = B + D + F - F which becomes C + A = B + D which is the same as eq.5a) A + C = B + D
Hint #2
In eq.3, replace A + C with B + D (from eq.5a): B + F = B + D Subtract B from each side of the above equation: B + F - B = B + D - B which makes F = D
Hint #3
eq.6 may be written as: B = (A + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × B = 4 × (A + C + D + F) ÷ 4 which becomes eq.6a) 4×B = A + C + D + F
Hint #4
eq.3 may be written as: A + C = B + F Add the left and right sides of eq.4 to the left and right sides of the above equation, respectively: A + C + A - C = B + F + C + D which becomes 2×A = B + F + C + D Subtract B from both sides: 2×A - B = B + F + C + D - B which becomes 2×A - B = F + C + D which may be written as eq.4a) 2×A - B = C + D + F
Hint #5
In eq.6a, substitute 2×A - B for C + D + F (from eq.4a): 4×B = A + 2×A - B which becomes 4×B = 3×A - B Add B to both sides of the above equation: 4×B + B = 3×A - B + B which makes 5×B = 3×A Divide both sides by 3: 5×B ÷ 3 = 3×A ÷ 3 which makes eq.6b) 1⅔×B = A
Hint #6
Substitute D for F, and 1⅔×B for A in eq.3: B + D = 1⅔×B + C Subtract B from both sides of the above equation: B + D - B = 1⅔×B + C - B which becomes eq.3a) D = ⅔×B + C
Hint #7
Substitute 1⅔×B for A, and ⅔×B + C for D (from eq.3a) in eq.4: 1⅔×B - C = C + ⅔×B + C which becomes 1⅔×B - C = 2×C + ⅔×B In the above equation, add C to both sides, and subtract ⅔×B from both sides: 1⅔×B - C + C - ⅔×B = 2×C + ⅔×B + C - ⅔×B which makes B = 3×C
Hint #8
Substitute (3×C) for B in eq.6b: 1⅔×(3×C) = A which makes 5×C = A
Hint #9
Substitute (3×C) for B in eq.3a: D = ⅔×(3×C) + C which becomes D = 2×C + C which makes D = 3×C and also makes F = D = 3×C
Hint #10
Substitute 5×C for A, and 3×C for F in eq.2: E = 5×C + 3×C which makes E = 8×C
Solution
Substitute 5×C for A, 3×C for B and D and F, and 8×C for E in eq.1: 5×C + 3×C + C + 3×C + 8×C + 3×C = 23 which simplifies to 23×C = 23 Divide both sides of the above equation by 23: 23×C ÷ 23 = 23 ÷ 23 which means C = 1 making A = 5×C = 5 × 1 = 5 B = D = F = 3×C = 3 × 1 = 3 E = 8×C = 8 × 1 = 8 and ABCDEF = 531383