Puzzle for August 20, 2022 ( )
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
eq.5 may be written as: C = (A + B + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (A + B + F) ÷ 3 which becomes eq.5a) 3×C = A + B + F
Hint #2
eq.6 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 5×E = A + B + C + D + F which may be written as eq.6a) 5×E = A + B + F + C + D
Hint #3
In eq.6a, replace A + B + F with 3×C (from eq.5a): 5×E = 3×C + C + D which becomes 5×E = 4×C + D Subtract 4×C from each side of the above equation: 5×E – 4×C = 4×C + D – 4×C which becomes eq.6b) 5×E – 4×C = D
Hint #4
In eq.6b, substitute (C + F) for E (from eq.2): 5×(C + F) = 4×C + D which becomes 5×C + 5×F = 4×C + D Subtract 4×C from each side of the equation above: 5×C + 5×F – 4×C = 4×C + D – 4×C which becomes eq.6c) C + 5×F = D
Hint #5
In eq.4, substitute C + 5×F for D (from eq.6c), and (C + F) for E (from eq.2): C + 5×F – (C + F) = B + (C + F) which becomes C + 5×F – C – F = B + C + F which becomes 4×F = B + C + F Subtract C and F from each side of the above equation: 4×F – C – F = B + C + F – C – F which becomes eq.4a) 3×F – C = B
Hint #6
Substitute 3×F – C for B (from eq.4a) into eq.5a: 3×C = A + 3×F – C + F which becomes 3×C = A + 4×F – C In the above equation, subtract 4×F from both sides, and add C to both sides: 3×C – 4×F + C = A + 4×F – C – 4×F + C which becomes eq.5b) 4×C – 4×F = A
Hint #7
Substitute (3×F – C) for B (from eq.4a), C + 5×F for D (from eq.6c), and 4×C – 4×F for A (from eq.5b) in eq.3: (3×F – C) + C + 5×F = 4×C – 4×F – (3×F – C) which becomes 8×F = 4×C – 4×F – 3×F + C which becomes 8×F = 5×C – 7×F Add 7×F to both sides of the above equation: 8×F + 7×F = 5×C – 7×F + 7×F which makes 15×F = 5×C Divide both sides by 5: 15×F ÷ 5 = 5×C ÷ 5 which makes 3×F = C
Hint #8
Substitute (3×F) for C in eq.5b: 4×(3×F) – 4×F = A which becomes 12×F – 4×F = A which makes 8×F = A
Hint #9
Substitute 3×F for C in eq.4a: 3×F – 3×F = B which makes 0 = B
Hint #10
Substitute 3×F for C in eq.6c: 3×F + 5×F = D which makes 8×F = D
Hint #11
Substitute 3×F for C in eq.2: E = 3×F + F which makes E = 4×F
Solution
Substitute 8×F for A and D, 0 for B, 3×F for C, and 4×F for E in eq.1: 8×F + 0 + 3×F + 8×F + 4×F + F = 24 which simplifies to 24×F = 24 Divide both sides of the above equation by 24: 24×F ÷ 24 = 24 ÷ 24 which means F = 1 making A = D = 8×F = 8 × 1 = 8 C = 3×F = 3 × 1 = 3 E = 4×F = 4 × 1 = 4 and ABCDEF = 803841