Puzzle for August 28, 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
Add B to both sides of eq.3: B + F + B = A – B + E + B which becomes eq.3a) 2×B + F = A + E Add A to both sides of eq.2: D + F + A = B + E + A which may be written as eq.2a) A + D + F = B + A + E
Hint #2
In eq.2a, replace A + E with 2×B + F (from eq.3a): A + D + F = B + 2×B + F which becomes A + D + F = 3×B + F Subtract F from each side of the equation above: A + D + F – F = 3×B + F – F which becomes eq.2b) A + D = 3×B
Hint #3
eq.5 may be written as: B = (A + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × (A + C + D) ÷ 3 which becomes 3×B = A + C + D which may be written as eq.5a) 3×B = A + D + C
Hint #4
In eq.5a, replace A + D with 3×B (from eq.2b): 3×B = 3×B + C Subtract 3×B from each side of the above equation: 3×B – 3×B = 3×B + C – 3×B which means 0 = C
Hint #5
In eq.4, substitute 2×B + F for A + E (from eq.3a), and 0 for C: 2×B + F + F – 0 = B + D – F + 0 which becomes 2×B + 2×F = B + D – F In the equation above, subtract B from both sides, and add F to both sides: 2×B + 2×F – B + F = B + D – F – B + F which becomes eq.4a) B + 3×F = D
Hint #6
Substitute B + 3×F for D (from eq.4a) in eq.2b: A + B + 3×F = 3×B Subtract B from each side of the above equation: A + B + 3×F – B = 3×B – B which becomes eq.2c) A + 3×F = 2×B
Hint #7
Substitute A + 3×F for 2×B (from eq.2c) into eq.3a: A + 3×F + F = A + E which becomes A + 4×F = A + E Subtract A from both sides of the equation above: A + 4×F – A = A + E – A which makes 4×F = E
Hint #8
eq.6 may be written as: E = (A + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + C + D + F) ÷ 4 which becomes 4×E = A + C + D + F which may be written as eq.6a) 4×E = A + D + C + F
Hint #9
In eq.6a, substitute (4×F) for E, 3×B for A + D (from eq.2b), and 0 for C: 4×(4×F) = 3×B + 0 + F which becomes 16×F = 3×B + F Subtract F from each side of the above equation: 16×F – F = 3×B + F – F which makes 15×F = 3×B Divide both sides by 3: 15×F ÷ 3 = 3×B ÷ 3 which makes 5×F = B
Hint #10
Substitute 5×F for B in eq.4a: 5×F + 3×F = D which makes 8×F = D
Hint #11
Substitute (5×F) for B in eq.2c: A + 3×F = 2×(5×F) which becomes A + 3×F = 10×F Subtract 3×F from both sides of the equation above: A + 3×F – 3×F = 10×F – 3×F which makes A = 7×F
Solution
Substitute 7×F for A, 5×F for B, 0 for C, 8×F for D, and 4×F for E in eq.1: 7×F + 5×F + 0 + 8×F + 4×F + F = 25 which simplifies to 25×F = 25 Divide both sides of the above equation by 25: 25×F ÷ 25 = 25 ÷ 25 which means F = 1 making A = 7×F = 7 × 1 = 7 B = 5×F = 5 × 1 = 5 D = 8×F = 8 × 1 = 8 E = 4×F = 4 × 1 = 4 and ABCDEF = 750841