Puzzle for January 23, 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 positive integer.
Scratchpad
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Hint #1
In eq.5, replace D with A + C (from eq.3): A + B + F = A + C – F In the above equation, subtract A from both sides, and add F to both sides: A + B + F – A + F = A + C – F – A + F which becomes eq.5a) B + 2×F = C
Hint #2
eq.5 may be written as: A + F + B = D – F In the equation above, replace A + F with C (from eq.4): C + B = D – F Add F to both sides: C + B + F = D – F + F which becomes eq.5b) C + B + F = D
Hint #3
In eq.5b, substitute B + 2×F for C (from eq.5a): B + 2×F + B + F = D which becomes eq.5c) 2×B + 3×F = D
Hint #4
Substitute E for A + B (from eq.2), and 2×B + 3×F for D (from eq.5c) in eq.5: E + F = 2×B + 3×F – F which becomes E + F = 2×B + 2×F Subtract F from each side of the equation above: E + F – F = 2×B + 2×F – F which becomes eq.5d) E = 2×B + F
Hint #5
Substitute 2×B + F for E (from eq.5d) in eq.2: A + B = 2×B + F Subtract B from each side of the above equation: A + B – B = 2×B + F – B which becomes eq.2a) A = B + F
Hint #6
Substitute B + F for A (from eq.2a), B + 2×F for C (from eq.5a), 2×B + 3×F for D (from eq.5c), and (2×B + F) for E (from eq.5d) in eq.6: B + F + (B × F) = B + 2×F + 2×B + 3×F – (2×B + F) which becomes B + F + (B × F) = 3×B + 5×F – 2×B – F which becomes B + F + (B × F) = B + 4×F Subtract B and F from each side of the above equation: B + F + (B × F) – B – F = B + 4×F – B – F which simplifies to B × F = 3×F Divide both sides by F: B × F ÷ F = 3×F ÷ F which makes B = 3
Hint #7
In eq.1, substitute B + F for A (from eq.2a), B + 2×F for C (from eq.5a), 2×B + 3×F for D (from eq.5c), and 2×B + F for E (from eq.5d): B + F + B + B + 2×F + 2×B + 3×F + 2×B + F + F = 29 which simplifies to eq.1a) 7×B + 8×F = 29
Solution
Substitute 3 for B in eq.1a: 7×3 + 8×F = 29 which becomes 21 + 8×F = 29 Subtract 21 from both sides of the equation above: 21 + 8×F – 21 = 29 – 21 which makes 8×F = 8 Divide both sides by 8: 8×F ÷ 8 = 8 ÷ 8 which means F = 1 making A = B + F = 3 + 1 = 4 (from eq.2a) C = B + 2×F = 3 + 2×1 = 3 + 2 = 5 (from eq.5a) D = 2×B + 3×F = 2×3 + 3×1 = 6 + 3 = 9 (from eq.5c) E = 2×B + F = 2×3 + 1 = 6 + 1 = 7 (from eq.5d) and ABCDEF = 435971