Puzzle for January 14, 2021 ( )
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 A + B with C + E (from eq.4): E + F = C + E + D which becomes eq.5a) F = C + D
Hint #2
In eq.3, replace F with C + D (from eq.5a): C + D – D = C – E which becomes C = C – E Subtract C from both sides of the equation above: C – C = C – E – C which makes 0 = –E which means 0 = E
Hint #3
In eq.2, replace F with C + D (from eq.5a): B + C = A + D + C + D which becomes eq.2a) B = A + 2×D
Hint #4
In eq.4, substitute 0 for E, and A + 2×D for B (from eq.2a): C + 0 = A + A + 2×D which becomes eq.4a) C = 2×A + 2×D
Hint #5
Substitute 2×A + 2×D for C (from eq.4a), 0 for E, and A + 2×D for B (from eq.2a) in eq.6: 2×A + 2×D – 0 = A + 2×D + D which becomes 2×A + 2×D = A + 3×D Subtract A and 2×D from each side of the above equation: 2×A + 2×D – A – 2×D = A + 3×D – A – 2×D which simplifies to A = D
Hint #6
Substitute A for D in in eq.2a: B = A + 2×A which becomes B = 3×A
Hint #7
Substitute A for D in in eq.4a: C = 2×A + 2×A which becomes C = 4×A
Hint #8
Substitute 0 for E, 3×A for B, and A for D in eq.5: 0 + F = A + 3×A + A which makes F = 5×A
Solution
Substitute 3×A for B, 4×A for C, A for D, 0 for E, and 5×A for F in eq.1: A + 3×A + 4×A + A + 0 + 5×A = 14 which simplifies to 14×A = 14 Divide both sides of the above equation by 14: 14×A ÷ 14 = 14 ÷ 14 which means A = 1 making B = 3×A = 3 × 1 = 3 C = 4×A = 4 × 1 = 4 D = A = 1 F = 5×A = 5 × 1 = 5 and ABCDEF = 134105