Puzzle for February 7, 2020 ( )
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 D and E to both sides of eq.3: C – E + D + E = A – D + D + E which becomes eq.3a) C + D = A + E In eq.2, replace A + E with C + D (from eq.3a): B + C = C + D Subtract C from both sides of the equation above: B + C – C = C + D – C which makes B = D
Hint #2
In eq.4, replace D with B: A + B – C = C + B + E In the equation above, subtract B from both sides, and add C to both sides: A + B – C – B + C = C + B + E – B + C which becomes eq.4a) A = 2×C + E
Hint #3
In eq.2, substitute 2×C + E for A (from eq.4a): B + C = 2×C + E + E which becomes B + C = 2×C + 2×E Subtract C from each side of the above equation: B + C – C = 2×C + 2×E – C which makes B = C + 2×E and also makes eq.2a) D = B = C + 2×E
Hint #4
Substitute E + F for D (from eq.5) in eq.3a: C + E + F = A + E Subtract C and E from both sides of the above equation: C + E + F – C – E = A + E – C – E which becomes eq.3b) F = A – C
Hint #5
Substitute A – C for F (from eq.3b) in eq.6: D + E = A + C + A – C which becomes eq.6a) D + E = 2×A
Hint #6
Substitute C + 2×E for D (from eq.2a), and (2×C + E) for A (from eq.4a) in eq.6a: C + 2×E + E = 2×(2×C + E) which becomes C + 3×E = 4×C + 2×E Subtract 2×E and C from each side of the equation above: C + 3×E – 2×E – C = 4×C + 2×E – 2×E – C which becomes E = 3×C
Hint #7
Substitute 3×C for E in eq.4a: A = 2×C + 3×C which means A = 5×C
Hint #8
Substitute 3×C for E, and (5×C) for A in eq.6a: D + 3×C = 2×(5×C) which becomes D + 3×C = 10×C Subtract 3×C from each side of the above equation: D + 3×C – 3×C = 10×C – 3×C which makes D = 7×C and also makes B = D = 7×C
Hint #9
Substitute 5×C for A in eq.3b: F = 5×C – C which makes F = 4×C
Solution
Substitute 5×C for A, 7×C for B and D, 3×C for E, and 4×C for F in eq.1: 5×C + 7×C + C + 7×C + 3×C + 4×C = 27 which simplifies to 27×C = 27 Divide both sides of the above equation by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 5×C = 5 × 1 = 5 B = D = 7×C = 7 × 1 = 7 E = 3×C = 3 × 1 = 3 F = 4×C = 4 × 1 = 4 and ABCDEF = 571734