Puzzle for July 3, 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
In eq.2, add F and B to both sides, and subtract C from each side: D – F + F + B – C = A – B + C + F + B – C which becomes B – C + D = A + F In eq.3, replace A + F with B – C + D: B + C + D = B – C + D Subtract B and D from each side of the above equation: B + C + D – B – D = B – C + D – B – D which simplifies to C = –C which makes C = 0
Hint #2
In eq.5, replace C with 0: 0 + D = A + E which becomes eq.5a) D = A + E
Hint #3
In eq.3, substitute 0 for C, and A + E for D (from eq.5a): B + 0 + A + E = A + F Subtract A from each side of the equation above: B + 0 + A + E – A = A + F – A which becomes eq.3a) B + E = F
Hint #4
Substitute B + E for F (from eq.3a) in eq.6: A – B + D = B + B + E which becomes A – B + D = 2×B + E Add B to both sides of the above equation: A – B + D + B = 2×B + E + B which becomes eq.6a) A + D = 3×B + E
Hint #5
Substitute A + E for D (from eq.5a) in eq.6a: A + A + E = 3×B + E Subtract E from both sides of the equation above: A + A + E – E = 3×B + E – E which becomes 2×A = 3×B Divide both sides by 2: 2×A ÷ 2 = 3×B ÷ 2 which makes eq.6b) A = 1½×B
Hint #6
Substitute 1½×B for A in eq.6a: 1½×B + D = 3×B + E Subtract 1½×B from each side of the equation above: 1½×B + D – 1½×B = 3×B + E – 1½×B which becomes eq.6c) D = 1½×B + E
Hint #7
Substitute 1½×B + E for D (from eq.6c), and B + E for F (from eq.3a) in eq.4: 1½×B + E + E = B – E + B + E which becomes 1½×B + 2×E = 2×B Subtract 1½×B from each side of the above equation: 1½×B + 2×E – 1½×B = 2×B – 1½×B which makes 2×E = ½×B Multiply both sides by 2: 2×E × 2 = ½×B × 2 which makes 4×E = B
Hint #8
Substitute (4×E) for B in eq.6b: A = 1½×(4×E) which makes A = 6×E
Hint #9
Substitute (4×E) for B in eq.6c: D = 1½×(4×E) + E which becomes D = 6×E + E which makes D = 7×E
Hint #10
Substitute 4×E for B in eq.3a: 4×E + E = F which makes 5×E = F
Solution
Substitute 6×E for A, 4×E for B, 0 for C, 7×E for D, and 5×E for F in eq.1: 6×E + 4×E + 0 + 7×E + E + 5×E = 23 which simplifies to 23×E = 23 Divide both sides of the above equation by 23: 23×E ÷ 23 = 23 ÷ 23 which means E = 1 making A = 6×E = 6 × 1 = 6 B = 4×E = 4 × 1 = 4 D = 7×E = 7 × 1 = 7 F = 5×E = 5 × 1 = 5 and ABCDEF = 640715