Puzzle for July 16, 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 C to both sides of eq.5: A + C + C = B – C + C which becomes eq.5a) A + 2×C = B In eq.2, replace F with A – C (from eq.3), and replace B with A + 2×C (from eq.5a): D + A – C = A + 2×C + C which becomes D + A – C = A + 3×C In the above equation, subtract A from both sides, and add C to both sides: D + A – C – A + C = A + 3×C – A + C which makes D = 4×C
Hint #2
In eq.4, replace B with A + 2×C (from eq.5a), and D with 4×C: A + 2×C + F = A + C + 4×C + E which becomes A + 2×C + F = A + 5×C + E Subtract A and 2×C from each side of the equation above: A + 2×C + F – A – 2×C = A + 5×C + E – A – 2×C which becomes eq.4a) F = 3×C + E
Hint #3
In eq.3, substitute 3×C + E for F (from eq.4a): 3×C + E = A – C Add C to both sides of the above equation: 3×C + E + C = A – C + C which becomes eq.3a) 4×C + E = A
Hint #4
Substitute 4×C for D, and 4×C + E for A in eq.6: B + 4×C – E = 4×C + E + E which becomes B + 4×C – E = 4×C + 2×E In the equation above, subtract 4×C from both sides, and add E to both sides: B + 4×C – E – 4×C + E = 4×C + 2×E – 4×C + E which simplifies to eq.6a) B = 3×E
Hint #5
Substitute 4×C + E for A (from eq.3a), and 3×E for B in eq.5: 4×C + E + C = 3×E – C which becomes 5×C + E = 3×E – C In the above equation, subtract E from each side, and add C to each side: 5×C + E – E + C = 3×E – C – E + C which becomes 6×C = 2×E Divide both sides by 2: 6×C ÷ 2 = 2×E ÷ 2 which makes 3×C = E
Hint #6
Substitute (3×C) for E in eq.6a: B = 3×(3×C) which makes B = 9×C
Hint #7
Substitute 3×C for E in eq.3a: 4×C + 3×C = A which makes 7×C = A
Hint #8
Substitute 3×C for E in eq.4a: F = 3×C + 3×C which makes F = 6×C
Solution
Substitute 7×C for A, 9×C for B, 4×C for D, 3×C for E, and 6×C for F in eq.1: 7×C + 9×C + C + 4×C + 3×C + 6×C = 30 which simplifies to 30×C = 30 Divide both sides of the equation above by 30: 30×C ÷ 30 = 30 ÷ 30 which means C = 1 making A = 7×C = 7 × 1 = 7 B = 9×C = 9 × 1 = 9 D = 4×C = 4 × 1 = 4 E = 3×C = 3 × 1 = 3 F = 6×C = 6 × 1 = 6 and ABCDEF = 791436