Puzzle for March 31, 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.4, replace A with B + C (from eq.3): C + F = B + C + B which becomes C + F = 2×B + C Subtract C from both sides of the above equation: C + F – C = 2×B + C – C which makes eq.4a) F = 2×B
Hint #2
In eq.6, replace A with B + C (from eq.3): E – B = B + C + B – C which becomes E – B = 2×B Add B to both sides of the above equation: E – B + B = 2×B + B which makes eq.6a) E = 3×B
Hint #3
In eq.2, substitute 2×B for F, and 3×B for E in eq.2: D – 2×B = B – 3×B which becomes D – 2×B = –2×B Add 2×B to both sides of the above equation: D – 2×B + 2×B = –2×B + 2×B which makes D = 0
Hint #4
Substitute 3×B for E, B + C for A (from eq.3), and 2×B for F in eq.5: B – C + 3×B = B + C + C + 2×B which becomes 4×B – C = 3×B + 2×C In the equation above, add C to both sides, and subtract 3×B from both sides: 4×B – C + C – 3×B = 3×B + 2×C + C – 3×B which makes B = 3×C
Hint #5
Substitute (3×C) for B in eq.4a: F = 2×(3×C) which makes F = 6×C
Hint #6
Substitute (3×C) for B in eq.6a: E = 3×(3×C) which makes E = 9×C
Hint #7
Substitute 3×C for B in eq.3: A = 3×C + C which makes A = 4×C
Solution
Substitute 4×C for A, 3×C for B, 0 for D, 9×C for E, and 6×C for F in eq.1: 4×C + 3×C + C + 0 + 9×C + 6×C = 23 which simplifies to 23×C = 23 Divide both sides of the equation above by 23: 23×C ÷ 23 = 23 ÷ 23 which means C = 1 making A = 4×C = 4 × 1 = 4 B = 3×C = 3 × 1 = 3 E = 9×C = 9 × 1 = 9 F = 6×C = 6 × 1 = 6 and ABCDEF = 431096