Puzzle for April 29, 2023 ( )
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 with B + C + E (from eq.2): B + C + E - C = C + E + F which becomes B + E = C + E + F Subtract E from each side of the equation above: B + E - E = C + E + F - E which becomes eq.5a) B = C + F
Hint #2
In eq.4, substitute (C + F) for B (from eq.5a): D - (C + F) = (C + F) + E which becomes D - C - F = C + F + E Add C and F to both sides of the above equation: D - C - F + C + F = C + F + E + C + F which becomes eq.4a) D = 2×C + 2×F + E
Hint #3
eq.6 may be written as: C + E = (A + B + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (C + E) = 3 × (A + B + F) ÷ 3 which becomes eq.6a) 3×C + 3×E = A + B + F
Hint #4
Add C to both sides of eq.5: A - C + C = C + E + F + C which becomes eq.5b) A = 2×C + E + F
Hint #5
In eq.6a, replace A with 2×C + E + F (from eq.5b), and B with C + F (from eq.5a): 3×C + 3×E = 2×C + E + F + C + F + F which becomes 3×C + 3×E = 3×C + E + 3×F Subtract 3×C and E from both sides of the above equation: 3×C + 3×E - 3×C - E = 3×C + E + 3×F - 3×C - E which simplifies to 2×E = 3×F Divide both sides by 2: 2×E ÷ 2 = 3×F ÷ 2 which makes eq.6b) E = 1½×F
Hint #6
In eq.5b, substitute 1½×F for E: A = 2×C + 1½×F + F which becomes eq.5c) A = 2×C + 2½×F
Hint #7
In eq.4a, substitute 1½×F for E: D = 2×C + 2×F + 1½×F which becomes eq.4b) D = 2×C + 3½×F
Hint #8
Substitute 2×C + 3½×F for D (from eq.4b), 2×C + 2½×F for A (from eq.5c), and 1½×F for E in eq.3: C + 2×C + 3½×F = 2×C + 2½×F + 1½×F which becomes 3×C + 3½×F = 2×C + 4×F Subtract 3½×F and 2×C from each side of the above equation: 3×C + 3½×F - 3½×F - 2×C = 2×C + 4×F - 3½×F - 2×C which makes C = ½×F Multiply both sides by 2: 2 × C = 2 × ½×F which makes 2×C = F
Hint #9
Substitute (2×C) for F in eq.5c: A = 2×C + 2½×(2×C) which becomes A = 2×C + 5×C which makes A = 7×C
Hint #10
Substitute 2×C for F in eq.5a: B = C + 2×C which makes B = 3×C
Hint #11
Substitute (2×C) for F in eq.4b: D = 2×C + 3½×(2×C) which becomes D = 2×C + 7×C which makes D = 9×C
Hint #12
Substitute (2×C) for F in eq.6b: E = 1½×(2×C) which makes E = 3×C
Solution
Substitute 7×C for A, 3×C for B and E, 9×C for D, and 2×C for F in eq.1: 7×C + 3×C + C + 9×C + 3×C + 2×C = 25 which simplifies to 25×C = 25 Divide both sides of the above equation by 25: 25×C ÷ 25 = 25 ÷ 25 which means C = 1 making A = 7×C = 7 × 1 = 7 B = E = 3×C = 3 × 1 = 3 D = 9×C = 9 × 1 = 9 F = 2×C = 2 × 1 = 2 and ABCDEF = 731932