Puzzle for May 20, 2022 ( )
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 B and E to both sides of eq.4: E – B + B + E = C + D – E + B + E which becomes eq.4a) 2×E = C + D + B Add C and D to both sides of eq.2: B – C + C + D = C – D + C + D which becomes B + D = 2×C which may be written as eq.2a) D + B = 2×C
Hint #2
In eq.4a, replace D + B with 2×C (from eq.2a): 2×E = C + 2×C which becomes 2×E = 3×C Divide both sides of the above equation by 2: 2×E ÷ 2 = 3×C ÷ 2 which makes eq.4b) E = 1½×C
Hint #3
Add A and E to both sides of eq.6: E + F – A + A + E = A + B + C – E + A + E which becomes 2×E + F = 2×A + B + C In the above equation, replace 2×E with C + D + B (from eq.4a): C + D + B + F = 2×A + B + C Subtract B and C from each side of the equation above: C + D + B + F – B – C = 2×A + B + C – B – C which simplifies to eq.6a) D + F = 2×A
Hint #4
Add D to both sides of eq.3: D – C + D = A – D + D which becomes 2×D – C = A In eq.6a, substitute (2×D – C) for A: D + F = 2×(2×D – C) which becomes D + F = 4×D – 2×C In the above equation, subtract D from both sides, and add 2×C to both sides: D + F – D + 2×C = 4×D – 2×C – D + 2×C which becomes eq.6b) F + 2×C = 3×D
Hint #5
In eq.6b, replace 2×C with D + B (from eq.2a): F + D + B = 3×D Subtract D from each side of the equation above: F + D + B – D = 3×D – D which becomes eq.6c) F + B = 2×D
Hint #6
Add D to both sides of eq.5: D – F + D = B – D + F + D which becomes 2×D – F = B + F Substitute F + B for 2×D (from eq.6c) in the above equation: F + B – F = B + F which becomes B = B + F Subtract B from each side: B – B = B + F – B which makes 0 = F
Hint #7
Substitute 0 for F in eq.6a: D + 0 = 2×A which makes D = 2×A
Hint #8
Substitute 0 for F, and (2×A) for D in eq.6b: 0 + 2×C = 3×(2×A) which makes 2×C = 6×A Divide both sides of the above equation by 2: 2×C ÷ 2 = 6×A ÷ 2 which makes C = 3×A
Hint #9
Substitute 0 for F, and (2×A) for D in eq.6c: 0 + B = 2×(2×A) which makes B = 4×A
Hint #10
Substitute (3×A) for C in eq.4b: E = 1½×(3×A) which makes E = 4½×A
Solution
Substitute 4×A for B, 3×A for C, 2×A for D, 4½×A for E, and 0 for F in eq.1: A + 4×A + 3×A + 2×A + 4½×A + 0 = 29 which simplifies to 14½×A = 29 Divide both sides of the above equation by 14½: 14½×A ÷ 14½ = 29 ÷ 14½ which means A = 2 making B = 4×A = 4 × 2 = 8 C = 3×A = 3 × 2 = 6 D = 2×A = 2 × 2 = 4 E = 4½×A = 4½ × 2 = 9 and ABCDEF = 286490