Puzzle for May 10, 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 E to each side of eq.4: B + D – E + E = C + E + F + E which becomes B + D = C + 2×E + F In the equation above, replace B + D with A + F (from eq.5): A + F = C + 2×E + F Subtract F from both sides: A + F – F = C + 2×E + F – F which becomes eq.4a) A = C + 2×E
Hint #2
In eq.6, replace A with C + 2×E (from eq.4a): C + D + E = C + 2×E + B – E which becomes C + D + E = C + E + B Subtract C and E from each side of the above equation: C + D + E – C – E = C + E + B – C – E which simplifies to D = B
Hint #3
In eq.5, substitute D for B: A + F = D + D which becomes eq.5a) A + F = 2×D Add A and C to both sides of eq.3: D – C + F + A + C = A + C + A + C which becomes D + F + A = 2×A + 2×C which may be written eq.3a) D + A + F = 2×(A + C)
Hint #4
In eq.3a, substitute 2×D for A + F (from eq.5a), and D – C + F for A + C (from eq.3): D + 2×D = 2×(D – C + F) which becomes 3×D = 2×D – 2×C + 2×F Subtract 2×D from both sides: 3×D – 2×D = 2×D – 2×C + 2×F – 2×D which becomes eq.3b) D = 2×F – 2×C
Hint #5
Substitute D for B in eq.4: D + D – E = C + E + F which becomes 2×D – E = C + E + F Subtract E from both sides of the equation above: 2×D – E – E = C + E + F – E which becomes eq.4b) 2×D – 2×E = C + F
Hint #6
Substitute (2×F – 2×C) for D (from eq.3b) in eq.4b: 2×(2×F – 2×C) – 2×E = C + F which becomes 4×F – 4×C – 2×E = C + F In the above equation, subtract 4×F from each side, and add 4×C to each side: 4×F – 4×C – 2×E – 4×F + 4×C = C + F – 4×F + 4×C which becomes –2×E = 5×C – 3×F Divide both sides by (–2): –2×E ÷ (–2) = (5×C – 3×F) ÷ (–2) which becomes eq.4c) E = –2½×C + 1½×F
Hint #7
Substitute (2×F – 2×C) for D (from eq.3b) in eq.5a: A + F = 2×(2×F – 2×C) which becomes A + F = 4×F – 4×C Subtract F from each side fo the equation above: A + F – F = 4×F – 4×C – F which becomes eq.5b) A = 3×F – 4×C
Hint #8
Substitute (–2½×C + 1½×F) for E (from eq.4c), 3×F – 4×C for A (from eq.5b), and 2×F – 2×C for D (from eq.3b) in eq.2: C + (–2½×C + 1½×F) + F = 3×F – 4×C + 2×F – 2×C which becomes –1½×C + 2½×F = 5×F – 6×C In the equation above, add 6×C to each side, and subtract 2½×F from both sides: –1½×C + 2½×F + 6×C – 2½×F = 5×F – 6×C + 6×C – 2½×F which becomes 4½×C = 2½×F Divide both sides by 2½: 4½×C ÷ 2½ = 2½×F ÷ 2½ which makes eq.2a) 1⅘×C = F
Hint #9
Substitute (1⅘×C) for F in eq.4c: E = –2½×C + 1½×(1⅘×C) which becomes E = –2½×C + 2.7×C which makes E = ⅕×C Multiply both sides by 5: 5×E = 5 × ⅕×C which makes 5×E = C
Hint #10
Substitute (5×E) for C in eq.2a: 1⅘×(5×E) = F which makes 9×E = F
Hint #11
Substitute (9×E) for F, and (5×E) for C in eq.5b: A = 3×(9×E) – 4×(5×E) which becomes A = 27×E – 20×E which makes A = 7×E
Hint #12
Substitute (9×E) for F, and (5×E) for C in eq.3b: D = 2×(9×E) – 2×(5×E) which becomes D = 18×E – 10×E which makes D = 8×E and also makes B = D = 8×E
Solution
Substitute 7×E for A, 8×E for B and D, 5×E for C, and 9×E for F in eq.1: 7×E + 8×E + 5×E + 8×E + E + 9×E = 38 which simplifies to 38×E = 38 Divide both sides of the above equation by 38: 38×E ÷ 38 = 38 ÷ 38 which means E = 1 making A = 7×E = 7 × 1 = 7 B = D = 8×E = 8 × 1 = 8 C = 5×E = 5 × 1 = 5 F = 9×E = 9 × 1 = 9 and ABCDEF = 785819