Puzzle for March 13, 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.
* AB, BC, and EF are 2-digit numbers (not A×B, B×C, or E×F).
Scratchpad
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Hint #1
eq.6 may be written as: E + F = (A + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × (E + F) = 3 × (A + C + D) ÷ 3 which becomes eq.6a) 3×(E + F) = A + C + D
Hint #2
In eq.6a, replace E + F with A + D – F (from eq.4): 3×(A + D – F) = A + C + D which becomes 3×A + 3×D – 3×F = A + C + D In the equation above, subtract A and D from both sides, and add 3×F to both sides: 3×A + 3×D – 3×F – A – D + 3×F = A + C + D – A – D + 3×F which becomes 2×A + 2×D = C + 3×F which may be written as eq.6b) 2×(A + D) = C + 3×F
Hint #3
In eq.3, add D to both sides, and subtract E from both sides: B – D + F + D – E = A + E + D – E which becomes eq.3a) B + F – E = A + D In eq.6b, substitute B + F – E for A + D (from eq.3a): 2×(B + F – E) = C + 3×F which becomes 2×B + 2×F – 2×E = C + 3×F Subtract 2×F and C from both sides of the equation above: 2×B + 2×F – 2×E – 2×F – C = C + 3×F – 2×F – C which becomes eq.6c) 2×B – 2×E – C = F
Hint #4
Subtract F from both sides of eq.3a: B + F – E – F = A + D – F which becomes eq.3b) B – E = A + D – F In eq.4, replace A + D – F with B – E (from eq.3b): B – E = E + F Add E to both sides of the above equation: B – E + E = E + F + E which becomes eq.4a) B = 2×E + F
Hint #5
Substitute 2×B – 2×E – C for F (from eq.6c) in eq.4a: B = 2×E + 2×B – 2×E – C which becomes B = 2×B – C In the equation above, subtract B from both sides, and add C to both sides: B – B + C = 2×B – C – B + C which makes C = B
Hint #6
Substitute B for C in eq.2: B + D = A + B Subtract B from each side of the equation above: B + D – B = A + B – B which makes D = A
Hint #7
Substitute A for D in eq.6b: 2×(A + A) = C + 3×F which becomes 2×(2×A) = C + 3×F which becomes 4×A = C + 3×F Subtract 3×F from both sides of the equation above: 4×A – 3×F = C + 3×F – 3×F which makes 4×A – 3×F = C and also makes eq.6d) 4×A – 3×F = C = B
Hint #8
Substitute 4×A – 3×F for B (from eq.6d) in eq.4a: 4×A – 3×F = 2×E + F Add 3×F to both sides of the above equation: 4×A – 3×F + 3×F = 2×E + F + 3×F which becomes 4×A = 2×E + 4×F Divide both sides by 4: 4×A ÷ 4 = (2×E + 4×F) ÷ 4 which becomes eq.4b) A = ½×E + F
Hint #9
eq.5 may be written as: 10×A + B + 10×E + F = 10×B + C – D – F Substitute B for C, and A for D in the equation above: 10×A + B + 10×E + F = 10×B + B – A – F which becomes 10×A + B + 10×E + F = 11×B – A – F Subtract B from both sides, and add A and F to both sides: 10×A + B + 10×E + F – B + A + F = 11×B – A – F – B + A + F which becomes eq.5a) 11×A + 10×E + 2×F = 10×B
Hint #10
Substitute (½×E + F) for A (from eq.4b), and (2×E + F) for B (from eq.4a) in eq.5a: 11×(½×E + F) + 10×E + 2×F = 10×(2×E + F) which becomes 5½×E + 11×F + 10×E + 2×F = 20×E + 10×F which becomes 15½×E + 13×F = 20×E + 10×F Subtract 15½×E and 10×F from both sides of the above equation: 15½×E + 13×F – 15½×E – 10×F = 20×E + 10×F – 15½×E – 10×F which becomes 3×F = 4½×E Divide both sides by 3: 3×F ÷ 3 = 4½×E ÷ 3 which makes F = 1½×E
Hint #11
Substitute 1½×E for F in eq.4b: A = ½×E + 1½×E which makes A = 2×E and also makes D = A = 2×E
Hint #12
Substitute 1½×E for F in eq.4a: B = 2×E + 1½×E which makes B = 3½×E and also makes C = B = 3½×E
Solution
Substitute 2×E for A and D, 3½×E for B and C, and 1½×E for F in eq.1: 2×E + 3½×E + 3½×E + 2×E + E + 1½×E = 27 which simplifies to 13½×E = 27 Divide both sides of the above equation by 13½: 13½×E ÷ 13½ = 27 ÷ 13½ which means E = 2 making A = D = 2×E = 2 × 2 = 4 B = C = 3½×E = 3½ × 2 = 7 F = 1½×E = 1½ × 2 = 3 and ABCDEF = 477423