Puzzle for March 19, 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
Subtract the left and right sides of eq.2 from the left and right sides of eq.4, respectively: C + F – (E + F) = A + B + D – (A + C) which becomes C + F – E – F = A + B + D – A – C which becomes C – E = B + D – C Add C to both sides of the equation above: C – E + C = B + D – C + C which becomes eq.4a) 2×C – E = B + D
Hint #2
Add C and D to both sides of eq.5: B – C + C + D = C – D + E + C + D which becomes B + D = 2×C + E In the above equation, replace B + D with 2×C – E (from eq.4a): 2×C – E = 2×C + E Subtract 2×C from both sides, and add E to both sides: 2×C – E – 2×C + E = 2×C + E – 2×C + E which makes 0 = 2×E which means 0 = E
Hint #3
In eq.3, substitute 0 for E: C – D = A – C + 0 which becomes eq.3a) C – D = A – C Add D and C to both sides of eq.3a: C – D + D + C = A – C + D + C which becomes eq.3b) 2×C = A + D
Hint #4
In eq.5, substitute A – C for C – D (from eq.3a), and 0 for E: B – C = A – C + 0 which becomes B – C = A – C Add C to both sides of the above equation: B – C + C = A – C + C which makes B = A
Hint #5
Multiply both sides of eq.2 by 2: 2×(E + F) = 2×(A + C) which becomes 2×E + 2×F = 2×A + 2×C Substitute 0 for E, and A + D for 2×C (from eq.3b) in the equation above: 2×0 + 2×F = 2×A + A + D which becomes 2×F = 3×A + D Subtract 3×A from both sides: 2×F – 3×A = 3×A + D – 3×A which becomes eq.2b) 2×F – 3×A = D
Hint #6
Substitute (2×F – 3×A) for D (from eq.2b), and A for B in eq.6: (2×F – 3×A) + F – A = A + A – (2×F – 3×A) which becomes 3×F – 4×A = 2×A – 2×F + 3×A which becomes 3×F – 4×A = 5×A – 2×F Add 4×A and 2×F to both sides of the above equation: 3×F – 4×A + 4×A + 2×F = 5×A – 2×F + 4×A + 2×F which makes eq.6a) 5×F = 9×A
Hint #7
Multiply both sides of eq.2 by 5: 5×(E + F) = 5×(A + C) which becomes 5×E + 5×F = 5×A + 5×C Substitute 0 for E, and 9×A for 5×F (from eq.6a) in the equation above: 5×0 + 9×A = 5×A + 5×C which becomes 9×A = 5×A + 5×C Subtract 5×A from each side: 9×A – 5×A = 5×A + 5×C – 5×A which becomes 4×A = 5×C Divide both sides by 4: 4×A ÷ 4 = 5×C ÷ 4 which makes A = 1¼×C and also makes B = A = 1¼×C
Hint #8
Substitute 0 for E, and 1¼×C for A in eq.2: 0 + F = 1¼×C + C which makes F = 2¼×C
Hint #9
Substitute 1¼×C for A in eq.3b: 2×C = 1¼×C + D Subtract 1¼×C from each side of the equation above: 2×C – 1¼×C = 1¼×C + D – 1¼×C which makes ¾×C = D
Solution
Substitute 1¼×C for A and B, ¾×C for D, 0 for E, and 2¼×C for F in eq.1: 1¼×C + 1¼×C + C + ¾×C + 0 + 2¼×C = 26 which simplifies to 6½×C = 26 Divide both sides of the above equation by 6½: 6½×C ÷ 6½ = 26 ÷ 6½ which means C = 4 making A = B = 1¼×C = 1¼ × 4 = 5 D = ¾×C = ¾ × 4 = 3 F = 2¼×C = 2¼ × 4 = 9 and ABCDEF = 554309