Puzzle for January 30, 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 E, A, and C to both sides of eq.4: C – E + E + A + C = D – A – C + E + A + C which becomes 2×C + A = D + E which may be written as eq.4a) A + 2×C = D + E Add C to both sides of eq.3: A + C + C = B + D + C which becomes eq.3a) A + 2×C = B + D + C
Hint #2
In eq.3a, replace A + 2×C with D + E (from eq.4a): D + E = B + D + C Subtract D from each side of the equation above: D + E – D = B + D + C – D which becomes eq.3b) E = B + C
Hint #3
In eq.2, replace E with B + C (from eq.3b): D + B + C = C + F Subtract C from each side of the equation above: D + B + C – C = C + F – C which becomes eq.2a) D + B = F
Hint #4
In eq.5, substitute B + C for E (from eq.3b), and D + B for F (from eq.2a): B + C + D + B – C = A + B + C which becomes 2×B + D = A + B + C Subtract B from each side of the above equation: 2×B + D – B = A + B + C – B which becomes eq.5a) B + D = A + C
Hint #5
Substitute B + D for A + C (from eq.5a) in eq.3: B + D = B + D – D which becomes B + D = B Subtract B from both sides of the above equation: B + D – B = B – B which makes D = 0
Hint #6
Substitute 0 for D in eq.2a: 0 + B = F which makes B = F
Hint #7
Substitute 0 for D in eq.4a: A + 2×C = 0 + E which becomes eq.4b) A + 2×C = E
Hint #8
Substitute 0 for D in eq.5a: B + 0 = A + C which makes B = A + C and also makes eq.5b) F = B = A + C
Hint #9
Substitute A + C for B and F (from eq.5b), 0 for D, and A + 2×C for E (from eq.4b) in eq.1: A + A + C + C + 0 + A + 2×C + A + C = 25 which becomes eq.1a) 4×A + 5×C = 25
Hint #10
Substitute (A + 2×C) for E (from eq.4b), and (A + C) for B and F (from eq.5b) in eq.6: (A × (A + 2×C)) + C = (A + C) × (A + C) which becomes A×A + A×2×C + C = A×A + A×C + C×A + C×C which is the same as A² + 2×A×C + C = A² + 2×A×C + C² Subtract A² and 2×A×C from both sides of the above equation: A² + 2×A×C + C – A² – 2×A×C = A² + 2×A×C + C² – A² – 2×A×C which simplifies to C = C² making C = 0 or C = 1
Hint #11
Check: C = 0 ... Substituting 0 for C in eq.1a would yield: 4×A + 5×0 = 25 which would make 4×A = 25 Dividing both sides of the above equation by 4 would yield: 4×A ÷ 4 = 25 ÷ 4 which would make A = 6¼ Since A must be an integer, then: A ≠ 6¼ which means C ≠ 0 and therefore makes C = 1
Solution
Substitute 1 for C in eq.1a: 4×A + 5×1 = 25 which becomes 4×A + 5 = 25 Subtract 5 from both sides of the above equation: 4×A + 5 – 5 = 25 – 5 which makes 4×A = 20 Divide both sides by 4: 4×A ÷ 4 = 20 ÷ 4 which means A = 5 making B = F = A + C = 5 + 1 = 6 (from eq.5b) E = A + 2×C = 5 + 2×1 = 5 + 2 = 7 (from eq.4b) and ABCDEF = 561076