Puzzle for October 20, 2019 ( )
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
In eq.3, replace A with B + D (from eq.4): D + E = B + D + C Subtract D from both sides of the equation above: D + E – D = B + D + C – D which becomes eq.3a) E = B + C
Hint #2
In eq.2, replace B + C with E (from eq.3a): E + F = E Subtract E from each side of the above equation: E + F – E = E – E which makes F = 0
Hint #3
In eq.5, replace F with 0: C – D = E – C – 0 Add C and D to both sides of the equation above: C – D + C + D = E – C – 0 + C + D which becomes 2×C = E + D which may be written as eq.5a) 2×C = D + E
Hint #4
In eq.3, substitute 2×C for D + E (from eq.5a): 2×C = A + C Subtract C from each side of the equation above: 2×C – C = A + C – C which makes C = A
Hint #5
Substitute C for A in eq.4: B + D = C Substitute B + D for C in eq.3a: E = B + B + D which becomes eq.3b) E = 2×B + D
Hint #6
Substitute A for C in eq.6: A × A = B + (D × E) Substitute (B + D) for A (from eq.4), and (2×B + D) for E (from eq.3b) in the above equation: (B + D) × (B + D) = B + (D × (2×B + D)) which becomes B×B + B×D + D×B + D×D = B + D×2×B + D×D which is equivalent to B² + 2×B×D + D² = B + 2×B×D + D² Subtract 2×B×D and D² from both sides: B² + 2×B×D + D² – 2×B×D – D² = B + 2×B×D + D² – 2×B×D – D² which simplifies to eq.6a) B² = B
Hint #7
To make eq.6a true, then: B = 0 and / or: B = 1 Check: B = 0 ... Substituting 0 for B in eq.3b would yield: E = 2×0 + D which would make E = D
Hint #8
Continue checking: B = 0 ... Substituting 0 for B in eq.3a would make: E = 0 + C which would mean E = C and would mean D = E = C = A
Hint #9
Finish checking: B = 0 ... Substituting A for C and D and E, and 0 for B and F in eq.1 would yield: A + 0 + A + A + A + 0 = 33 which would make 4×A = 33 Dividing both sides of the above equation by 4 would yield: 4×A ÷ 4 = 33 ÷ 4 which would make A = 8¼ Since A must be an integer, then: B ≠ 0 which means B = 1
Hint #10
Substitute 1 for B in eq.4: 1 + D = A which also means eq.4a) C = A = 1 + D
Hint #11
Substitute 1 for B in eq.3b: E = 2×1 + D which means eq.3c) E = 2 + D
Solution
Substitute 1 + D for A and C (from eq.4a), 1 for B, 2 + D for E (from eq.3c), and 0 for F in eq.1: 1 + D + 1 + 1 + D + D + 2 + D + 0 = 33 which simplifies to 4×D + 5 = 33 Subtract 5 from both sides of the equation above: 4×D + 5 – 5 = 33 – 5 which makes 4×D = 28 Divide both sides by 4: 4×D ÷ 4 = 28 ÷ 4 which means D = 7 making A = C = 1 + D = 1 + 7 = 8 (from eq.4a) E = 2 + D = 2 + 7 = 9 (from eq.3c) and ABCDEF = 818790