Puzzle for December 17, 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 positive integer.
Scratchpad
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Hint #1
In eq.2, replace A with E + F (from eq.1): E = E + F + D – E which becomes E = F + D which may be written as eq.2a) E = D + F
Hint #2
In eq.3, replace D + F with E (from eq.2a): C + E = B + E Subtract E from each side of the above equation: C + E – E = B + E – E which makes C = B
Hint #3
In eq.4, substitute D + F for E (from eq.2a), and B for C: D + D + F – F = B + B which becomes 2×D = 2×B Divide both sides of the above equation by 2: 2×D ÷ 2 = 2×B ÷ 2 which makes D = B
Hint #4
Substitute B for C in eq.6: (A × B) – B = B × E which may be written as B × (A – 1) = B × E Divide both sides of the equation above by B: B × (A – 1) ÷ B = B × E ÷ B which becomes eq.6a) A – 1 = E
Hint #5
Substitute A – 1 for E (from eq.6a) in eq.1: A = A – 1 + F Subtract A from both sides of the above equation: A – A = A – 1 + F – A which becomes 0 = –1 + F Add 1 to both sides: 0 + 1 = –1 + F + 1 which makes 1 = F
Hint #6
Substitute B for C and D, 1 for F, and A – 1 for E (from eq.6a) in eq.3: B + B + 1 = B + A – 1 which becomes 2×B + 1 = B + A – 1 In the above equation, subtract B from each side, and add 1 to each side: 2×B + 1 – B + 1 = B + A – 1 – B + 1 which makes eq.3a) B + 2 = A
Hint #7
Substitute B + 2 for A (from eq.3a) in eq.6a: B + 2 – 1 = E which makes eq.6b) B + 1 = E
Solution
Substitute B for C and D, 1 for F, B + 2 for A (from eq.3a), and B + 1 for E (from eq.6b) in eq.5: B + B + B – 1 = B + 2 + B + 1 + 1 which becomes 3×B – 1 = 2×B + 4 In the equation above, add 1 to both sides, and subtract 2×B from both sides: 3×B – 1 + 1 – 2×B = 2×B + 4 + 1 – 2×B which makes B = 5 and makes A = B + 2 = 5 + 2 = 7 (from eq.3a) D = C = B = 5 E = B + 1 = 5 + 1 = 6 (from eq.6b) and ABCDEF = 755561