Puzzle for January 9, 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 non-negative integer.
Scratchpad
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Hint #1
Multiply both sides of eq.3 by (–1): (–1) × (C – D) = (–1) × (B – E) which becomes –C + D = –B + E which may be written as eq.3a) D – C = E – B
Hint #2
In eq.4, replace D – C with E – B (from eq.3a): E – A + B = E – B In the above equation, subtract E from both sides: E – A + B – E = E – B – E which becomes –A + B = –B Add A and B to both sides: –A + B + A + B = –B + A + B which simplifies to 2×B = A
Hint #3
In eq.2, replace A with 2×B: D = 2×B + B which makes D = 3×B
Hint #4
Add D and E to both sides of eq.3: C – D + D + E = B – E + D + E which becomes C + E = B + D In eq.5, substitute B + D for C + E: A + D + F = B + B + D which becomes A + D + F = 2×B + D Subtract D from both sides of the equation above: A + D + F – D = 2×B + D – D which becomes eq.5a) A + F = 2×B
Hint #5
Substitute A for 2×B in eq.5a: A + F = A Subtract A from both sides: A + F – A = A – A which makes F = 0
Hint #6
Add C and E to both sides of eq.6: B – C + C + E = C – E + C + E which becomes eq.6a) B + E = 2×C
Hint #7
eq.5 may be re-written as: A + D + F = C + B + E Substitute 2×B for A, 3×B for D, 0 for F, and 2×C for B + E (from eq.6a) in the above equation: 2×B + 3×B + 0 = C + 2×C which becomes 5×B = 3×C Divide both sides by 3: 5×B ÷ 3 = 3×C ÷ 3 which makes 1⅔×B = C
Hint #8
Substitute (1⅔×B) for C in eq.6a: B + E = 2×(1⅔×B) which becomes B + E = 3⅓×B Subtract B from each side: B + E – B = 3⅓×B – B which makes E = 2⅓×B
Solution
Substitute 2×B for A, 1⅔×B for C, 3×B for D, 2⅓×B for E, and 0 for F in eq.1: 2×B + B + 1⅔×B + 3×B + 2⅓×B + 0 = 30 which simplifies to 10×B = 30 Divide both sides of the equation above by 10: 10×B ÷ 10 = 30 ÷ 10 which means B = 3 making A = 2×B = 2 × 3 = 6 C = 1⅔×B = 1⅔ × 3 = 5 D = 3×B = 3 × 3 = 9 E = 2⅓×B = 2⅓ × 3 = 7 and ABCDEF = 635970