Puzzle for June 12, 2021 ( )
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 the left and right sides of eq.3 to the left and right sides of eq.2, respectively: D + F + (C + D) = A + C + (B + E) which becomes 2×D + F + C = A + C + B + E which may be written as eq.2a) 2×D + C + F = A + B + C + E
Hint #2
In eq.2a, replace C + F with A + D + E (from eq.6): 2×D + A + D + E = A + B + C + E which becomes 3×D + A + E = A + B + C + E Subtract A and E from both sides of the above equation: 3×D + A + E – A – E = A + B + C + E – A – E which becomes eq.2b) 3×D = B + C
Hint #3
Add D to both sides of eq.4: F – D + D = B + D + D which becomes eq.4a) F = B + 2×D In eq.2, replace F with B + 2×D (from eq.4a): D + B + 2×D = A + C which becomes eq.2c) B + 3×D = A + C
Hint #4
In eq.2c, substitute B + C for 3×D (from eq.2b): B + B + C = A + C which becomes 2×B + C = A + C Subtract C from each side of the above equation: 2×B + C – C = A + C – C which makes 2×B = A
Hint #5
Subtract the the left and right sides of eq.2 from the left and right sides of eq.6, respectively: C + F – (D + F) = A + D + E – (A + C) which becomes C + F – D – F = A + D + E – A – C which becomes C – D = D + E – C In the above equation, subtract D from both sides, and add C to both sides: C – D – D + C = D + E – C – D + C which becomes eq.6a) 2×C – 2×D = E
Hint #6
Substitute (B + 2×D) for F (from eq.4a), and 2×C – 2×D for E (from eq.6a) in eq.5: B + (B + 2×D) = C + 2×C – 2×D – (B + 2×D) which becomes 2×B + 2×D = 3×C – 2×D – B – 2×D which becomes 2×B + 2×D = 3×C – 4×D – B Add 4×D and B to both sides of the above equation: 2×B + 2×D + 4×D + B = 3×C – 4×D – B + 4×D + B which becomes eq.5a) 3×B + 6×D = 3×C
Hint #7
Divide both sides of eq.5a by 3: (3×B + 6×D) ÷ 3 = 3×C ÷ 3 which makes eq.5b) B + 2×D = C Substitute C for B + 2×D (from eq.5b) in eq.4a: F = C
Hint #8
Substitute B + 2×D for C (from eq.5b) in eq.2b: 3×D = B + B + 2×D which becomes 3×D = 2×B + 2×D Subtract 2×D from each side of the equation above: 3×D – 2×D = 2×B + 2×D – 2×D which makes D = 2×B
Hint #9
Substitute (2×B) for D in eq.5b: B + 2×(2×B) = C which becomes B + 4×B = C which becomes 5×B = C and also makes F = C = 5×B
Hint #10
Substitute (5×B) for C, and (2×B) for D in eq.6a: 2×(5×B) – 2×(2×B) = E which becomes 10×B – 4×B = E which makes 6×B = E
Solution
Substitute 2×B for A and D, 5×B for C and F, and 6×B for E in eq.1: 2×B + B + 5×B + 2×B + 6×B + 5×B = 21 which simplifies to 21×B = 21 Divide both sides of the above equation by 21: 21×B ÷ 21 = 21 ÷ 21 which means B = 1 making A = D = 2×B = 2 × 1 = 2 C = F = 5×B = 5 × 1 = 5 E = 6×B = 6 × 1 = 6 and ABCDEF = 215265