Puzzle for September 30, 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
Add C to both sides of eq.3: A + C + C = D + F – C + C which becomes eq.3a) A + 2×C = D + F eq.5 may be re-written as: eq.5a) D + F + E = A + B + C – F
Hint #2
In eq.5a, replace D + F with A + 2×C (from eq.3a), and replace B + C with A + E (from eq.1): A + 2×C + E = A + A + E – F which becomes A + 2×C + E = 2×A + E – F Subtract A and E from both sides of the above equation: A + 2×C + E – A – E = 2×A + E – F – A – E which becomes eq.5b) 2×C = A – F
Hint #3
Subtract F from both sides of eq.3a: A + 2×C – F = D + F – F which is equivalent to A – F + 2×C = D In the above equation, replace A – F with 2×C (from eq.5b): 2×C + 2×C = D which makes 4×C = D
Hint #4
In eq.4, substitute 4×C for D in eq.4: C + F = 4×C Subtract C from both sides of the equation above: C + F – C = 4×C – C which makes F = 3×C
Hint #5
Substitute 4×C for D in eq.2: C + 4×C = E which makes 5×C = E
Hint #6
Substitute 3×C for F in eq.5b: 2×C = A – 3×C Add 3×C to both sides of the above equation: 2×C + 3×C = A – 3×C + 3×C which makes 5×C = A
Hint #7
Substitute 5×C for A and E in eq.1: B + C = 5×C + 5×C which becomes B + C = 10×C Subtract C from each side of the equation above: B + C – C = 10×C – C which makes B = 9×C
Solution
Substitute 5×C for E and A, 3×C for F, and 9×C for B in eq.6: 5×C × 3×C = 5×C + 9×C + C which becomes 15×C×C = 15×C Divide both sides of the above equation by 15×C: 15×C×C ÷ 15×C = 15×C ÷ 15×C which makes C = 1 making A = E = 5×C = 5 × 1 = 5 B = 9×C = 9 × 1 = 9 D = 4×C = 4 × 1 = 4 F = 3×C = 3 × 1 = 3 and ABCDEF = 591453