Puzzle for February 26, 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 positive integer.
* BC and EF are 2-digit numbers (not B×C or E×F).
Scratchpad
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Hint #1
Add C to both sides of eq.2: C + F + C = A – C + C which becomes eq.2a) 2×C + F = A In eq.1, replace A with 2×C + F (from eq.2a): 2×C + F + C = D + F which becomes 3×C + F = D + F Subtract F from both sides of the equation above: 3×C + F – F = D + F – F which makes eq.1a) 3×C = D
Hint #2
In eq.3, replace D with 3×C, and B with C × F (from eq.4): 3×C × F = C × F × C Divide both sides of the equation above by C × F: 3×C × F ÷ (C × F) = C × F × C ÷ (C × F) which makes 3 = C and makes D = 3×C = 3 × 3 = 9
Hint #3
In eq.3, substitute 9 for D, and 3 for C: 9 × F = B × 3 Divide both sides of the above equation by 3: 9 × F ÷ 3 = B × 3 ÷ 3 which makes 3×F = B
Hint #4
Substitute 3 for C in eq.2a: 2×3 + F = A which becomes eq.2b) 6 + F = A
Hint #5
eq.5 may be written as: 10×B + C = D × E In the above equation, substitute (3×F) for B, 3 for C, and 9 for D: 10×(3×F) + 3 = 9 × E which becomes 30×F + 3 = 9×E Divide both sides of the above equation by 9: (30×F + 3) ÷ 9 = 9×E ÷ 9 which becomes eq.5a) (30×F + 3) ÷ 9 = E
Hint #6
eq.6 may be written as: 10×E + F = A × D In the above equation, substitute ((30×F + 3) ÷ 9) for E (from eq.5a), (6 + F) for A (from eq.2b), and 9 for D: 10×((30×F + 3) ÷ 9) + F = (6 + F) × 9 which becomes ((300×F + 30) ÷ 9) + F = 54 + 9×F Subtract F from both sides: ((300×F + 30) ÷ 9) + F – F = 54 + 9×F – F which becomes eq.6a) (300×F + 30) ÷ 9 = 54 + 8×F
Hint #7
Multiply both sides of eq.6a by 9: (300×F + 30) ÷ 9 × 9 = (54 + 8×F) × 9 which becomes 300×F + 30 = 486 + 72×F Subtract 30 and 72×F from both sides of the equation above: 300×F + 30 – 30 – 72×F = 486 + 72×F – 30 – 72×F which becomes 228×F = 456 Divide both sides by 228: 228×F ÷ 228 = 456 ÷ 228 which makes F = 2 making B = 3×F = 3 × 2 = 6
Hint #8
Substitute 2 for F into eq.2b: 6 + 2 = A which makes 8 = A
Solution
Substitute 2 for F into eq.5a: (30×2 + 3) ÷ 9 = E which becomes (60 + 3) ÷ 9 = E which becomes 63 ÷ 9 = E which makes 7 = E and makes ABCDEF = 863972