Puzzle for December 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.
* BC is a 2-digit number (not B×C).
Scratchpad
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Hint #1
In eq.2, replace A with C + F (from eq.3): C + F + C = B + F which becomes 2×C + F = B + F Subtract F from each side of the equation above: 2×C + F – F = B + F – F which becomes 2×C = B
Hint #2
In eq.1, replace B with 2×C: 2×C + C = D which makes 3×C = D
Hint #3
Add F and E to both sides of eq.4: A – B – F + F + E = B – E + F + E which becomes A – B + E = B + F Substitute A – B + E for B + F in eq.2: A + C = A – B + E In the equation above, subtract A from both sides, and add B to both sides: A + C – A + B = A – B + E – A + B which becomes eq.2b) C + B = E
Hint #4
Substitute 2×C for B in eq.2b: C + 2×C = E which makes 3×C = E
Hint #5
Substitute 2×C for B, and 3×C for D and E in eq.5: 2×C + (C × F) = 3×C + 3×C which becomes 2×C + (C × F) = 6×C Subtract 2×C from both sides of the equation above: 2×C + (C × F) – 2×C = 6×C – 2×C which becomes C × F = 4×C Divide both sides by C: C × F ÷ C = 4×C ÷ C which makes F = 4
Hint #6
eq.6 may be written as: D × E = 10×B + C + D + E Substitute 3×C for D and E, and (2×C) for B in the equation above: 3×C × 3×C = 10×(2×C) + C + 3×C + 3×C which becomes 9×C² = 20×C + 7×C which becomes 9×C² = 27×C Divide both sides by 9×C: 9×C² ÷ 9×C = 27×C ÷ 9×C which makes C = 3 making B = 2×C = 2×3 = 6 D = E = 3×C = 3×3 = 9
Solution
Substitute 3 for C, and 4 for F in eq.3: 3 + 4 = A which makes 7 = A and makes ABCDEF = 763994