Puzzle for September 22, 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
In eq.5, replace F with E – A (from eq.3), and replace C with A – B (from eq.2): E + E – A = A + B + A – B which becomes 2×E – A = 2×A Add A to both sides of the above equation: 2×E – A + A = 2×A + A which makes 2×E = 3×A Divide both sides by 2: 2×E ÷ 2 = 3×A ÷ 2 which means E = 1½×A
Hint #2
In eq.3, replace E with 1½×A: F = 1½×A – A which makes F = ½×A
Hint #3
In eq.6, substitute 1½×A for E, and ½×A for F: A + 1½×A = C × ½×A which becomes 2½×A = C × ½×A Divide both sides of the above equation by ½×A: 2½×A ÷ ½×A = C × ½×A ÷ ½×A which makes 5 = C
Hint #4
Substitute 5 for C in eq.2: A – B = 5 In the equation above, add B to both sides, and subtract 5 from both sides: A – B + B – 5 = 5 + B – 5 which makes eq.2a) A – 5 = B
Hint #5
Substitute ½×A for F, and (A – 5) for B (from eq.2a) in eq.4: D – A = ½×A – (A – 5) which is equivalent to D – A = ½×A – A + 5 which becomes D – A = –½×A + 5 Add A to both sides of the equation above: D – A + A = –½×A + 5 + A which makes eq.4a) D = ½×A + 5
Solution
Substitute A – 5 for B (from eq.2a), 5 for C, ½×A + 5 for D (from eq.4a), 1½×A for E, and ½×A for F in eq.1: A + A – 5 + 5 + ½×A + 5 + 1½×A + ½×A = 32 which simplifies to 4½×A + 5 = 32 Subtract 5 from both sides of the above equation: 4½×A + 5 – 5 = 32 – 5 which makes 4½×A = 27 Divide both sides by 4½: 4½×A ÷ 4½ = 27 ÷ 4½ which means A = 6 making B = A – 5 = 6 – 5 = 1 (from eq.2a) D = ½×A + 5 = ½×6 + 5 = 3 + 5 = 8 (from eq.4a) E = 1½×A = 1½×6 = 9 F = ½×A = ½×6 = 3 and ABCDEF = 615893