Puzzle for August 7, 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 F to both sides of eq.3: C + D – F + F = A + E + F which becomes C + D = A + E + F In eq.4, replace C + D with A + E + F: A + B = A + E + F – A which becomes eq.4a) A + B = E + F
Hint #2
Add A to both sides of eq.2: D + F + A = B + E + A which may be written as A + D + F = A + B + E In the above equation, replace A + B with E + F (from eq.4a): A + D + F = E + F + E which becomes A + D + F = 2×E + F Subtract F from both sides of the above equation: A + D + F – F = 2×E + F – F which becomes eq.2a) A + D = 2×E
Hint #3
Add E to both sides of eq.5: B + D – E + E = C – D + E + E which becomes B + D = C – D + 2×E In the above equation, substitute A + D for 2×E (from eq.2a): B + D = C – D + A + D which becomes eq.5a) B + D = C + A
Hint #4
Subtract the left and right sides of eq.5a from the left and right sides of eq.4, respectively: A + B – (B + D) = C + D – A – (C + A) which is equivalent to A + B – B – D = C + D – A – C – A which becomes A – D = D – 2×A In the above equation, add D and 2×A to both sides of the equation above: A – D + D + 2×A = D – 2×A + D + 2×A which becomes 3×A = 2×D Divide both sides by 2: 3×A ÷ 2 = 2×D ÷ 2 which makes 1½×A = D
Hint #5
Substitute 1½×A for D in eq.2a: A + 1½×A = 2×E which makes 2½×A = 2×E Divide both sides of the above equation by 2: 2½×A ÷ 2 = 2×E ÷ 2 which makes 1¼×A = E
Hint #6
Add A to both sides of eq.6: C + F – A + A = B – C + D + E + A which becomes C + F = B – C + D + E + A which may be written as C + F = A + B – C + D + E Substitute E + F for A + B (from eq.4a) into the above equation: C + F = E + F – C + D + E which becomes C + F = 2×E + F – C + D Subtract F from both sides, and add C to both sides: C + F – F + C = 2×E + F – C + D – F + C which becomes eq.6a) 2×C = 2×E + D
Hint #7
Substitute (1¼×A) for E, and 1½×A for D in eq.6a: 2×C = 2×(1¼×A) + 1½×A which becomes 2×C = 2½×A + 1½×A which makes 2×C = 4×A Divide both sides of the above equation by 2: 2×C ÷ 2 = 4×A ÷ 2 which makes C = 2×A
Hint #8
Substitute 2×A for C, 1½×A for D, and 1¼×A for E in eq.3: 2×A + 1½×A – F = A + 1¼×A which becomes 3½×A – F = 2¼×A In the above equation, add F to both sides, and subtract 2¼×A from both sides: 3½×A – F + F – 2¼×A = 2¼×A + F – 2¼×A which makes 1¼×A = F
Hint #9
Substitute 1½×A for D, and 1¼×A for F and E in eq.2: 1½×A + 1¼×A = B + 1¼×A Subtract 1¼×A from both sides of the equation above: 1½×A + 1¼×A – 1¼×A = B + 1¼×A – 1¼×A which makes 1½×A = B
Solution
Substitute 1½×A for B and D, 2×A for C, and 1¼×A for E and F in eq.1: A + 1½×A + 2×A + 1½×A + 1¼×A + 1¼×A = 34 which simplifies to 8½×A = 34 Divide both sides of the above equation by 8½: 8½×A ÷ 8½ = 34 ÷ 8½ which means A = 4 making B = D = 1½×A = 1½ × 4 = 6 C = 2×A = 2 × 4 = 8 E = F = 1¼×A = 1¼ × 4 = 5 and ABCDEF = 468655