Puzzle for January 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 non-negative integer.
Scratchpad
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Hint #1
Add E to each side of eq.5: E + F + E = A + C – E + E which becomes eq.5a) 2×E + F = A + C In eq.5a, replace F with C + E (from eq.3): 2×E + C + E = A + C which becomes C + 3×E = A + C Subtract C from both sides of the equation above: C + 3×E – C = A + C – C which makes 3×E = A
Hint #2
Subtract the left and right sides of eq.5 from the left and right sides of eq.6, respectively: B – E + F – (E + F) = A + C + E – (A + C – E) which is equivalent to B – E + F – E – F = A + C + E – A – C + E which simplifies to B – 2×E = 2×E Add 2×E to both sides of the above equation: B – 2×E + 2×E = 2×E + 2×E which becomes B = 4×E
Hint #3
In eq.2, substitute 3×E for A, and 4×E for B: C + F = 3×E + 4×E which becomes C + F = 7×E Subtract C from both sides of the above equation: C + F – C = 7×E – C which becomes eq.2a) F = 7×E – C
Hint #4
Substitute 7×E – C for F (from eq.2a) in eq.3: 7×E – C = C + E In the above equation, add C to both sides, and subtract E from each side: 7×E – C + C – E = C + E + C – E which becomes 6×E = 2×C Divide both sides by 2: 6×E ÷ 2 = 2×C ÷ 2 which makes 3×E = C
Hint #5
Substitute 3×E for C in eq.2a: F = 7×E – 3×E which makes F = 4×E
Hint #6
Substitute 3×E for A, and 4×E for F in eq.4: D – 3×E = 4×E – D Add 3×E and D to both sides of the above equation: D – 3×E + 3×E + D = 4×E – D + 3×E + D which becomes 2×D = 7×E Divide both sides by 2: 2×D ÷ 2 = 7×E ÷ 2 which makes D = 3½×E
Solution
Substitute 3×E for A and C, 4×E for B and F, and 3½×E for D in eq.1: 3×E + 4×E + 3×E + 3½×E + E + 4×E = 37 which simplifies to 18½×E = 37 Divide both sides of the above equation by 18½: 18½×E ÷ 18½ = 37 ÷ 18½ which makes E = 2 making A = C = 3×E = 3 × 2 = 6 B = F = 4×E = 4 × 2 = 8 D = 3½×E = 3½ × 2 = 7 and ABCDEF = 686728