Puzzle for September 25, 2019 ( )
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
Subtract E from both sides of eq.6: F – E = A + E – E which becomes F – E = A In eq.2, replace A with F – E, and replace B with E + F (from eq.5): E + F + E = F – E + D Add E to each side, and subtract F from each side of the above equation: E + F + E + E – F = F – E + D + E – F which simplifies to eq.2a) 3×E = D
Hint #2
In eq.3, replace D with 3×E, and replace B with E + F (from eq.5): C + 3×E = E + F + E + F which becomes C + 3×E = 2×E + 2×F Subtract 3×E from both sides of the above equation: C + 3×E – 3×E = 2×E + 2×F – 3×E which becomes eq.3a) C = 2×F – E
Hint #3
In eq.3a, substitute (A + E) for F (from eq.6): C = 2×(A + E) – E which is the same as C = 2×A + 2×E – E which becomes eq.3b) C = 2×A + E
Hint #4
Substitute 3×E for D, and 2×A + E for C (from eq.3b) in eq.4: 3×E = A + 2×A + E Subtract E from both sides of the above equation: 3×E – E = A + 2×A + E – E which means 2×E = 3×A Divide both sides by 2: 2×E ÷ 2 = 3×A ÷ 2 which makes E = 1½×A
Hint #5
Substitute (1½×A) for E in eq.2a: 3×(1½×A) = D which means 4½×A = D
Hint #6
Substitute 1½×A for E in eq.3b: C = 2×A + 1½×A which makes C = 3½×A
Hint #7
Substitute 1½×A for E in eq.6: F = A + 1½×A which makes F = 2½×A
Hint #8
Substitute 1½×A for E, and 2½×A for F in eq.5: 1½×A + 2½×A = B which makes 4×A = B
Solution
Substitute 4×A for B, 3½×A for C, 4½×A for D, 1½×A for E, and 2½×A for F in eq.1: A + 4×A + 3½×A + 4½×A + 1½×A + 2½×A = 34 which simplifies to 17×A = 34 Divide both sides of the equation above by 17: 17×A ÷ 17 = 34 ÷ 17 which means A = 2 making B = 4×A = 4 × 2 = 8 C = 3½×A = 3½× × 2 = 7 D = 4½×A = 4½× × 2 = 9 E = 1½×A = 1½× × 2 = 3 F = 2½×A = 2½× × 2 = 5 and ABCDEF = 287935