Puzzle for May 29, 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 D to both sides of eq.5: A + D + D = C – D + F + D which becomes eq.5a) A + 2×D = C + F Add B to both sides of eq.6: C + F – B + B = A + B + E + B which becomes eq.6a) C + F = A + 2×B + E
Hint #2
In eq.5a, replace C + F with A + 2×B + E (from eq.6a): A + 2×D = A + 2×B + E Subtract A from each side of the above equation: A + 2×D – A = A + 2×B + E – A which becomes eq.5b) 2×D = 2×B + E
Hint #3
In eq.3, subtract D and F from both sides, and add E to both sides: E + F – D – F + E = B + D – E – D – F + E which simplifies to eq.3a) 2×E – D = B – F Subtract F, C, and E from both sides of eq.4: D + F – F – C – E = B + C + E – F – C – E which simplifies to eq.4a) D – C – E = B – F
Hint #4
In eq.3a, replace B – F with D – C – E (from eq.4a): 2×E – D = D – C – E Add D, C, and E to both sides of the equation above: 2×E – D + D + C + E = D – C – E + D + C + E which simplifies to eq.3b) 3×E + C = 2×D
Hint #5
In eq.3b, substitute 2×B + E for 2×D (from eq.5b): 3×E + C = 2×B + E Subtract E from each side of the equation above: 3×E + C – E = 2×B + E – E which becomes eq.3c) 2×E + C = 2×B
Hint #6
Substitute B + E for C (from eq.2) in eq.3c: 2×E + B + E = 2×B which becomes 3×E + B = 2×B Subtract B from both sides of the equation above: 3×E + B – B = 2×B – B which makes 3×E = B
Hint #7
Substitute (3×E) for B in eq.3c: 2×E + C = 2×(3×E) which becomes 2×E + C = 6×E Subtract 2×E from both sides of the above equation: 2×E + C – 2×E = 6×E – 2×E which makes C = 4×E
Hint #8
Substitute 4×E for C in eq.3b: 3×E + 4×E = 2×D which becomes 7×E = 2×D Divide both sides of the above equation by 2: 7×E ÷ 2 = 2×D ÷ 2 which makes 3½×E = D
Hint #9
Substitute 3×E for B, and 3½×E for D in eq.3: E + F = 3×E + 3½×E – E which becomes E + F = 5½×E Subtract E from both sides of the above equation: E + F – E = 5½×E – E which makes F = 4½×E
Hint #10
Substitute (3½×E) for D, 4×E for C, and 4½×E for F in eq.5a: A + 2×(3½×E) = 4×E + 4½×E which becomes A + 7×E = 8½×E Subtract 7×E from both sides of the above equation: A + 7×E – 7×E = 8½×E – 7×E which makes A = 1½×E
Solution
Substitute 1½×E for A, 3×E for B, 4×E for C, 3½×E for D, and 4½×E for F in eq.5a: 1½×E + 3×E + 4×E + 3½×E + E + 4½×E = 35 which simplifies to 17½×E = 35 Divide both sides of the equation above by 17½: 17½×E ÷ 17½ = 35 ÷ 17½ which means E = 2 making A = 1½×E = 1½ × 2 = 3 B = 3×E = 3 × 2 = 6 C = 4×E = 4 × 2 = 8 D = 3½×E = 3½ × 2 = 7 F = 4½×E = 4½ × 2 = 9 and ABCDEF = 368729