Puzzle for March 29, 2022 ( )
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
In eq.6, replace B + D with C + F (from eq.3): C + E = C + F + F which becomes C + E = C + 2×F Subtract C from each side of the above equation: C + E – C = C + 2×F – C which makes E = 2×F
Hint #2
In eq.2, replace E with 2×F: 2×F = A + F Subtract F from both sides of the above equation: 2×F – F = A + F – F which makes F = A
Hint #3
Subtract D from each side of eq.3: B + D – D = C + F – D which becomes eq.3a) B = C + F – D Subtract E from each side of eq.5: B + E – E = C + D – E which becomes eq.5a) B = C + D – E
Hint #4
In eq.5a, substitute C + F – D for B (from eq.3a), and 2×F for E: C + F – D = C + D – 2×F In the equation above, subtract C from both sides, and add D and 2×F to both sides: C + F – D – C + D + 2×F = C + D – 2×F – C + D + 2×F which simplifies to 3×F = 2×D Divide both sides by 2: 3×F ÷ 2 = 2×D ÷ 2 which makes 1½×F = D
Hint #5
Substitute 1½×F for D, 2×F for E, and F for A in eq.4: 1½×F + 2×F = F + B which becomes 3½×F = F + B Subtract F from each side of the above equation: 3½×F – F = F + B – F which makes 2½×F = B
Hint #6
Substitute 2½×F for B, and 1½×F for D in eq.3: 2½×F + 1½×F = C + F which becomes 4×F = C + F Subtract F from each side of the equation above: 4×F – F = C + F – F which makes 3×F = C
Solution
Substitute F for A, 2½×F for B, 3×F for C, 1½×F for D, and 2×F for E in eq.1: F + 2½×F + 3×F + 1½×F + 2×F + F = 22 which simplifies to 11×F = 22 Divide both sides of the above equation by 11: 11×F ÷ 11 = 22 ÷ 11 which means F = 2 making A = F = 2 B = 2½×F = 2½ × 2 = 5 C = 3×F = 3 × 2 = 6 D = 1½×F = 1½ × 2 = 3 E = 2×F = 2 × 2 = 4 and ABCDEF = 256342