Puzzle for November 26, 2023 ( )
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.3, replace A with F - D (from eq.2): D + F = F - D + E In the above equation, subtract F from both sides, and add D to both sides: D + F - F + D = F - D + E - F + D which simplifies to 2×D = E
Hint #2
eq.6 may be re-written as: B = (C + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × B = 4 × (C + D + E + F) ÷ 4 which becomes eq.6a) 4×B = C + D + E + F
Hint #3
In eq.1, replace C + D + E + F with 4×B (from eq.6a): A + B + 4×B = 11 which becomes eq.1a) A + 5×B = 11 Subtract 5×B from each side of eq.1a: A + 5×B - 5×B = 11 - 5×B which becomes eq.1b) A = 11 - 5×B
Hint #4
Add B, F, and E to both sides of eq.5: C + D - B - F + B + F + E = A + B - E + F + B + F + E which becomes C + D + E = A + 2×B + 2×F In the above equation, substitute 11 - 5×B for A (from eq.1b): C + D + E = 11 - 5×B + 2×B + 2×F which becomes eq.5a) C + D + E = 11 - 3×B + 2×F
Hint #5
Substitute 11 - 3×B + 2×F for C + D + E (from eq.5a) into eq.6a: 4×B = 11 - 3×B + 2×F + F which becomes 4×B = 11 - 3×B + 3×F Add 3×B to both sides of the equation above: 4×B + 3×B = 11 - 3×B + 3×F + 3×B which becomes eq.6b) 7×B = 11 + 3×F
Hint #6
Substitute A + 5×B for 11 (from eq.1a) in eq.6b: 7×B = A + 5×B + 3×F Subtract 5×B and 3×F from each side of the above equation: 7×B - 5×B - 3×F = A + 5×B + 3×F - 5×B - 3×F which becomes eq.6c) 2×B - 3×F = A
Hint #7
Substitute 2×D for E, and (2×B - 3×F) for A (from eq.6c) in eq.4: 2×D + F = B - (2×B - 3×F) which becomes 2×D + F = B - 2×B + 3×F which becomes 2×D + F = -B + 3×F In the equation above, subtract F from both sides, and add B to both sides: 2×D + F - F + B = -B + 3×F - F + B which becomes eq.4a) 2×D + B = 2×F
Hint #8
Substitute 2×B - 3×F for A (from eq.6c) into eq.2: 2×B - 3×F = F - D Add 3×F and D to both sides of the above equation: 2×B - 3×F + 3×F + D = F - D + 3×F + D which becomes 2×B + D = 4×F which may be written as eq.2a) 2×B + D = 2×(2×F)
Hint #9
Substitute B + 2×D for 2×F (from eq.4a) into eq.2a: 2×B + D = 2×(B + 2×D) which becomes 2×B + D = 2×B + 4×D Subtract 2×B and D from each side of the equation above: 2×B + D - 2×B - D = 2×B + 4×D - 2×B - D which becomes 0 = 3×D which makes 0 = D and also makes E = 2×D = 2×0 = 0
Hint #10
Substitute 0 for D in eq.4a: 2×0 + B = 2×F which becomes 0 + B = 2×F which makes B = 2×F
Hint #11
Substitute 0 for D in eq.2: A = F - 0 which makes A = F
Hint #12
Substitute F for A, and (2×F) for B in eq.1a: F + 5×(2×F) = 11 which becomes F + 10×F = 11 which makes 11×F = 11 Divide both sides of the above equation by 11: 11×F ÷ 11 = 11 ÷ 11 which makes F = 1 making A = F = 1 B = 2×F = 2 × 1 = 2
Solution
Substitute 0 for D and E, 2 for B, and 1 for F in eq.5a: C + 0 + 0 = 11 - 3×2 + 2×1 which becomes C = 11 - 6 + 2 which makes C = 7 and makes ABCDEF = 127001