Puzzle for December 14, 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.6, replace C - D with D - E (from eq.5): A + D - E = D + E + F In the above equation, subtract D from both sides, and add E to both sides: A + D - E - D + E = D + E + F - D + E which becomes eq.6a) A = 2×E + F
Hint #2
In eq.4, replace A with 2×E + F (from eq.6a): D + F = 2×E + F - F which becomes eq.4a) D + F = 2×E
Hint #3
Add F and B to both sides of eq.2: B - F + F + B = D - B + F + B which becomes eq.2a) 2×B = D + F
Hint #4
In eq.2a, substitute 2×E for D + F (from eq.4a): 2×B = 2×E Divide both sides of the above equation by 2: 2×B ÷ 2 = 2×E ÷ 2 which makes B = E
Hint #5
Substitute B for E in eq.3: C + B = A + B Subtract B from each side of the equation above: C + B - B = A + B - B which makes C = A
Hint #6
Substitute A for C in eq.6: A + A - D = D + E + F which becomes 2×A - D = D + E + F Add D to both sides of the equation above: 2×A - D + D = D + E + F + D which becomes eq.6b) 2×A = 2×D + E + F
Hint #7
Subtract F from both sides of eq.4a: D + F - F = 2×E - F which becomes eq.4b) D = 2×E - F
Hint #8
Substitute (2×E + F) for A (from eq.6a), and (2×E - F) for D (from eq.4b) in eq.6b: 2×(2×E + F) = 2×(2×E - F) + E + F which becomes 4×E + 2×F = 4×E - 2×F + E + F which becomes 4×E + 2×F = 5×E - F In the above equation, subtract 4×E from both sides, and add F to both sides: 4×E + 2×F - 4×E + F = 5×E - F - 4×E + F which makes 3×F = E and also makes 3×F = E = B
Hint #9
Substitute (3×F) for E in eq.4b: D = 2×(3×F) - F which becomes D = 6×F - F which makes D = 5×F
Hint #10
Substitute (3×F) for E in eq.6a: A = 2×(3×F) + F which becomes A = 6×F + F which makes A = 7×F and also makes C = A = 7×F
Solution
Substitute 7×F for A and C, 3×F for B and E, and 5×F for D in eq.1: 7×F + 3×F + 7×F + 5×F + 3×F + F = 26 which simplifies to 26×F = 26 Divide both sides of the above equation by 26: 26×F ÷ 26 = 26 ÷ 26 which means F = 1 making A = C = 7×F = 7×1 = 7 B = E = 3×F = 3×1 = 3 D = 5×F = 5×1 = 5 and ABCDEF = 737531