Puzzle for May 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
Subtract F from both sides of eq.3: A + B – F – F = C + D + F – F which becomes eq.3a) A + B – 2×F = C + D Add D, A, and F to both sides of eq.4: A – D + E + D + A + F = C – A – F + D + A + F which becomes eq.4a) 2×A + E + F = C + D
Hint #2
In eq.4a, replace C + D with A + B – 2×F (from eq.3a): 2×A + E + F = A + B – 2×F In the equation above, subtract A from both sides, and add 2×F to both sides: 2×A + E + F – A + 2×F = A + B – 2×F – A + 2×F which becomes eq.4b) A + E + 3×F = B
Hint #3
eq.6 may be written as: A + E + F = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × (A + E + F) = 3 × (B + C + D) ÷ 3 which becomes eq.6a) 3×A + 3×E + 3×F = B + C + D
Hint #4
In eq.6a, replace B with A + E + 3×F (from eq.4b), and replace C + D with 2×A + E + F (from eq.4a): 3×A + 3×E + 3×F = A + E + 3×F + 2×A + E + F which becomes 3×A + 3×E + 3×F = 3×A + 2×E + 4×F Subtract 3×A, 2×E, and 3×F from both sides of the above equation: 3×A + 3×E + 3×F – 3×A – 2×E – 3×F = 3×A + 2×E + 4×F – 3×A – 2×E – 3×F which simplifies to E = F
Hint #5
Subtract the left and right sides of eq.5 from the left and right sides of eq.3, respectively: A + B – F – (A + B – D) = C + D + F – (D + E + F) which becomes A + B – F – A – B + D = C + D + F – D – E – F which becomes eq.3b) –F + D = C – E
Hint #6
In eq.3b, substitute F for E: –F + D = C – F Add F to both sides of the equation above: –F + D + F = C – F + F which makes D = C
Hint #7
In eq.4b, substitute F for E: A + F + 3×F = B which becomes eq.4c) A + 4×F = B
Hint #8
Substitute F for E, and D for C in eq.4a: 2×A + F + F = D + D which becomes 2×A + 2×F = 2×D Divide both sides of the above equation by 2: (2×A + 2×F) ÷ 2 = 2×D ÷ 2 which becomes eq.4d) A + F = D
Hint #9
Substitute A + F for D (from eq.4d), A + 4×F for B (from eq.4c), and F for E in eq.2: A + A + F = A + 4×F + F + F which becomes 2×A + F = A + 6×F Subtract F and A from both sides of the equation above: 2×A + F – F – A = A + 6×F – F – A which makes A = 5×F
Hint #10
Substitute 5×F for A in eq.4c: 5×F + 4×F = B which makes 9×F = B
Hint #11
Substitute 5×F for A in eq.4d: 5×F + F = D which makes 6×F = D and also makes 6×F = D = C
Solution
Substitute 5×F for A, 9×F for B, 6×F for C and D, and F for E in eq.1: 5×F + 9×F + 6×F + 6×F + F + F = 28 which simplifies to 28×F = 28 Divide both sides of the above equation by 28: 28×F ÷ 28 = 28 ÷ 28 which means F = 1 making A = 5×F = 5 × 1 = 5 B = 9×F = 9 × 1 = 9 C = D = 6×F = 6 × 1 = 6 E = F = 1 and ABCDEF = 596611