Puzzle for October 31, 2019 ( )
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
Help Area
Hint #1
Add D, E, and F to both sides of eq.4: C – E + D + E + F = B – D – F + D + E + F which becomes C + D + F = B + E In eq.5, replace B + E with C + D + F: C + D + F = C – D + F Subtract both C and F from each side of the above equation: C + D + F – C – F = C – D + F – C – F which simplifies to D = –D which means D = 0
Hint #2
Add F to both sides of eq.6: A – F + F = B + F + F which becomes eq.6a) A = B + 2×F In eq.3, replace A with B + 2×F (from eq.6a): B + C = B + 2×F + E + F which becomes B + C = B + E + 3×F Subtract B from both sides of the equation above: B + C – B = B + E + 3×F – B which becomes eq.3a) C = E + 3×F
Hint #3
In eq.5, substitute E + 3×F for C (from eq.3a), and 0 for D: B + E = E + 3×F – 0 + F which becomes B + E = E + 4×F Subtract E from both sides of the above equation: B + E – E = E + 4×F – E which makes B = 4×F
Hint #4
Substitute 4×F for B in eq.6a: A = 4×F + 2×F which makes A = 6×F
Hint #5
Substitute 6×F for A, and 0 for D in eq.2: C + 0 + F = 6×F – E Subtract F from each side of the above equation: C + 0 + F – F = 6×F – E – F which becomes eq.2a) C = 5×F – E
Hint #6
Substitute 5×F – E for C (from eq.2a) in eq.3a: 5×F – E = E + 3×F In the above equation, add E to both sides, and subtract 3×F from both sides: 5×F – E + E – 3×F = E + 3×F + E – 3×F which simplifies to 2×F = 2×E Divide both sides by 2: 2×F ÷ 2 = 2×E ÷ 2 which makes F = E
Hint #7
Substitute F for E in eq.3a: C = F + 3×F which makes C = 4×F
Solution
Substitute 6×F for A, 4×F for B and C, 0 for D, and F for E in eq.1: 6×F + 4×F + 4×F + 0 + F + F = 16 which simplifies to 16×F = 16 Divide both sides of the equation above by 16: 16×F ÷ 16 = 16 ÷ 16 which means F = 1 making A = 6×F = 6×1 = 6 B = C = 4×F = 4×1 = 4 E = F = 1 and ABCDEF = 644011