Puzzle for September 26, 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 both A and C to each side of eq.2: B – C + A + C = D – A + A + C which becomes B + A = D + C which may be written as A + B = C + D In eq.4, replace A + B with C + D: C + E = C + D Subtract C from both sides of the equation above: C + E – C = C + D – C which makes E = D
Hint #2
In eq.3, replace D with E: A + C = B + E + E which becomes A + C = B + 2×E Subtract the left and right sides of eq.4 from the left and right sides of the equation above: A + C – (C + E) = B + 2×E – (A + B) which may be written as A + C – C – E = B + 2×E – A – B which becomes A – E = 2×E – A Add A and E to both sides of the above equation: A – E + A + E = 2×E – A + A + E which makes 2×A = 3×E Divide both sides by 2: 2×A ÷ 2 = 3×E ÷ 2 which means A = 1½×E
Hint #3
Subtract D from both sides of eq.5: D + F – D = B – D which becomes F = B – D In the above equation, substitute E for D: eq.5a) F = B – E
Hint #4
Substitute E for D, 1½×E for A, and B – E for F (from eq.5a) in eq.6: B – E – E = 1½×E – B + B – E which becomes B – 2×E = ½×E Add 2×E to both sides of the above equation: B – 2×E + 2×E = ½×E + 2×E which makes B = 2½×E
Hint #5
Substitute 2½×E for B in eq.5a: F = 2½×E – E which makes F = 1½×E
Hint #6
Substitute 1½×E for A, and 2½×E for B in eq.4: C + E = 1½×E + 2½×E which becomes C + E = 4×E Subtract E from both sides of the above equation: C + E – E = 4×E – E which makes C = 3×E
Solution
Substitute 1½×E for A and F, 2½×E for B, 3×E for C, and E for D into eq.1: 1½×E + 2½×E + 3×E + E + E + 1½×E = 21 which simplifies to 10½×E = 21 Divide both sides of the equation above by 10½: 10½×E ÷ 10½ = 21 ÷ 10½ = 2 which means E = 2 making A = F = 1½×E = 1½ × 2 = 3 B = 2½×E = 2½ × 2 = 5 C = 3×E = 3 × 2 = 6 D = E = 2 and ABCDEF = 356223