Puzzle for March 26, 2020 ( )
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 C with B + D (from eq.4): B + B + D = A + E which becomes eq.3a) 2×B + D = A + E In eq.5, subtract B from both sides, and add E to both sides: D – E – B + E = A + B – B + E which becomes eq.5a) D – B = A + E
Hint #2
In eq.5a, replace A + E with 2×B + D (from eq.3a): D – B = 2×B + D In the equation above, add B to both sides, and subtract D from both sides: D – B + B – D = 2×B + D + B – D which makes 0 = 3×B which means 0 = B
Hint #3
In eq.4, replace B with 0: C = 0 + D which makes C = D
Hint #4
In eq.3, replace B with 0: 0 + C = A + E which becomes eq.3b) C = A + E
Hint #5
Add E to each side of eq.6: A – E + F + E = C + E + E which becomes A + F = C + 2×E In the equation above, substitute A + E for C (from eq.3b): A + F = A + E + 2×E which becomes A + F = A + 3×E Subtract A from both sides: A + F – A = A + 3×E – A which makes F = 3×E
Hint #6
Add F to both sides of eq.2: F – D + F = A – F + F which becomes 2×F – D = A Substitute (3×E) for F, and C for D in the equation above: 2×(3×E) – C = A which becomes eq.2a) 6×E – C = A
Hint #7
Substitute (A + E) for C (from eq.3b) in eq.2a: 6×E – (A + E) = A which is equivalent to 6×E – A – E = A which becomes 5×E – A = A Add A to both sides of the equation above: 5×E – A + A = A + A which makes 5×E = 2×A Divide both sides by 2: 5×E ÷ 2 = 2×A ÷ 2 which means 2½×E = A
Hint #8
Substitute 2½×E for A in eq.3b: C = 2½×E + E which makes C = 3½×E and which also makes D = C = 3½×E
Solution
Substitute 2½×E for A, 0 for B, 3½×E for C and D, and 3×E for F in eq.1: 2½×E + 0 + 3½×E + 3½×E + E + 3×E = 24 which simplifies to 12×E = 24 Divide both sides of the above equation by 12: 12×E ÷ 12 = 24 ÷ 12 which means E = 2 making A = 2½×E = 2½ × 2 = 5 C = D = 3½×E = 3½ × 2 = 7 F = 3×E = 3 × 2 = 6 and ABCDEF = 507726