Puzzle for March 7, 2021 ( )
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 F to both sides of eq.4: D + F + F = C – F + F which becomes eq.4a) D + 2×F = C In eq.6, replace C with D + 2×F (from eq.4a): A – B + D = D + 2×F + F which becomes A – B + D = D + 3×F Subtract D from both sides of the equation above: A – B + D – D = D + 3×F – D which becomes eq.6a) A – B = 3×F
Hint #2
In eq.5, replace A – B with 3×F (from eq.6a): B + D – F = 3×F Add F to both sides of the equation above: B + D – F + F = 3×F + F which becomes eq.5a) B + D = 4×F
Hint #3
In eq.3, substitute 4×F for B + D (from eq.5a), and D + 2×F for C (from eq.4a): 4×F + E = A + D + 2×F – D which becomes 4×F + E = A + 2×F Subtract 2×F from both sides of the equation above: 4×F + E – 2×F = A + 2×F – 2×F which becomes eq.3a) 2×F + E = A
Hint #4
Substitute 2×F + E for A (from eq.3a) in eq.6a: 2×F + E – B = 3×F In the equation above, add B to both sides, and subtract 3×F from both sides: 2×F + E – B + B – 3×F = 3×F + B – 3×F which becomes eq.6b) E – F = B
Hint #5
Substitute E – F for B (from eq.6b) in eq.5a: E – F + D = 4×F In the equation above, subtract E from each side, and add F to each side: E – F + D – E + F = 4×F – E + F which becomes eq.5b) D = 5×F – E
Hint #6
Substitute 5×F – E for D (from eq.5b) in eq.4a: 5×F – E + 2×F = C which becomes eq.4c) 7×F – E = C
Hint #7
Substitute 7×F – E for C (from eq.4c), 5×F – E for D (from eq.5b), and 2×F + E for A (from eq.3a) in eq.2: 7×F – E + 5×F – E = 2×F + E + E which becomes 12×F – 2×E = 2×F + 2×E In the equation above, add 2×E to both sides, and subtract 2×F from both sides: 12×F – 2×E + 2×E – 2×F = 2×F + 2×E + 2×E – 2×F which simplifies to 10×F = 4×E Divide both sides by 4: 10×F ÷ 4 = 4×E ÷ 4 which makes 2½×F = E
Hint #8
Substitute 2½×F for E in eq.3a: 2×F + 2½×F = A which makes 4½×F = A
Hint #9
Substitute 2½×F for E in eq.6b: 2½×F – F = B which makes 1½×F = B
Hint #10
Substitute 2½×F for E in eq.4c: 7×F – 2½×F = C which makes 4½×F = C
Hint #11
Substitute 2½×F for E in eq.5b: D = 5×F – 2½×F which makes D = 2½×F
Solution
Substitute 4½×F for A and C, 1½×F for B, and 2½×F for D and E in eq.1: 4½×F + 1½×F + 4½×F + 2½×F + 2½×F + F = 33 which simplifies to 16½×F = 33 Divide both sides of the equation above by 16½: 16½×F ÷ 16½ = 33 ÷ 16½ which means F = 2 making A = C = 4½×F = 4½ × 2 = 9 B = 1½×F = 1½ × 2 = 3 D = E = 2½×F = 2½ × 2 = 5 and ABCDEF = 939552