Puzzle for October 16, 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
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Hint #1
Add E and A to both sides of eq.2: F – E + E + A = C + E – A + E + A which becomes F + A = C + 2×E which may be written as eq.2a) A + F = C + 2×E Add D to both sides of eq.6: A – D + F + D = B + C + D + E + D which becomes eq.6a) A + F = B + C + 2×D + E
Hint #2
In eq.2a, replace A + F with B + C + 2×D + E (from eq.6a): B + C + 2×D + E = C + 2×E Subtract C and E from both sides of the above equation: B + C + 2×D + E – C – E = C + 2×E – C – E which becomes eq.2b) B + 2×D = E
Hint #3
Add F and C to both sides of eq.5: B + C + D – F + F + C = A – C + F + C which becomes eq.5a) B + 2×C + D = A + F In eq.6a, replace A + F with B + 2×C + D (from eq.5a), and E with B + 2×D (from eq.2b): B + 2×C + D = B + C + 2×D + B + 2×D which becomes B + 2×C + D = 2×B + C + 4×D Subtract B, C, and D from both sides of the equation above: B + 2×C + D – B – C – D = 2×B + C + 4×D – B – C – D which simplifies to eq.6b) C = B + 3×D
Hint #4
Substitute B + 3×D for C (from eq.6b), and B + 2×D for E (from eq.2b) in eq.3: B + 3×D + B + 2×D = B + D + F which becomes 2×B + 5×D = B + D + F Subtract B and D from each side of the equation above: 2×B + 5×D – B – D = B + D + F – B – D which becomes eq.3a) B + 4×D = F
Hint #5
Substitute (B + 3×D) for C (from eq.6b), and (B + 4×D) for F (from eq.3a) in eq.5: B + (B + 3×D) + D – (B + 4×D) = A – (B + 3×D) which becomes 2×B + 4×D – B – 4×D = A – B – 3×D which becomes B = A – B – 3×D Add B and 3×D to both sides of the equation above: B + B + 3×D = A – B – 3×D + B + 3×D which becomes eq.5b) 2×B + 3×D = A
Hint #6
Substitute B + 2×D for E (from eq.2b), B + 4×D for F (from eq.3a), 2×B + 3×D for A (from eq.5b), and B + 3×D for C (from eq.6b) in eq.4: D + B + 2×D + B + 4×D = 2×B + 3×D + B + 3×D which becomes 2×B + 7×D = 3×B + 6×D Subtract 2×B and 6×D from both sides of the equation above: 2×B + 7×D – 2×B – 6×D = 3×B + 6×D – 2×B – 6×D which makes D = B
Hint #7
Substitute B for D in eq.5b: 2×B + 3×B = A which makes 5×B = A
Hint #8
Substitute B for D in eq.6b: C = B + 3×B which makes C = 4×B
Hint #9
Substitute B for D in eq.2b: B + 2×B = E which makes 3×B = E
Hint #10
Substitute B for D in eq.3a: B + 4×B = F which makes 5×B = F
Solution
Substitute 5×B for A and F, 4×B for C, B for D, and 3×B for E in eq.1: 5×B + B + 4×B + B + 3×B + 5×B = 19 which simplifies to 19×B = 19 Divide both sides of the above equation by 19: 19×B ÷ 19 = 19 ÷ 19 which makes B = 1 making A = F = 5×B = 5 × 1 = 5 C = 4×B = 4 × 1 = 4 D = B = 1 E = 3×B = 3 × 1 = 3 and ABCDEF = 514135