Puzzle for May 28, 2022 ( )
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 D and A to both sides of eq.4: A – D + D + A = F – A + D + A which becomes 2×A = F + D which is the same as eq.4a) 2×A = D + F In eq.2, replace D + F with 2×A (from eq.4a): 2×A = A + E Subtract A from each side of the equation above: 2×A – A = A + E – A which makes A = E
Hint #2
eq.6 may be written as: D – B = (A + B + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (D – B) = 4 × (A + B + D + E) ÷ 4 which becomes 4×D – 4×B = A + B + D + E Subtract B and D from both sides: 4×D – 4×B – B – D = A + B + D + E – B – D which becomes eq.6a) 3×D – 5×B = A + E
Hint #3
In eq.6a, substitute D + F for A + E (from eq.2): 3×D – 5×B = D + F Subtract D from each side of the above equation: 3×D – 5×B – D = D + F – D which becomes eq.6b) 2×D – 5×B = F
Hint #4
In eq.6a, replace E with A: 3×D – 5×B = A + A which becomes 3×D – 5×B = 2×A Divide both sides of the above equation by 2: (3×D – 5×B) ÷ 2 = 2×A ÷ 2 which makes 1½×D – 2½×B = A and also makes eq.6c) 1½×D – 2½×B = A = E
Hint #5
Substitute 1½×D – 2½×B for A (from eq.6c) and 2×D – 5×B for F (from eq.6b) in eq.3: 1½×D – 2½×B + C = B + 2×D – 5×B which becomes 1½×D – 2½×B + C = 2×D – 4×B In the above equation, subtract 1½×D from both sides, and add 2½×B to both sides: 1½×D – 2½×B + C – 1½×D + 2½×B = 2×D – 4×B – 1½×D + 2½×B which simplifies to eq.3a) C = ½×D – 1½×B
Hint #6
Substitute 1½×D – 2½×B for E (from eq.6c), 2×D – 5×B for F (from eq.6b), and ½×D – 1½×B for C (from eq.3a) in eq.5: 1½×D – 2½×B + 2×D – 5×B – D = B + ½×D – 1½×B + D which becomes 2½×D – 7½×B = 1½×D – ½×B In the above equation, add 7½×B to both sides, and subtract 1½×D from both sides: 2½×D – 7½×B + 7½×B – 1½×D = 1½×D – ½×B + 7½×B – 1½×D which simplifies to D = 7×B
Hint #7
Substitute (7×B) for D in eq.3a: C = ½×(7×B) – 1½×B which becomes C = 3½×B – 1½×B which makes C = 2×B
Hint #8
Substitute (7×B) for D in eq.6b: 2×(7×B) – 5×B = F which becomes 14×B – 5×B = F which makes 9×B = F
Hint #9
Substitute (7×B) for D in eq.6c: 1½×(7×B) – 2½×B = A = E which becomes 10½×B – 2½×B = A = E which makes 8×B = A = E
Solution
Substitute 8×B for A and E, 2×B for C, 7×B for D, and 9×B for F in eq.1: 8×B + B + 2×B + 7×B + 8×B + 9×B = 35 which simplifies to 35×B = 35 Divide both sides of the above equation by 35: 35×B ÷ 35 = 35 ÷ 35 which means B = 1 making A = E = 8×B = 8 × 1 = 8 C = 2×B = 2 × 1 = 2 D = 7×B = 7 × 1 = 7 F = 9×B = 9 × 1 = 9 and ABCDEF = 812789