Puzzle for February 24, 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
eq.6 may be written as: A + D – C = B + C + F In the equation above, replace A + D with B + C + E + F (from eq.3): B + C + E + F – C = B + C + F which becomes B + E + F = B + C + F Subtract B and F from both sides: B + E + F – B – F = B + C + F – B – F which simplifies to eq.6a) E = C
Hint #2
In eq.2, replace E with C: C + C = A which makes eq.2a) 2×C = A
Hint #3
Substitute 2×C for A, and C for E in eq.5: B + C + D = 2×C + C Subtract B and C from each side of the above equation: B + C + D – B – C = 2×C + C – B – C which becomes eq.5a) D = 2×C – B
Hint #4
Substitute 2×C – B for D (from eq.5a) in eq.4: 2×C – B – B = B + C which becomes 2×C – 2×B = B + C In the equation above, add 2×B to each side, and subtract C from each side: 2×C – 2×B + 2×B – C = B + C + 2×B – C which makes C = 3×B and also makes E = C = 3×B
Hint #5
Substitute (3×B) for C in eq.5a: D = 2×(3×B) – B which becomes D = 6×B – B which makes D = 5×B
Hint #6
Substitute (3×B) for C in eq.2a: 2×(3×B) = A which makes 6×B = A
Hint #7
Substitute 3×B for C and E, 6×B for A, and 5×B for D in eq.3: B + 3×B + 3×B + F = 6×B + 5×B which becomes 7×B + F = 11×B Subtract 7×B from both sides of the above equations: 7×B + F – 7×B = 11×B – 7×B which makes F = 4×B
Solution
Substitute 6×B for A, 3×B for C and E, 5×B for D, and 4×B for F in eq.1: 6×B + B + 3×B + 5×B + 3×B + 4×B = 22 which simplifies to 22×B = 22 Divide both sides of the equation above by 22: 22×B ÷ 22 = 22 ÷ 22 which means B = 1 making A = 6×B = 6 × 1 = 6 C = E = 3×B = 3 × 1 = 3 D = 5×B = 5 × 1 = 5 F = 4×B = 4 × 1 = 4 and ABCDEF = 613534