Puzzle for September 19, 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
Subtract A and D from both sides of eq.3: A + B – A – D = C + D – A – D which becomes eq.3a) B – D = C – A Subtract A and F from both sides of eq.4: C + F – A – F = A + D – A – F which becomes eq.4a) C – A = D – F
Hint #2
In eq.4a, replace C – A with B – D (from eq.3a): B – D = D – F Add D and F to both sides of the above equation: B – D + D + F = D – F + D + F which becomes eq.4b) B + F = 2×D
Hint #3
eq.5 may be written as: E = (B + D + F) ÷ 3 Multiply both sides of the equation above by 3: E × 3 = ((B + D + F) ÷ 3) × 3 which becomes E × 3 = B + D + F which may be written as eq.5a) 3×E = B + F + D
Hint #4
In eq.5a, replace B + F with 2×D (from eq.4b): 3×E = 2×D + D which becomes 3×E = 3×D Divide both sides of the equation above by 3: 3×E ÷ 3 = 3×D ÷ 3 which makes E = D
Hint #5
In eq.2, substitute D for E: D + D = A + F which becomes eq.2a) 2×D = A + F
Hint #6
Substitute B + F for 2×D (from eq.4b) in eq.2a: B + F = A + F Subtract F from each side of the equation above: B + F – F = A + F – F which makes B = A
Hint #7
eq.6 may be written as: C – F = (A + B + D + E) ÷ 4 Multiply both sides of the equation above by 4: (C – F) × 4 = ((A + B + D + E) ÷ 4) × 4 which becomes (C – F) × 4 = A + B + D + E which may be written as eq.6a) 4×C – 4×F = A + B + D + E
Hint #8
Substitute A for B, and D for E in eq.6a: 4×C – 4×F = A + A + D + D which becomes 4×C – 4×F = 2×A + 2×D which may be written as eq.6b) 4×C – 4×F = 2×(A + D)
Hint #9
Substitute C + F for A + D (from eq.4) in eq.6b: 4×C – 4×F = 2×(C + F) which may be written as 4×C – 4×F = 2×C + 2×F In the above equation, add 4×F to both sides, and subtract 2×C from both sides: 4×C – 4×F + 4×F – 2×C = 2×C + 2×F + 4×F – 2×C which becomes 2×C = 6×F Divide both sides by 2: 2×C ÷ 2 = 6×F ÷ 2 which makes C = 3×F
Hint #10
Substitute 3×F for C in eq.4a: 3×F – A = D – F Add F to both sides of the equation above: 3×F – A + F = D – F + F which becomes eq.4c) 4×F – A = D
Hint #11
Substitute (4×F – A) for D (from eq.4c) into eq.2a: 2×(4×F – A) = A + F which becomes 8×F – 2×A = A + F In the equation above, add 2×A to both sides, and subtract F from both sides: 8×F – 2×A + 2×A – F = A + F + 2×A – F which becomes 7×F = 3×A Divide both sides by 3: 7×F ÷ 3 = 3×A ÷ 3 which makes 2⅓×F = A and also makes B = A = 2⅓×F
Hint #12
Substitute 2⅓×F for A in eq.4c: 4×F – 2⅓×F = D which makes 1⅔×F = D and also makes E = D = 1⅔×F
Solution
Substitute 2⅓×F for A and B, 3×F for C, and 1⅔×F for D and E in eq.1: 2⅓×F + 2⅓×F + 3×F + 1⅔×F + 1⅔×F + F = 36 which simplifies to 12×F = 36 Divide both sides of the above equation by 12: 12×F ÷ 12 = 36 ÷ 12 which means F = 3 making A = B = 2⅓×F = 2⅓ × 3 = 7 C = 3×F = 3 × 3 = 9 D = E = 1⅔×F = 1⅔ × 3 = 5 and ABCDEF = 779553