Puzzle for April 12, 2020 ( )
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.
* BC and DE are 2-digit numbers (not B×C or D×E).
Scratchpad
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Hint #1
eq.5 may be written as: F + 10×B + C = 10×D + E Subtract F, 10×B, and E from both sides of the above equation: F + 10×B + C – F – 10×B – E = 10×D + E – F – 10×B – E which becomes eq.5a) C – E = 10×D – 10×B – F
Hint #2
Add F to both sides of eq.2: C – E – F + F = A + D + F which becomes C – E = A + D + F In the above equation, replace A with D + B (from eq.4): C – E = D + B + D + F which becomes eq.2a) C – E = B + 2×D + F
Hint #3
In eq.5a, replace C – E with B + 2×D + F (from eq.2a): B + 2×D + F = 10×D – 10×B – F In the above equation, subtract 2×D from each side, and add 10×B and F to both sides: B + 2×D + F – 2×D + 10×B + F = 10×D – 10×B – F – 2×D + 10×B + F which simplifies to eq.5b) 11×B + 2×F = 8×D
Hint #4
Add B to both sides of eq.3: D – B + B = B – F + B which becomes D = 2×B – F In eq.5b, substitute (2×B – F) for D: 11×B + 2×F = 8×(2×B – F) which becomes 11×B + 2×F = 16×B – 8×F In the above equation, add 8×F to each side, and subtract 11×B from each side: 11×B + 2×F + 8×F – 11×B = 16×B – 8×F + 8×F – 11×B which becomes 10×F = 5×B Divide both sides by 5: 10×F ÷ 5 = 5×B ÷ 5 which makes 2×F = B
Hint #5
In eq.5b, substitute (2×F) for B: 11×(2×F) + 2×F = 8×D which is the same as 22×F + 2×F = 8×D which means 24×F = 8×D Divide both sides of the equation above by 8: 24×F ÷ 8 = 8×D ÷ 8 which makes 3×F = D
Hint #6
Substitute 3×F for D, and 2×F for B in eq.4: 3×F + 2×F = A which makes 5×F = A
Hint #7
Substitute 2×F for B, and (3×F) for D in eq.2a: C – E = 2×F + 2×(3×F) + F which is equivalent to C – E = 2×F + 6×F + F which becomes C – E = 9×F Add E to both sides of the equation above: C – E + E = 9×F + E which becomes eq.2b) C = E + 9×F
Hint #8
Substitute 5×F for A, 2×F for B, E + 9×F for C (from eq.2b), and 3×F for D in eq.1: 5×F + 2×F + E + 9×F + 3×F + E + F = 20 which becomes 2×E + 20×F = 20 Divide both sides of the above equation by 2: (2×E + 20×F) ÷ 2 = 20 ÷ 2 which becomes E + 10×F = 10 Subtract 10×F from both sides: E + 10×F – 10×F = 10 – 10×F which becomes eq.1a) E = 10 – 10×F
Solution
To make eq.1a true, check several possible values for F and E: If F = 0 then E = 10 – 10×0 = 10 – 0 = 10 If F = 1 then E = 10 – 10×1 = 10 – 10 = 0 If F = 2 then E = 10 – 10×2 = 10 – 20 = –10 If F < 2 then E < –10 Since E must be a non-negative one-digit integer, then E = 0 which makes F = 1 making A = 5×F = 5×1 = 5 B = 2×F = 2×1 = 2 C = E + 9×F = 0 + 9×1 = 9 (from eq.2b) D = 3×F = 3×1 = 3 and ABCDEF = 529301