Puzzle for July 20, 2019 ( )
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.
* AB and CD are 2-digit numbers (not A×B or C×D).
Scratchpad
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Hint #1
Add both A and B to each side of eq.3: D - A + A + B = A - B + A + B which becomes eq.3a) D + B = 2×A
Hint #2
Add A to both sides of eq.5: D + E - A + A = A + F + A which becomes D + E = 2×A + F In the above equation, replace 2×A with D + B (from eq.3a): D + E = D + B + F Subtract D from both sides: D + E - D = D + B + F - D which becomes eq.5a) E = B + F
Hint #3
In eq.2, replace E with B + F (from eq.5a): B + C = B + F + F Subtract B from both sides: B + C - B = B + F + F - B which makes C = 2×F
Hint #4
In eq.2, replace E with C + D (from eq.4): B + C = C + D + F Subtract C from both sides of the above equation: B + C - C = C + D + F - C which simplifies to eq.2a) B = D + F
Hint #5
eq.6 may be written as: 10×A + B - (10×C + D) = D + E which is the same as 10×A + B - 10×C - D = D + E Add D to each side of the equation above: 10×A + B - 10×C - D + D = D + E + D which becomes 10×A + B - 10×C = 2×D + E which may be written as eq.6a) 5×(2×A) + B - 10×C = 2×D + E
Hint #6
Substitute (D + B) for 2×A (from eq.3a), 2×F for C, and B + F for E (from eq.5a) in eq.6a: 5×(D + B) + B - 10×(2×F) = 2×D + B + F which becomes 5×D + 5×B + B - 20×F = 2×D + B + F In the equation above, add 20×F to each side, and subtract both B and 2×D from each side: 5×D + 5×B + B - 20×F + 20×F - B - 2×D = 2×D + B + F + 20×F - B - 2×D which simplifies to eq.6b) 3×D + 5×B = 21×F
Hint #7
Substitute (D + F) for B (from eq.2a) in eq.6b: 3×D + 5×(D + F) = 21×F which is equivalent to 3×D + 5×D + 5×F = 21×F Subtract 5×F from each side of the equation above: 3×D + 5×D + 5×F - 5×F = 21×F - 5×F which becomes 8×D = 16×F Divide both sides by 8: 8×D ÷ 8 = 16×F ÷ 8 which makes D = 2×F
Hint #8
Substitute 2×F for D in eq.2a: B = 2×F + F which makes B = 3×F
Hint #9
Substitute 2×F for D in eq.5a: E = 3×F + F which makes E = 4×F
Hint #10
Substitute 2×F for D, and 3×F for B in eq.3a: 2×F + 3×F = 2×A which becomes 5×F = 2×A Divide both sides of the equation above by 2: 5×F ÷ 2 = 2×A ÷ 2 which means 2½×F = A
Solution
Substitute 2½×F for A, 3×F for B, 2×F for C and D, and 4×F for E in eq.1: 2½×F + 3×F + 2×F + 2×F + 4×F + F = 29 which simplifies to 14½×F = 29 Divide both sides by 14½: 14½×F ÷ 14½ = 29 ÷ 14½ which means F = 2 making A = 2½×F = 2½ × 2 = 5 B = 3×F = 3 × 2 = 6 C = D = 2×F = 2 × 2 = 4 E = 4×F = 4 × 2 = 8 and ABCDEF = 564482