Puzzle for July 8, 2023 ( )
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 DE are 2-digit numbers (not A×B or D×E).
Scratchpad
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Hint #1
Add B, F, and C to both sides of eq.5: E - B - F + B + F + C = A - C - E + B + F + C which becomes E + C = A - E + B + F which may be written as eq.5a) E + C = A + B + F - E
Hint #2
In eq.5a, replace A + B with E + F (from eq.2): E + C = E + F + F - E which becomes eq.5b) E + C = 2×F
Hint #3
Add B and F to both sides of eq.4: C + F - B + B + F = B + E - F + B + F which becomes eq.4a) C + 2×F = 2×B + E
Hint #4
In eq.4a, replace 2×F with E + C (from eq.5b): C + E + C = 2×B + E which becomes 2×C + E = 2×B + E Subtract E from both sides of the equation above: 2×C + E - E = 2×B + E - E which makes 2×C = 2×B Divide both sides by 2: 2×C ÷ 2 = 2×B ÷ 2 which makes C = B
Hint #5
Subtract C from each side of eq.5b: E + C - C = 2×F - C which becomes eq.5c) E = 2×F - C
Hint #6
In eq.5a, substitute (2×F - C) for E (from eq.5c), and C for B: (2×F - C) + C = A + C + F - (2×F - C) which becomes 2×F = A + C + F - 2×F + C which becomes 2×F = A + 2×C - F In the above equation, subtract 2×C from both sides, and add F to both sides: 2×F - 2×C + F = A + 2×C - F - 2×C + F which becomes eq.5d) 3×F - 2×C = A
Hint #7
Substitute C for B, (2×F - C) for E (from eq.5c), and 3×F - 2×C for A (from eq.5d) in eq.3: C + D - (2×F - C) = 3×F - 2×C + C - D which becomes C + D - 2×F + C = 3×F - C - D which becomes 2×C + D - 2×F = 3×F - C - D In the above equation, subtract 2×C from both sides, and add 2×F and D to both sides: 2×C + D - 2×F - 2×C + 2×F + D = 3×F - C - D - 2×C + 2×F + D which simplifies to 2×D = 5×F - 3×C Divide both sides by 2: 2×D ÷ 2 = (5×F - 3×C) ÷ 2 which becomes eq.3a) D = 2½×F - 1½×C
Hint #8
eq.6 may be written as: 10×A + B = A + 10×D + E Subtract A from each side of the equation above: 10×A + B - A = A + 10×D + E - A which becomes eq.6a) 9×A + B = 10×D + E
Hint #9
Substitute (3×F - 2×C) for A (from eq.5d), C for B, (2½×F - 1½×C) for D (from eq.3a), and 2×F - C for E (from eq.5c) in eq.6a: 9×(3×F - 2×C) + C = 10×(2½×F - 1½×C) + 2×F - C which becomes 27×F - 18×C + C = 25×F - 15×C + 2×F - C which becomes 27×F - 17×C = 27×F - 16×C In the above equation, subtract 27×F from both sides, and add 17×C to both sides: 27×F - 17×C - 27×F + 17×C = 27×F - 16×C - 27×F + 17×C which makes 0 = C and also makes 0 = C = B
Hint #10
Substitute 0 for C in eq.5c: E = 2×F - 0 which makes E = 2×F
Hint #11
Substitute 0 for C in eq.5d: 3×F - 2×0 = A which becomes 3×F - 0 = A which makes 3×F = A
Hint #12
Substitute 0 for C in eq.3a: D = 2½×F - 1½×0 which becomes D = 2½×F - 0 which makes D = 2½×F
Solution
Substitute 3×F for A, 0 for B and C, 2½×F for D, and 2×F for E in eq.1: 3×F + 0 + 0 + 2½×F + 2×F + F = 17 which simplifies to 8½×F = 17 Divide both sides of the above equation by 8½: 8½×F ÷ 8½ = 17 ÷ 8½ which means F = 2 making A = 3×F = 3 × 2 = 6 D = 2½×F = 2½ × 2 = 5 E = 2×F = 2 × 2 = 4 and ABCDEF = 600542