Puzzle for June 30, 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 positive integer.
* BC, DE, and EF are 2-digit numbers (not B×C, D×E, or E×F).
Scratchpad
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Hint #1
Add D to both sides of eq.2: E - D + D = F + D which becomes eq.2a) E = F + D
Hint #2
In eq.4, replace E with F + D (from eq.2a): C + D - F = F + D - B In the equation above, subtract D from both sides: C + D - F - D = F + D - B - D which becomes C - F = F - B Add B and F to both sides: C - F + B + F = F - B + B + F which becomes C + B = 2×F which may be written as eq.4a) B + C = 2×F
Hint #3
In eq.6, replace B + C with 2×F (from eq.4a): D × F = 2×F Divide both sides of the above equation by F: D × F ÷ F = 2×F ÷ F which makes D = 2
Hint #4
In eq.2a, substitute 2 for D: eq.2b) E = F + 2 Subtract 2 from each side of eq.2b: E - 2 = F + 2 - 2 which makes eq.2c) E - 2 = F
Hint #5
Substitute (E - 2) for F (from eq.2c) in eq.4a: B + C = 2×(E - 2) which becomes B + C = 2×E - 4 Subtract C from each side of the above equation: B + C - C = 2×E - 4 - C which becomes eq.4b) B = 2×E - 4 - C
Hint #6
eq.5 may be written as: 10×B + C + 10×D + E = A + 10×E + F In the above equation, substitute 2 for D, and E - 2 for F (from eq.2c): 10×B + C + 10×2 + E = A + 10×E + E - 2 which becomes 10×B + C + 20 + E = A + 11×E - 2 Subtract E from each side, and add 2 to each side: 10×B + C + 20 + E - E + 2 = A + 11×E - 2 - E + 2 which becomes eq.5a) 10×B + C + 22 = A + 10×E
Hint #7
Subtract C from both sides of eq.3: A + C - C = B + E - C which becomes A = B + E - C Substitute B + E - C for A in eq.5a: 10×B + C + 22 = B + E - C + 10×E Subtract B from both sides, and add C to both sides: 10×B + C + 22 - B + C = B + E - C + 10×E - B + C which becomes eq.5b) 9×B + 2×C + 22 = 11×E
Hint #8
Substitute (2×E - 4 - C) for B (from eq.4b) in eq.5b: 9×(2×E - 4 - C) + 2×C + 22 = 11×E which is equivalent to 18×E - 36 - 9×C + 2×C + 22 = 11×E which becomes 18×E - 7×C - 14 = 11×E Add 7×C to both sides, and subtract 11×E from both sides: 18×E - 7×C - 14 + 7×C - 11×E = 11×E + 7×C - 11×E which becomes 7×E - 14 = 7×C Divide both sides by 7: (7×E - 14) ÷ 7 = 7×C ÷ 7 which makes eq.5c) E - 2 = C
Hint #9
Substitute (E - 2) for C (from eq.5c) in eq.4b: B = 2×E - 4 - (E - 2) which is the same as B = 2×E - 4 - E + 2 which simplifies to eq.5d) B = E - 2 and which also means B = C
Hint #10
Substitute B for C in eq.3: A + B = B + E Subtract B from both sides of the above equation: A + B - B = B + E - B which makes A = E
Solution
Substitute E for A, E - 2 for B (from eq.5d) and for C (from eq.5c) and for F (from eq.2c), and 2 for D in eq.1: E + E - 2 + E - 2 + 2 + E + E - 2 = 41 which simplifies to 5×E - 4 = 41 Add 4 to each side: 5×E - 4 + 4 = 41 + 4 which makes 5×E = 45 Divide both sides by 5: 5×E ÷ 5 = 45 ÷ 5 which means E = 9 making A = E = 9 B = C = F = E - 2 = 9 - 2 = 7 (from eq.5d, eq.5c, and eq.2c) and ABCDEF = 977297