Puzzle for March 31, 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 EF are 2-digit numbers (not A×B or E×F).
Scratchpad
Help Area
Hint #1
Add B to both sides of eq.2: C + B = A - B + B which becomes eq.2a) C + B = A Add (B - F) to both sides of eq.3: B + F + (B - F) = E - B + (B - F) which becomes eq.3a) 2×B = E - F
Hint #2
In eq.4, replace E - F with 2×B (from eq.3a), and replace A with C + B (from eq.2a): 2×B = C + B - D Replace C with B + E (from eq.5): 2×B = B + E + B - D which becomes 2×B = 2×B + E - D Add D to each side, and subtract 2×B from each side of the above equation: 2×B + D - 2×B = 2×B + E - D + D - 2×B which simplifies to D = E
Hint #3
In eq.4, substitute D for E: D - F = A - D Add (D + F - A) to each side of the above equation: D - F + (D + F - A) = A - D + (D + F - A) which becomes eq.4a) 2×D - A = F
Hint #4
eq.6 may be written as: 10×A + B = B + C + D + 10×E + F Subtract B from each side of the above equation: 10×A + B - B = C + B + D + 10×E + F - B which becomes 10×A = C + D + 10×E + F Substitute B + E for C (from eq.5): 10×A = B + E + D + 10×E + F which becomes eq.6a) 10×A = B + D + 11×E + F
Hint #5
Multiply both sides of eq.6a by 2: 2 × 10×A = 2 × (B + D + 11×E + F) which becomes 20×A = 2×B + 2×D + 22×E + 2×F Substitute E - F for 2×B (from eq.3a): 20×A = E - F + 2×D + 22×E + 2×F which becomes eq.6b) 20×A = 2×D + 23×E + F
Hint #6
Substitute D for E, and 2×D - A for F (from eq.4a) in eq.6b: 20×A = 2×D + 23×D + 2×D - A which becomes 20×A = 27×D - A Add A to both sides: 20×A + A = 27×D - A + A which makes 21×A = 27×D Divide both sides by 3: 21×A ÷ 3 = 27×D ÷ 3 which makes eq.6c) 7×A = 9×D
Hint #7
Multiply both sides of eq.4a by 7: 7 × (2×D - A) = 7×F which becomes 14×D - 7×A = 7×F Substitute 9×D for 7×A (from eq.6c): 14×D - 9×D = 7×F which becomes 5×D = 7×F Divide both sides by 5: 5×D ÷ 5 = 7×F ÷ 5 which makes D = 1⅖×F and which also makes E = D = 1⅖×F
Hint #8
In eq.3a, substitute 1⅖×F for E: 2×B = 1⅖×F - F which becomes 2×B = ⅖×F Divide each side by 2: 2×B ÷ 2 = ⅖×F ÷ 2 which makes B = ⅕×F
Hint #9
Substitute 1⅖×F for D in eq.4a: 2×1⅖×F - A = F which becomes 2⅘×F - A = F Add (A - F) to both sides: 2⅘×F - A + (A - F) = F + (A - F) which simplifies to 1⅘×F = A
Hint #10
Substitute 1⅘×F for A, and ⅕×F for B in eq.2: C = 1⅘×F - ⅕×F which makes C = 1⅗×F
Solution
Substitute 1⅘×F for A, ⅕×F for B, 1⅗×F for C, and 1⅖×F for D and E in eq.1: 1⅘×F + ⅕×F + 1⅗×F + 1⅖×F + 1⅖×F + F = 37 which simplifies to 7⅖×F = 37 Divide both sides by 7⅖: 7⅖×F ÷ 7⅖ = 37 ÷ 7⅖ which means F = 5 making A = 1⅘×F = 1⅘ × 5 = 9 B = ⅕×F = ⅕ × 5 = 1 C = 1⅗×F = 1⅗ × 5 = 8 D = E = 1⅖×F = 1⅖ × 5 = 7 and ABCDEF = 918775