Puzzle for March 10, 2021 ( )
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 is a 2-digit number (not B×C).
Scratchpad
Help Area
Hint #1
eq.5 may be re-written as: A + C – D = D + B + F In the above equation, replace B + F with D + E (from eq.4): A + C – D = D + D + E which becomes A + C – D = 2×D + E Add D to both sides: A + C – D + D = 2×D + E + D which becomes eq.5a) A + C = 3×D + E
Hint #2
In eq.1, replace A + C with 3×D + E (from eq.5a): E + F = 3×D + E + D which becomes E + F = 4×D + E Subtract E from each side of the above equation: E + F – E = 4×D + E – E which makes F = 4×D
Hint #3
In eq.2, substitute 4×D for F: C = D + 4×D which makes C = 5×D
Hint #4
In eq.3, replace A + C with 3×D + E (from eq.5a): B – D = 3×D + E – E which becomes B – D = 3×D Add D to both sides of the above equation: B – D + D = 3×D + D which makes B = 4×D
Hint #5
Substitute 5×D for C, and 4×D for B and F into eq.5: A + 5×D – D = 4×D + D + 4×D which becomes A + 4×D = 9×D Subtract 4×D from each side of the equation above: A + 4×D – 4×D = 9×D – 4×D which makes A = 5×D
Solution
eq.6 may be written as: 10×B + C = (B + C) × C Substitute (4×D) for B, and 5×D for C in the above euation: 10×(4×D) + 5×D = ((4×D) + 5×D) × 5×D which becomes 40×D + 5×D = 9×D × 5×D which becomes 45×D = 45×D² Divide both sides by 45×D: 45×D ÷ 45×D = 45×D² ÷ 45×D which makes 1 = D making A = C = 5×D = 5 × 1 = 5 B = F = 4×D = 4 × 1 = 4 E = 7×D = 7 × 1 = 7 and ABCDEF = 545174