Puzzle for April 2, 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 is a 2-digit number (not A×B).
Scratchpad
Help Area
Hint #1
Add C to both sides of eq.3: D + F - B + C = A + B - C + C which becomes eq.3a) D + F - B + C = A + B
Hint #2
In eq.4, replace A + B with D + F - B + C (from eq.3a): D + F - B + C + C = D + E + F which becomes D + F - B + 2×C = D + E + F Subtract D and F from each side of the equation above: D + F - B + 2×C - D - F = D + E + F - D - F which becomes -B + 2×C = E which may be written as eq.4a) 2×C - B = E
Hint #3
Add C to both sides of eq.2: F - C + C = C - B + C which becomes eq.2a) F = 2×C - B In eq.4a, replace 2×C - B with F (from eq.2a): F = E
Hint #4
In eq.4, substitute 2×C - B for E (from eq.4a) and for F (from eq.2a): A + B + C = D + 2×C - B + 2×C - B which becomes A + B + C = D + 4×C - 2×B In the above equation, subtract C from both sides, and add 2×B to both sides: A + B + C - C + 2×B = D + 4×C - 2×B - C + 2×B which becomes eq.4b) A + 3×B = D + 3×C
Hint #5
eq.5 may be written as: E = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + B + C + D) ÷ 4 which becomes eq.5a) 4×E = A + B + C + D
Hint #6
Substitute (2×C - B) for E (from eq.4a) into eq.5a: 4×(2×C - B) = A + B + C + D which becomes 8×C - 4×B = A + B + C + D Subtract B and C from each side of the equation above: 8×C - 4×B - B - C = A + B + C + D - B - C which becomes eq.5b) 7×C - 5×B = A + D
Hint #7
Subtract 3×B from each side of eq.4b: A + 3×B - 3×B = D + 3×C - 3×B which becomes A = D + 3×C - 3×B Substitute D + 3×C - 3×B for A in eq.5b: 7×C - 5×B = D + 3×C - 3×B + D which becomes eq.5c) 7×C - 5×B = 2×D + 3×C - 3×B
Hint #8
In eq.5c, add 5×B to both sides, and subtract 3×C from both sides: 7×C - 5×B + 5×B - 3×C = 2×D + 3×C - 3×B + 5×B - 3×C which becomes 4×C = 2×D + 2×B Divide both sides of the above equation by 2: 4×C ÷ 2 = (2×D + 2×B) ÷ 2 which becomes eq.5d) 2×C = D + B
Hint #9
Substitute D + B for 2×C in eq.2a: F = D + B - B which makes F = D
Hint #10
Substitute F for E and D in eq.5a: 4×F = A + B + C + F Subtract F from both sides of the above equation: 4×F - F = A + B + C + F - F which becomes eq.5e) 3×F = A + B + C
Hint #11
Substitute 3×F for A + B + C (from eq.5e), and F for D and E in eq.1: 3×F + F + F + F = 42 which makes 6×F = 42 Divide both sides of the above equation by 6: 6×F ÷ 6 = 42 ÷ 6 which makes F = 7 and also makes D = F = E = 7
Hint #12
Substitute 7 for E and F in eq.6: AB = 7 × 7 which makes AB = 49 which means A = 4 and B = 9
Solution
Substitute 7 for F, 4 for A, and 9 for B in eq.5e: 3×7 = 4 + 9 + C which becomes 21 = 13 + C Subtract 13 from each side of the above equation: 21 - 13 = 13 + C - 13 which makes 8 = C and makes ABCDEF = 498777