Puzzle for February 5, 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.
Scratchpad
Help Area
Hint #1
Add D and A to both sides of eq.3: B - D + D + A = D - A + D + A which becomes eq.3a) A + B = 2×D
Hint #2
eq.6 may be written as: D = (A + B + C + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 × (A + B + C + F) ÷ 4 which becomes eq.6a) 4×D = A + B + C + F
Hint #3
eq.6a may be written as: 4×D = A + C + B + F In the above equation, replace B + F with A + C (from eq.2): 4×D = A + C + A + C which becomes 4×D = 2×A + 2×C Divide both sides by 2: 4×D ÷ 2 = (2×A + 2×C) ÷ 2 which becomes eq.6b) 2×D = A + C
Hint #4
In eq.3a, replace 2×D with A + C (from eq.6b): A + B = A + C Subtract A from each side of the equation above: A + B - A = A + C - A which makes B = C
Hint #5
In eq.2, substitute B for C: A + B = B + F Subtract B from each side of the equation above: A + B - B = B + F - B which makes A = F
Hint #6
eq.5 may be written as: F = (B + C + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × F = 3 × (B + C + E) ÷ 3 which becomes eq.5a) 3×F = B + C + E
Hint #7
In eq.4, replace D - A with B - D (from eq.3): C + B - D = E + F Add D to both sides of the equation above: C + B - D + D = E + F + D which becomes C + B = E + F + D which may be written as eq.4a) B + C = D + E + F
Hint #8
In eq.5a, replace B + C with D + E + F (from eq.4a): 3×F = D + E + F + E which becomes 3×F = D + 2×E + F Subtract F from both sides of the equation above: 3×F - F = D + 2×E + F - F which becomes eq.5b) 2×F = D + 2×E
Hint #9
Subtract A from both sides of eq.3a: A + B - A = 2×D - A which becomes eq.3b) B = 2×D - A
Hint #10
In eq.5b, substitute A for F: 2×A = D + 2×E Subtract D from each side of the equation above: 2×A - D = D + 2×E - D which becomes eq.5c) 2×A - D = 2×E
Hint #11
Substitute B for C, and A for F in eq.4a: B + B = D + E + A which becomes 2×B = D + E + A Multiply both sides of the above equation by 2: 2 × (2×B) = 2 × (D + E + A) which becomes eq.4b) 4×B = 2×D + 2×E + 2×A
Hint #12
Substitute (2×D - A) for B (from eq.3b), and 2×A - D for 2×E (from eq.5c) in eq.4b: 4×(2×D - A) = 2×D + 2×A - D + 2×A which becomes 8×D - 4×A = D + 4×A In the above equation, add 4×A to both sides, and subtract D from both sides: 8×D - 4×A + 4×A - D = D + 4×A + 4×A - D which makes 7×D = 8×A Divide both sides by 8: 7×D ÷ 8 = 8×A ÷ 8 which makes ⅞×D = A and also makes ⅞×D = A = F
Hint #13
Substitute ⅞×D for A in eq.3b: B = 2×D - ⅞×D which makes B = 1⅛×D and also makes C = B = 1⅛×D
Hint #14
Substitute (⅞×D) for A in eq.5c: 2×(⅞×D) - D = 2×E which becomes 1¾×D - D = 2×E which makes ¾×D = 2×E Divide both sides of the above equation by 2: ¾×D ÷ 2 = 2×E ÷ 2 which makes ⅜×D = E
Solution
Substitute ⅞×D for A and F, 1⅛×D for B and C, and ⅜×D for E in eq.1: ⅞×D + 1⅛×D + 1⅛×D + D + ⅜×D + ⅞×D = 43 which simplifies to 5⅜×D = 43 Divide both sides of the above equation by 5⅜: 5⅜×D ÷ 5⅜ = 43 ÷ 5⅜ which means D = 8 making A = F = ⅞×D = ⅞ × 8 = 7 B = C = 1⅛×D = 1⅛ × 8 = 9 E = ⅜×D = ⅜ × 8 = 3 and ABCDEF = 799837